NEET Chemistry

The s Block Elements

Question 1. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.

Reason R: The deep blue solution is due to the formation of amide.

In the light of the above statements, choose the correct answer from the options given below:

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true and R is the correct explanation of A.
4. Both A and R are true but R is not the correct explanation of A.

Answer: 1. A is true but R is false.

Sodium metal dissolves in liquid ammonia giving a deep blue solution which is conducting in nature. \(\mathrm{Na}+(x+y) \mathrm{NH}_3 \rightarrow\left[\mathrm{Na}\left(\mathrm{NH}_3\right)_x\right]^{+}+\left[e\left(\mathrm{NH}_3\right)_y\right]^{-}\)

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus, imparts blue colour to the solution.

The solution is paramagnetic and on standing, slowly liberates hydrogen gas resulting in the formation of amide. \(\mathrm{Na}_{(a m)}^{+}+e^{-}+\mathrm{NH}_{3(0)} \rightarrow \mathrm{NaNH}_{2(a m)}+1 / 2 \mathrm{H}_{2(\mathrm{~g})}\)

(where ‘am’ denotes solution in ammonia.)

Question 2. Identify the incorrect statement from the following.

1. Alkali metals react with water to form their hydroxides.
2. The oxidation number of K in KO₂ is +4.
3. The ionisation enthalpy of alkali metals decreases from top to bottom in the group.
4. Lithium is the strongest reducing agent among the alkali metals.

Answer: 2. The oxidation number of K in KO₂ is +4.

The superoxide species (KO₂) is represented as O^–₂, since the compound is neutral, therefore the oxidation state of potassium is +1. Remaining all the given statements are correct

Question 3. Match List-1 with List-2.

Choose the correct answer from the options given below:

1. (1) -(D), (2) – (A), (3) – (C), (4) – (B)
2. (1) -(C), (2) – (D), (3) – (B), (4) – (A)
3. (1) -(A), (2) – (C), (3) – (D), (4) – (B)
4. (1) -(B), (2) – (C), (3) – (A), (4) – (D)

Answer: 4. (1) -(B), (2) – (C), (3) – (A), (4) – (D)

Question 4. Ionic mobility of which of the following alkali metal ions is lowest when the aqueous solution of their salts is put under an electric field?

1. K
2. Rb
3. Li
4. Na

Asnwer: 3. Li

The hydration enthalpy of alkali metal ions decreases with an increase in ionic sizes i.e., Li⁺>Na⁺>K>Rb⁺>Cs⁺

Hence, lithium having a maximum degree of hydration will be the least mobile.

The order of ionic mobility is \(\left.\left[\mathrm{Li}_{(a q)}\right]^{+}<\left[\mathrm{Na}_{(a q)}\right)\right]^{+}<\left[\mathrm{K}_{(a q)}\right]^{+}<\left[\mathrm{Rb}_{(a q)}\right]^{+}\)

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Question 5. Which one of the alkali metals, forms only, the normal oxide, ₂O on heating in the air?

1. Rb
2. K
3. Li
4. Na

Answer: 3. Li

When alkali metals are heated in an atmosphere of oxygen, the alkali metals ignite and form oxides. On combustion Li forms Li₂O; sodium gives the peroxide Na₂O₂ and potassium and rubidium give superoxide (MO₂).

Question 6. The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins follows the order

1. Li⁺ < K⁺ < Na⁺ < Rb⁺
2. Rb⁺ < K⁺ < Na⁺ < Li⁺
3. K⁺ < Na⁺ < Rb⁺ < Li⁺
4. Na⁺ < Li⁺ < K⁺ < Rb⁺

Answer: 2. Rb⁺ < K⁺ < Na⁺ < Li⁺

The order of decreasing hydration enthalpy of alkali metal ions is: Li⁺ > Na⁺ > K⁺ > Rb⁺

Thus, the ease of adsorption of hydrated ions is in the order: Rb⁺< K⁺< Na⁺< Li⁺

Question 7. The sequence of ionic mobility in aqueous solution is

1. Rb⁺ > K⁺ > Cs⁺ > Na⁺
2. Na⁺ > K⁺ > Rb⁺ > Cs⁺
3. K⁺ > Na⁺ > Rb⁺ > Cs⁺
4. Cs⁺ > Rb⁺ > K⁺ > Na⁺

Answer: 4. Cs⁺ > Rb⁺ > K⁺ > Na+

The smaller the size of the cation, the higher the hydration and its reflective size will increase hence mobility in an aqueous solution will decrease. Hence, the correct sequence of ionic mobility in an aqueous solution of the given cations is Cs⁺>Rb⁺>K⁺>Na⁺.

Question 8. When a substance (A) reacts with water it produces a combustible gas (B) and a solution of substance (C) in water. When another substance (D) reacts with this solution of (C), it also produces the same gas (B) on warming but (D) can produce gas (B) on reaction with dilute sulphuric acid at room temperature. Substance (A) imparts a deep golden-yellow colour to the smokeless flame of the Bunsen burner. Then (A), (B), (C) and (D) respectively are

1. \(\mathrm{Ca}, \mathrm{H}_2, \mathrm{Ca}(\mathrm{OH})_2, \mathrm{Sn}\)
2. \(\mathrm{K}, \mathrm{H}_2, \mathrm{KOH}, \mathrm{Al}\)
3. \(\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}\)
4. \(\mathrm{CaC}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{Ca}(\mathrm{OH})_2, \mathrm{Fe}\)

Answer: 3. \(\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}\)

Only ‘Na’ imparts a golden colour to Bunsen flame, therefore, A = Na, B = H₂, C = NaOH, D = Zn.

⇒ \(\underset{\text{A}}{2 \mathrm{Na}}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text{A}}{2 \mathrm{NaOH}}+\underset{\text{B}}{\mathrm{H}_2}\)

⇒ \(\underset{\text{D}}{\mathrm{Zn}}+\underset{\text{C}}{2 \mathrm{NaOH}} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\underset{\text{B}}{\mathrm{H}_2}\)

⇒ \(\underset{\text{D}}{\mathrm{Zn}}+\mathrm{H}_2 \mathrm{SO}_4 \text { (dil.) } \rightarrow \mathrm{ZnSO}_4+\underset{\text{B}}{\mathrm{H}_2}\)

Question 9. Which one of the following properties of alkali metals increases in magnitude as the atomic number rises?

1. Ionic radius
2. Melting point
3. Electronegativity
4. First ionization energy

Answer: 1. Ionic radius

In a group, the ionic radius increases with an increase in atomic number whereas the m.pt. decreases down in a group due to the weakening of metallic bonds. Similarly, electronegativity and ionization energy also decrease down the group.

Question 10. In the case of alkali metals, the covalent character decreases in the order

1. MF > MCI > MBr > MI
2. MF > MCI > Ml > MBr
3. MI > MBr > MCI > MF
4. MCI > MI > MBr > MF

Answer: 3. MI > MBr > MCI > MF

Alkali metals are highly electropositive and halogens are electronegative. Thus, for the halides of a given alkali metal, the covalent character decreases with an increase in the electronegativity of halogens.

∴ The order of the covalent character of halides is

MI> MBr > MCI> MF

Question 11. The alkali metals form salt-like hydrides by direct synthesis at elevated temperatures. The thermal stability of these hydrides decreases in which of the following orders?

1. NaH > LiH > KH > RbH > CsH
2. LiH > NaH > KH > RbH > CsH
3. CsH > RbH > KH > NaH > LiH
4. KH > NaH > LiH > CsH > RbH

Answer: 2. LiH > NaH > KH > RbH > CsH

The ionic character of the bonds in hydrides increases from LiH to CsH due to the weakening of the M-H bond so, the thermal stability of these hydrides decreases in the order of LiH > NaH > KH > RbH > CsH.

Question 12. Which compound will show the highest lattice energy?

1. RbF
2. CsF
3. NaF
4. KF

Answer: 3. NaF

With the same anion, the smaller the size of the cation, the higher is the lattice energy. Therefore, NaF will show the highest lattice energy among the given compounds.

Question 13. Which of the alkali metal chloride (MCI) forms its dihydrate salt (MCl.2H₂O) easily?

1. LiCl
2. CsCl
3. RbCl
4. KCl

Answer: 1. LiCl

LiCl is deliquescent and crystallises from aqueous solution as hydrates, LiCl.2H₂O.

Question 14. Crude sodium chloride obtained by crystallisation of brine solution does not contain

1. \(\mathrm{MgSO}_4\)
2. \(\mathrm{Na}_2 \mathrm{SO}_4\)
3. \(\mathrm{MgCl}_2\)
4. \(\mathrm{CaSO}_4\)

Answer: 1. \(\mathrm{MgSO}_4\)

Crude sodium chloride, generally obtained by the crystallisation of brine solution contains sodium sulphate (Na₂SO₄), calcium sulphate (CaSO₄), calcium chloride (CaCl₂) and magnesium chloride (MgCl₂) as impurities. Crude sodium chloride does not contain MgSO₄

Question 15. In the Castner-Kellner cell for the production of sodium hydroxide

1. Brine is electrolyzed using graphite electrodes
2. Molten sodium chloride is electrolysed
3. Sodium amalgam is formed at a mercury cathode
4. Brine is electrolyzed with pt electrodes.

Answer: 3. Sodium amalgam is formed at a mercury cathode

In the Castner-Kellner cell, sodium amalgam is formed at the mercury cathode.

A brine solution is electrolysed using a mercury cathode and a carbon anode.

Question 16. Which of the following statements is incorrect?

1. Pure sodium metal dissolves in liquid ammonia to give a blue solution.
2. NaOH reacts with glass to give sodium silicate.
3. Aluminium reacts with excess NaOH to give Al(OH)₃.
4. NaHCO₃ on heating gives Na₂CO₃.

Answer: 3. Aluminium reacts with excess NaOH to give Al(OH)₃.

Al reacts with NaOH to give sodium aluminate

Question 17. In which of the following processes, fused sodium hydroxide is electrolysed at a 330 °C temperature for extraction of sodium?

1. Castner’s process
2. Down’s process
3. Cyanide process
4. Both (2) and (3).

Answer: 1. Castner’s process

In Castner’s process, for the production of sodium metal, sodium hydroxide (NaOH) is electroplated at a temperature of 330°C.

Question 18. Which of the following is known as a fusion mixture?

1. Mixture of Na₂CO₃ + NaHCO₃
2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)
3. Mixture of \(\mathrm{K}_2 \mathrm{CO}_3+\mathrm{Na}_2 \mathrm{CO}_3\)
4. \(\mathrm{NaHCO}_3\)

Answer: 3. Mixture of \(\mathrm{K}_2 \mathrm{CO}_3+\mathrm{Na}_2 \mathrm{CO}_3\)

K₂CO₃ and Na₂CO₃ mixture is called a fusion mixture.

Question 19. Washing soda has a formula

1. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 7 \mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 3 \mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{Na}_2 \mathrm{CO}_3\)

Answer: 2. \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O}\)

Na₂CO₃.10H₂O is washing soda.

Question 20. The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

1. Iron
2. Copper
3. Calcium
4. Potassium

Answer: 4. Potassium

Potassium ions are then the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and, with solider, are responsible for the transmission of nerve signals.

Question 21. The function of the sodium pump is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a constituent of this pump?

1. K⁺
2. Fe²⁺
3. Ca²⁺
4. Mg²⁺

Answer: 1. K⁺

Question 22. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³, the simplest formula for this compound is

1. \(\mathrm{Mg}_2 X_3\)
2. \(\mathrm{Mg} X_2\)
3. \(\mathrm{Mg}_2 X\)
4. \(\mathrm{Mg}_3 X_2\)

Answer: 4. \(\mathrm{Mg}_3 X_2\)

The electronic configuration of X is 1s², 2s² 2p³.

So, the valency of X will be 3.

Magnesium ion = Mg²⁺

Formula: Mg₃X₂

Question 23. The electronic configuration of calcium atoms may be written as

1. \([\mathrm{Ne}] 4 p^2\)
2. \([\mathrm{Ar}] 4 s^2\)
3. \([\mathrm{Ne}] 4 s^2\)
4. \([\mathrm{Ar}] 4 p^2\)

Answer: 2. \([\mathrm{Ar}] 4 s^2\)

⇒ \({ }_{20} \mathrm{Ca} \longrightarrow 1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2\)

⇒ \({ }_{18} \mathrm{Ar} \longrightarrow 1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6\)

Hence, \({ }_{20} \mathrm{Ca} \longrightarrow[\mathrm{Ar}] 4 s^2\)

Question 24. Compared with the alkaline earth metals, the alkali metals exhibit

1. Smaller ionic radii
2. Highest boiling points
3. Greater hardness
4. Lower ionization energies.

Answer: 4. Lower ionization energies.

The alkali metals are larger in size and have smaller nuclear charge thus they have lower ionization energy in comparison to alkaline earth metals.

Question 25. Which of the following atoms will have the smallest size?

1. Mg
2. Na
3. Be
4. Li

Answer: 3. Be

The atomic size decreases within a period from left to right, therefore Li > Be and Na > Mg. The size increases in a group from top to bottom. Hence, the size of Na is greater than Li. Overall order Na > Mg > Li > Be. Thus, Be has the smallest size.

Question 26. The structures of beryllium chloride in solid state and vapour phase are

1. Chain in both
2. Chain and dimer, respectively
3. Linear in both
4. Dimer and linear, respectively.

Answer: 2. Chain and dimer, respectively

In the vapour phase, BeCl₂ is found in dimer form.

While in a solid state, it is found as a polymer (chain structure)

Question 27. Among the following alkaline earth metal halides one which is covalent and soluble in organic solvents is

1. Beryllium chloride
2. Calcium chloride
3. Strontium chloride
4. Magnesium chloride

Answer: 1. Beryllium chloride

Except for beryllium halides, all other halides of alkaline earth metals are ionic in nature. Beryllium halides are essentially covalent and soluble in organic solvents.

Question 28. HCl was passed through a solution of CaCl₂, MgCl₂ and NaCl. Which of the following compound(s) crystallise(s)?

1. Both MgCl₂ and CaCl₂
2. Only NaCl
3. Only MgCl₂
4. NaCl, MgCl₂ and CaCl₂

Answer: 2. Only NaCl

CaCl₂ and MgCI₂ are more soluble than NaCl. Thus, when HCl was passed through a solution containing CaCl₂, MgCI₂, and NaCl, only NaCl crystallised.

Question 29. Which of the following is an amphoteric hydroxide?

1. Be(OH)₂
2. Sr(OH)₂
3. Ca(OH)₂
4. Mg(OH)₂

Answer: 1. Be(OH)₂

Be(OH)₂ is amphoteric in nature as it reacts with acid and alkali.

⇒ \(\mathrm{Be}(\mathrm{OH})_2+2 \mathrm{OH}^{-} \rightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-}\)

⇒ \(\mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl}+2 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right] \mathrm{Cl}_2\)

Question 30. Among CaH₂, BeH₂, BaH₂, the order of ionic character is

1. BeH₂ < CaH₂ < BaH₂
2. CaH₂ < BeH₂ < BaH₂
3. BeH₂ < BaH₂ < CaH₂
4. BaH₂ < B₂ < CaH₂

Answer: 1. BeH₂ < CaH₂ < BaH₂

BeH₂ < CaH₂< BaH₂

On moving down the group, the metallic character of metals increases. So, the ionic character of metal hydrides increases.

Hence, BeH₂ will be the least ionic

Question 31. On heating which of the following releases CO₂ most easily?

1. Na₂CO₃
2. MgCO₃
3. CaCO₃
4. K₂CO₃

Answer: 2. MgCO₃

The stability of carbonates increases down the group with an increase in the size of the metal ions. Also, the alkali metal carbonates are more stable than alkaline earth metal carbonates. Hence, MgCO₃ is the least stable and it releases CO₂ most easily.

⇒ \(\mathrm{MgCO}_3\)\(\underrightarrow{\Delta}\)\(\mathrm{MgO}+\mathrm{CO}_2\)

Question 32. Solubility of the alkaline earth metal sulphates in water decreases in the sequence

1. Sr > Ca > Mg > Ba
2. Ba > Mg > Sr > Ca
3. Mg > Ca > Sr > Ba
4. Ca > Sr > Ba > Mg

Answer: 3. Mg > Ca > Sr > Ba

The solubility of alkaline earth metal sulphates decreases down the group because hydration energy decreases.

Question 33. Which of the following compounds has the lowest melting point?

1. CaCl₂
2. CaBr₂
3. Cal₂
4. CaF₂

Answer: 3. Cal₂

As the covalent character in the compound increases and the ionic character decreases, the melting point of the compound decreases. So, CaI₂ has the highest covalent character and lowest melting point.

Question 34. Which of the following alkaline earth metal sulphates has a hydration enthalpy higher than the lattice enthalpy?

1. CaSO₄
2. BeSO₄
3. BaSO₄
4. SrSO₄

Answer: 2. BeSO₄

The hydration enthalpy of BeSO₄ is higher than its lattice energy. Within group 2, the hydration energy decreases down the group while lattice energy is almost the same.

Question 35. Which one of the following compounds is a peroxide?

1. KO₂
2. BaO₂
3. MnO₂
4. NO₂

Answer: 2. BaO₂

BaO₂ has peroxide linkage

Question 36. Property of the alkaline earth metals that increases with their atomic number

1. Solubility of their hydroxides in water
2. Solubility of their sulphates in water
3. Ionization energy
4. Electronegativity

Answer: 1. Solubility of their hydroxides in water

The solubility of an ionic compound depends on two factors:

1. Lattice energy, and
2. Hydration energy

In the case of alkaline earth metal hydroxides, the lattice energy decreases as we move down the group. This decrease is more than the decrease in the hydration energy down the group.

Question 37. Which of the following oxides is not expected to react with sodium hydroxide?

1. CaO
2. SiO₂
3. BeO
4. B₂O₃

Answer: 1. CaO

CaO being a basic oxide does not react with NaOH, however, SiO₂ (acidic oxide), BeO (amphoteric oxide) and B₂O₃ (acidic oxide) react with NaOH

Question 38. The correct order of increasing thermal stability of K₂CO₃, MgCO₃, CaCO₃ and BeCO₃ is

1. BeCO₃ < MgCO₃ < CaCO₃ < K₂CO₃
2. MgCO₃ < BeCO₃ < CaCO₃ < K₂CO₃
3. K₂CO₃ < MgCO₃ < CaCO₃ < BeCO₃
4. BeCO₃ < MgCO₃ < K₂CO₃ < CaCO₃

Answer: 1. BeCO₃ < MgCO₃ < CaCO₃ < K₂CO₃

In all cases, for a particular set of group 1 or group 2 compounds, the thermal stability increases down the group as the ionic radius of the cation increases, and its polarising power decreases.

Group 1 compounds tend to be more thermally stable than group 2 compounds because group 1 cation has a smaller charge and a larger ionic radius, and so, a lower polarising power, particularly when adjacent metals on the same period are compared.

Hence, the order of increasing thermal stability of K₂CO₃, MgCO₃, CaCO₃ and BeCO₃ is BeCO₃ < MgCO₃ < CaCO₃ < K₂CO₃

Question 39. In which of the following the hydration energy is higher than the lattice energy?

1. MgSO₄
2. RaSO₄
3. SrSO₄
4. BaSO₄

Answer: 1. MgSO₄

When hydration energy exceeds lattice energy, the compound becomes soluble in water. The solubility of alkaline earth metal sulphates decreases in the order:

The solubilities of BeSO₄ and MgSO₄ are due to the high energy of solvation of smaller Be²⁺ and Mg²⁺ ions.

Question 40. The solubility in water of sulphate down the Be group is Be > Mg > Ca > Sr > Ba. This is due to

1. Decreasing lattice energy
2. The high heat of solvation for smaller ions like Be²⁺
3. Increase in melting points
4. Increasing molecular weight.

Answer: 2. High heat of solvation for smaller ions like Be²⁺

As we move down the group from BeSO₄ to BaSO₄ the enthalpy of hydration of the positive ion becomes smaller due to increasing in ionic size. Salts of heavier metal ions are 1ess soluble than those of lighter ions.

Question 41. All the following substances react with water. The pair that gives the same gaseous product is

1. K and KO₂
2. Na and Na₂O₂
3. Ca and CaH₂
4. Ba and BaO₂

Answer: 3. Ca and CaH₂

The pair which gives the same gaseous product is Ca and CaH₂.

⇒ \(\mathrm{Ca}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2\)

⇒ \(\mathrm{CaH}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{H}_2\)

Whereas, \(\mathrm{K}\) gives \(\mathrm{H}_2\) while \(\mathrm{KO}_2\) gives \(\mathrm{O}_2\) and \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(2 \mathrm{~K}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_2\)

⇒ \(2 \mathrm{KO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{O}_2+\mathrm{H}_2 \mathrm{O}_2\)

Similarly, \(\mathrm{Na}\) gives \(\mathrm{H}_2\), while \(\mathrm{Na}_2 \mathrm{O}_2\) gives \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2\)

⇒ \(\mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2\)

Likewise Ba gives \(\mathrm{H}_2\) while \(\mathrm{BaO}_2\) gives \(\mathrm{H}_2 \mathrm{O}_2\).

⇒ \(\mathrm{Ba}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ba}(\mathrm{OH})_2+\mathrm{H}_2\)

⇒ \(\mathrm{BaO}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ba}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{O}_2\)

Question 42. Which of the following statements is false?

1. Strontium decomposes water more readily than beryllium.
2. Barium carbonate melts at a higher temperature than calcium carbonate.
3. Barium hydroxide is more soluble in water than magnesium hydroxide.
4. Beryllium hydroxide is more basic than barium hydroxide.

Answer: 4. Beryllium hydroxide is more basic than barium hydroxide.

Beryllium hydroxide although amphoteric, is however less basic than barium hydroxide.

Question 43. In context with beryllium, which one of the following statements is incorrect?

1. It is rendered passive by nitric acid.
2. It forms Be₂C.
3. Its salts rarely hydrolyse.
4. Its hydride is electron-deficient and polymeric.

Answer: 3. Its salts rarely hydrolyse.

Due to the very small size of Be²⁺, beryllium salts are readily hydrolysed because of high hydration energy. \(\mathrm{BeCl}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl}\)

Question 44. The suspension of slaked lime in water is known as

1. Lime water
2. Quicklime
3. Milk of lime
4. Aqueous solution of slaked lime.

Answer: 3. Milk of lime

⇒ \(\underset{\text { Quick lime }}{\mathrm{CaO}}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Slaked lime }}{\mathrm{Ca}(\mathrm{OH})_2+\text { Heat }}\)

This process is known as slaking of lime. The paste of lime in water (i.e., suspension) is called milk of lime while the filtered and clear solution is known as lime water

Question 45. The product obtained as a result of a reaction of nitrogen with CaC₂ is

1. CaCN₃
2. Ca₂CN
3. Ca(CN)₂
4. CaCN

Answer: 3. Ca(CN)₂

Read Ca(CN)₂ as CaCN₂

⇒ \(\mathrm{CaC}_2+\mathrm{N}_2 \longrightarrow \underset{\text{Slaked lime}}{\mathrm{CaCN}_2}+\mathrm{C}\)

Question 46. Which one of the following is present as an active ingredient in bleaching powder for bleaching action?

1. CaOCl₂
2. Ca(OCl)₂
3. CaO₂Cl
4. CaCl₂

Answer: 2. Ca(OCl)₂

The active ingredient in bleaching powder for bleaching action is Ca(OCl)₂.

Question 47. Match List-1 with List-2 for the compositions of substances and select the correct answer using the code given:

1. (1)-(C), (2)-(D), (3)-(A), (4)-(B)
2. (1)-(B), (2)-(C), (3)-(D), (4)-(A)
3. (1)-(A), (2)-(B), (3)-(C), (4)-(E)
4. (1)-(D), (2)-(C), (3)-(B), (4)-(A)

Answer: 2. (1)-(B), (2)-(C), (3)-(D), (4)-(A)

Question 48. The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO₂ is bubbled through an aqueous solution of B, and C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound is

1. CaCO₃
2. Na₂CO₃
3. K₂CO₃
4. CaSO_4•2H₂O

Answer: 1. CaCO₃

The reactions can be summarised as follows:

Question 49. Which of the following represents calcium chlorite?

1. Ca(ClO₃)₂
2. Ca(ClO₂)₂
3. CaClO₂
4. Ca(ClO₄)₂

Answer: 2. Ca(ClO₂)₂

Since the valency of calcium is 2 and a chlorite ion is CIO^–₂, therefore calcium chlorite is Ca(ClO₂)₂.

Question 50. Identify the correct statement.

1. Plaster of Paris can be obtained by hydration of gypsum.
2. Plaster of Paris is obtained by partial oxidation of gypsum.
3. Gypsum contains a lower percentage of calcium than Plaster of Paris.
4. Gypsum is obtained by heating Plaster of Paris.

Answer: 3. Gypsum contains a lower percentage of calcium than Plaster of Paris.

Gypsum is CaSO₄.2H₂O and plaster of Paris is (CaSO₄)₂.H₂O. Therefore, gypsum contains a lower percentage of calcium than Plaster of Paris.

Question 51. Bleaching powder is obtained by the action of chlorine gas and

1. Dilute solution of Ca(OH)₂
2. Concentrated solution of Ca(OH)₂
3. Dry CaO
4. Dry-slaked lime.

Answer: 4. Dry slaked lime.

Cl₂ gas reacts with dry slaked lime, Ca(OH)₂ to give bleaching powder.

⇒ \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{Cl}_2\) \(\underrightarrow{\Delta}\) \(\mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{O}\)

Question 52. Which one of the following statements is correct?

1. The bone in the human body is an inert and unchanging substance.
2. Mg plays roles in neuromuscular function and interneuronal transmission.
3. The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g.
4. All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor.

Answer: 3. The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g.

Bones are not the inert and unchanging substances. Calcium plays an important role in neuromuscular function and interneuronal transmission.

All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor.

Question 53. Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is

1. Sr
2. Be
3. Mg
4. Ca

Answer: 3. Mg

All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor

Question 54. Which of the following statements is false?

1. Ca²⁺ ions are not important in maintaining the regular beating of the heart.
2. Mg²⁺ ions are important in the green parts of the plants.
3. Mg²⁺ ions form a complex with ATP.
4. Ca²⁺ ions are important in blood clotting.

Answer: 1. Ca²⁺ ions are not important in maintaining the regular beating of the heart.

Ca²⁺ ions are required to trigger the contraction of muscles and to maintain the regular beating of the heart

Question 55. Which of the following metal ions play an important role in muscle contraction?

1. K⁺
2. Na⁺
3. Mg²⁺
4. Ca²⁺

Answer: 4. Ca²⁺

Calcium is an essential element for the contraction of muscles.

The p Block Elements

Question 1. Taking stability as the factor, which one of the following represents the correct relationship?

1. AlCl > AlCl₃
2. TlI > TlI₃
3. TICI₃ > TlCl
4. InI₃ > InI

Answer: 2. TlI > TlI₃

As we move down the group, the stability of the +3 oxidation state decreases while that of the +1 oxidation state increases. Thus, the stability order for the +1 oxidation state is Al < Ga < In< Tl and the +3 oxidation state is Al > Ga > In > TI.

Hence, the stability order is, AlC3 > AICI, TII > TlI₃, TlCl > TlCl₃ and InI > InI₃

Question 2. BF₃ is a planar and electron-deficient compound. Hybridization and the number of electrons around the central atom, respectively are

1. sp² and 8
2. sp³ and 4
3. sp³ and 6
4. sp² and 6

Answer: 4. sp² and 6

BF₃ is a sp²-hybridised planar molecule. It forms 3σ-bonds with 3F-atoms, hence has six electrons around it.

Question 3. The correct order of atomic radii in group 13 elements is

1. B < Al < In < Ga < Tl
2. B < Al < Ga < In < l
3. B < Ga < Al < Tl < In
4. B < Ga < Al < In < Tl

Answer: 4. B < Ga < Al < In < Tl

Question 4. AlF₃ is soluble in HF only in the presence of KF. It is due to the formation of

1. \(\mathrm{K}_3\left[\mathrm{AlF}_3 \mathrm{H}_3\right]\)
2. \(\mathrm{K}_3\left[\mathrm{AlF}_6\right]\)
3. \(\mathrm{AlH}_3\)
4. \(\mathrm{K}\left[\mathrm{AlF}_3 \mathrm{H}\right]\)

Answer: 2. \(\mathrm{K}_3\left[\mathrm{AlF}_6\right]\)

AlF₃ is insoluble in anhydrous HF because the F^– ions are not available in hydrogen-bonded HF molecules but, it becomes soluble in the presence of a small amount of KF due to the formation of complex, K₃[AlF₃].

⇒ \(\mathrm{AlF}_3+3 \mathrm{KF} \rightarrow \mathrm{K}_3\left[\mathrm{AlF}_6\right]\)

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Question 5. The stability of the +1 oxidation state among Al, Ga, In, and Tl increases in the sequence

1. Al < Ga < In < Tl
2. Tl < In < Ga < Al
3. In < Tl < Ga < Al
4. Ga < In < Al < Tl

Answer: 1. Al < Ga < In < Tl

In group 13 elements, the stability of the +3 oxidation state decreases down the group while that of the +1 oxidation state increases due to the inert pair effect. Hence, the stability of the +1 oxidation state increases in the sequence: Al <Ga<In<Tl.

Question 6. Aluminum (3) chloride forms a dimer because aluminum

1. Belongs to 3rd group
2. Can have a higher coordination number
3. Cannot form a trimer
4. Has high ionization energy.

Answer: 2. Can have a higher coordination number

AlCl₃ forms a dimer, as Al due to the presence of 3d-orbitals can expand its covalency from four to six. Also, dimerization enables Al atoms to complete their octets.

Question 7. Which one of the following elements is unable to form \(\mathrm{MF}_6^{3-} \text {. }\) ion?

1. Ga
2. Al
3. B
4. In

Answer: 3. B

Boron does not have vacant d-orbitals in its valence shell, so it cannot expand its covalency beyond 4 i.e., ‘B’ cannot form the ions like \(\mathrm{MF}_6^{3-} \text {. }\)

Question 8. The tendency of BF₃, BCl_3, and BBr₃ to behave as Lewis acid decreases in the sequence

1. BCl₃ > BF₃ > BBr₃
2. BBr₃ > BCl₃ > BF₃
3. BBr₃ > BF₃ > BCl₃
4. BF₃ > BCl₃ > BBr₃

Answer: 2. BBr3 > BCl₃ > BF3

The relative Lewis acid character of boron trihalides is found to follow the following order, BBr₃ > BCl₃ > BF₃, but the expected order on the basis of electronegativity of the halogens (electronegativity of halogens decreases from F to I) should be, BF₃, > BCl₃ > BBr₃

This anomaly is explained on the basis of the relative tendency. of the halogen atom to back donate its unutilized electrons to the vacant p-orbital of the boron atom. In BF₃, boron has a vacant 2p orbital and each fluorine has fully filled unutilized 2p-orbitals.

Fluorine transfers two electrons to the vacant 2p-orbital of boron, thus forming pπ – pπ bond

This type of bond has some double-bond character and is known as dative or back bonding. All the three bond lengths are the same. It is possible when a double bond is delocalized. The delocalization may be represented as :

The tendency to back donate decreases from F to I as the energy level difference between B and halogen atoms increases from F to L So, the order of Lewis acid strength is BF₃ < BCl₃ < BBr₃

Question 9. Boron compounds behave as Lewis acids, because of their

1. Ionisation property
2. Electron deficient nature
3. Acidic nature
4. Covalent nature.

Answer: 2. Electron-deficient nature

Lewis acids are those substances that can accept a pair of electrons and boron compounds usually are deficient in electrons.

Question 10. Which of the following statements is not correct about diborane?

1. There are two 3-center-2-electron bonds.
2. The four terminal B – H bonds are two center two-electron bonds.
3. The four terminal hydrogen atoms and the two boron atoms lie in one plane.
4. Both the boron atoms are sp² hybridized.

Answer: 4. Both the boron atoms are sp² hybridized.

In diborane (B₂H₆), each boron (B) atom uses sp³ hybrid orbitals for bonding.

Question 11. Boric acid is an acid because its molecule

1. Contains replaceable H⁺ ion
2. Gives up a proton
3. Accepts OH^– from water releasing a proton
4. Combines with a proton from the water molecule.

Answer: 3. Accepts OH^– from water releasing a proton

Boric acid behaves as a Lewis acid, by accepting a pair of electrons from the OH^– ion of water thereby releasing a proton.

Question 12. Which of the following structures is similar to graphite

1. B₄C
2. B₂H₆
3. BN
4. B

Answer: 3. BN

BN is known as inorganic graphite and has a structure similar to graphite.

Question 13. The type of hybridisation of boron in diborane is

1. sp³-hybridisation
2. sp²-hybridisation
3. sp-hybridization
4. sp³d²-hybridisation.

Answer: 1. sp³-hybridisation

Each ‘B’ atom in diborane (B₂H₆) is sp³-hybridised’ Of the 4-hybrid orbitals, three have one electron each, while the 4th is empty.

Two orbitals of each form o bond with two ‘H’- atoms, with one of the remaining hybrid orbital (either filled or empty), ls orbital of ‘H’ atom and one of the hybrid orbitals of other ‘B’ atom overlap to form three centered two-electron bond. So there exist two such types of three-centered bonds.

Question 14. Which of the following statements about H₃BO₃ is not correct?

1. It has a layer structure in which planar BO₃ units are joined by hydrogen bonds.
2. It does not act as a proton donor but acts as a Lewis acid by accepting hydroxyl ions.
3. It is a strong tribasic acid.
4. It is prepared by acidifying an aqueous solution of borax.

Answer: 3. It is a strong tribasic acid.

H₃BO₃ is a weak monobasic acid. \(\mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}\)

Question 15. Which of the following is an incorrect statement?

1. SnF₄ is ionic in nature.
2. PbF₄ is covalent in nature.
3. SiCl₄ is easily hydrolyzed.
4. GeX₄(X = F, Cl, Br, I) is more stable than GeX₂.

Answer: 2. PbF₄ is covalent in nature.

Generally, the halides of group-4 elements are covalent in nature. PbF₄ and SnF₄ are exceptions which are ionic in nature.

Question 16. Which of the following species is not stable?

1. [SiCl₆]^2-
2. [SiF₆]^2-
3. [GeCl₆]^2-
4. [Sn(OH)₆]^2-

Answer: 1. [SiCl₆]^2-

[SiCl₆]^2- is not stable due to steric hindrance by large-sized Cl atoms.

Question 17. It is because of the inability of ns² electrons of the valence shell to participate in bonding that

1. Sn²⁺ is oxidizing while Pb⁴⁺ is reducing
2. Sn²⁺+ and Pb²⁺ are both oxidizing and reducing
3. Sn⁴⁺ is reducing while Pb⁴⁺ is oxidizing
4. Sn²⁺ is reducing while Pb⁴⁺ is oxidizing.

Answer: 4. Sn²⁺ is reducing while Pb⁴⁺ is oxidizing.

The inertness of s-subshell electrons towards bond formation is known as the inert pair effect. This effect increases down the group thus, for Sn, the +4 oxidation state is more stable whereas, for Pb, the +2 oxidation state is more stable, i.e., Sn²⁺ is reducing while Pb⁴⁺ is oxidizing.

Question 18. Which of the following oxidation states is the most characteristic of lead and tin respectively?

1. +2, +4
2. +4, +4
3. +2, +2
4. +4, +2

Answer: 1. +2, +4

When ns² electrons of the outermost shell do not participate in bonding then these ns² electrons are called inert pair and the effect is called the inert pair effect.  Due to this inert pair effect, Ge, Sn, and Pb of group 14 have a tendency to form Loth +a and +2 ions.

Now the inert pair effect increases down the group, hence the stability of M²⁺ ions increases, and M⁴⁺ ions decrease down the group. For this reason, Pb²⁺ is more stable than Pb⁴⁺ and Sn⁴⁺ is more stable than Sn²⁺

Question 19. Carbon and silicon belong to the (4) group. The maximum coordination number of carbon in commonly occurring compounds is 4, whereas that of silicon is 6. This is due to

1. Availability of low lying d-orbitals in silicon
2. The large size of the silicon
3. More electropositive nature of silicon
4. Both (2) and (3).

Answer: 1. Availability of low-lying d-orbitals in silicon

Carbon has no d-orbitals, while silicon contains d-orbitals in its valence shell which can be used for bonding purposes.

Question 20. Match List-1 with List-2

Choose the correct answer from the options given below:

1. (1) – (C), (2) – (1), (3) – (D), (4) – (B)
2. (1) – (C), (2) – (D), (3) – (A), (4) – (B)
3. (1) – (B), (2) – (D), (3) – (A), (4) – (C)
4. (1) – (D), (2) – (A), (3) – (B), (4) – (C)

Answer: 1. (1) – (C), (2) – (1), (3) – (D), (4) – (B)

Question 21. Choose the correct statement.

1. Diamond and graphite have a two-dimensional network.
2. Diamond is covalent and graphite is ionic.
3. Diamond is sp³ hybridized and graphite is sp² hybridized.
4. Both diamond and graphite are used as lubricants.

Answer: 3. Diamond is sp³ hybridized and graphite is sp² hybridized.

In diamond, each carbon atom undergoes sp³ hybridization and is linked to four other carbon atoms by using hybridized orbitals in a tetrahedral fashion. In graphite, each carbon atom in a hexagonal ring undergoes sp² hybridization and makes three sigma bonds with three neighboring carbon atoms. The fourth electron forms a π-bond.

Question 22. Which of the following does not show electrical conduction?

1. Diamond
2. Graphite
3. Potassium
4. Sodium

Answer: 1. Diamond

Except for diamonds other three conduct electricity. Potassium and sodium are metallic conductors, while graphite is a non-metallic conductor.

Question 23. The percentage of lead in lead pencils is

1. 80
2. 20
3. Zero
4. 70

Answer: 3. Zero

The lead pencil contains graphite and clay. It does not contain lead.

Question 24. In graphite, electrons are

1. Localized on each C-atom
2. Localized on every third C-atom
3. Spread out between the structure
4. Present in antibonding orbital.

Answer: 3. Spread out between the structure

In graphite, each carbon atom undergoes sp²-hybridisation and is covalently bonded to three other .urbor. atoms by single bonds. The fourth electron forms π-bond. The electrons are delocalized over the whole sheet.

i.e., electrons are spread out between the structure.

Question 25. Which of the following types of forces bind together the carbon atoms in a diamond?

1. Ionic
2. Covalent
3. Dipolar
4. VanderWaals

Answer: 2. Covalent

In diamond, each carbon atom is sp³ hybridized and thus, forms covalent bonds with four other carbon atoms lying at the corners of a regular tetrahedron.

Question 26. Which of the following is an insulator?

1. Graphite
2. Aluminium
3. Diamond
4. Silicon

Answer: 3. Diamond

All the above are conductors except diamond. Diamond is an insulator.

Question 27. Identify the correct statements from the following:

1. CO₂(g) is used as a refrigerant for ice cream and frozen food.
2. The structure of C60 contains twelve six-carbon rings and twenty-five carbon rings.
3. ZSM-5, a type of zeolite, is used to convert alcohol into gasoline.
4. CO is a colorless and odorless gas.

1. (1), (2), and (3) only
2. (1) and (3) only
3. (2) and (3) only
4. (3) and (4) only

Answer: 4. (3) and (4) only

1. Solid CO₂ (dry ice) is used as a refrigerant for ice cream and frozen food.
2. The structure of C₆₀ contains twenty-six-membered rings and twelve five-membered rings.
3. 3 and 4 are correct statements

Question 28. Which of the following compounds is used in cosmetic surgery?

1. Silica
2. Silicates
3. Silicones
4. Zeolites

Answer: 3. Silicones

Silicones being biocompatible are used in surgical and cosmetic Plants.

Question 29. Which of these is not a monomer for a high molecular mass silicone polymer?

1. Me₃SiCl
2. PhSiCl₃
3. MeSiCl₃
4. Me₂SiCl₃

Answer: 1. Me₃SiCl

It can form only dimer.

Question 30. The basic structural unit of silicates is

1. \(\mathrm{SiO}_3^{2-}\)
2. \(\mathrm{SiO}_4^{2-}\)
3. \(\mathrm{SiO}^{-}\)
4. \(\mathrm{SiO}_4^{4-}\)

Answer: 4. \(\mathrm{SiO}_4^{4-}\)

SiO^4-₄ orthosilicate is the basic unit of silicates.

Question 31. Which statement is wrong?

1. Beryl is an example of cyclic silicate.
2. Mg₂SiO₄ is orthosilicate.
3. The basic structural unit in silicates is the SiO₄ tetrahedron.
4. Feldspars are not aluminosilicates.

Answer: 4. Feldspars are not aluminosilicates.

Feldspars are three-dimensional aluminosilicates.

Question 32. Name the two types of the structure of silicate in which one oxygen atom of [SiO₄]^4- is shared.

1. Linear chain silicate
2. Sheet silicate
3. Borosilicate
4. Three dimensional

Answer: 3. Pyrosilicate

Borosilicate contains two units of SiO₄^4- joined along a corner containing oxygen-atom

Question 33. The straight-chain polymer is formed by

1. Hydrolysis of CH₃SiCl₃ followed by condensation polymerization
2. Hydrolysis of (CH₃)₄Si by addition polymerisation
3. Hydrolysis of (CH₃)₂SiCl₂ followed by condensation polymerisation
4. Hydrolysis of (CH₃)₃SiCl followed

Answer: 3. Hydrolysis of (CH₃)₂SiCl₂ followed by condensation polymerization

Hydrolysis of substituted chlorosilanes yields corresponding silanols which undergo polymerization. Out of the given chlorosilanes, only (CH₃)₂SiCl₂ will give linear polymer on hydrolysis followed by polymerisation.

Question 34. Which of the following anions is present in the chain structure of silicates?

1. \(\left(\mathrm{Si}_2 \mathrm{O}_5^{2-}\right)_n\)
2. \(\left(\mathrm{SiO}_3^{2-}\right)_n\)
3. \(\mathrm{SiO}_4^{4-}\)
4. \(\mathrm{Si}_2 \mathrm{O}_7^{6-}\)

Answer: 2. \(\left(\mathrm{SiO}_3^{2-}\right)_n\)

Chain silicates are formed by sharing two oxygen atoms by each tetrahedron. Anions of chain silicate have two general formula

1. \(\left(\mathrm{SiO}_3\right)_n^{2 n-}\)
2. \(\left(\mathrm{Si}_4 \mathrm{O}_{11}\right)_n^{6 n-}\)

Question 35. Which one of the following statements about the zeolite is false?

1. They are used as cation exchangers.
2. They have an open structure which enables them to take up small molecules.
3. Zeolites are aluminosilicates having a three-dimensional network.
4. Some of the SiO₄^4- units are replaced by AlO₄^5- and AlO₆^9- ions in zeolites.

Answer: 4. Some of the SiO₄^4- units are replaced by AlO₄^5- and AlO₆^9- ions in zeolites.

In zeolites, some of the Si⁴⁺ ions may be replaced by Al³⁺ ions. This results in an unbalanced anionic charge To maintain electrical neutrality, positive ions must be introduced.

Question 36. The substance used as a smoke screen in warfare is

1. SiCl₄
2. PH₃
3. PCl₅
4. Acetylene.

Answer: 1. SiCl₄

SiCl₄ gets hydrolysed in moist air and gives white fumes which are used as a smoke screen in warfare.

Hydrocarbons

Question 1. The dihedral angle of the least stable conformer of ethane is

1. 0°
2. 120°
3. 180°
4. 60°

Answer: 1. 0°

The dihedral angle of the least stable conformer of ethane is 0°. The magnitude of torsional strain depends upon the angle of rotation about the C – C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, the staggered form has the least torsional strain, and the eclipsed form has the maximum torsional strain.

Question 2.

Consider the above reaction and identify the missing reagent/chemical.

1. DIBAL-H
2. B₂H₆
3. Red Phosphorus
4. CaO

Answer: 4. CaO

Question 3. Which of the following alkanes cannot be made in good yield by the Wurtz reaction?

1. n-Hexane
2. 2, 3-Dimethylbutane
3. n-Heptane
4. n-Butane

Answer: 3. n-Heptane

Wurtz reaction is used for the preparation of higher alkanes containing an even number of C-atoms. Thus this reaction cannot be used for the preparation of n-heptane.

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Question 4. The alkane that gives only one monochloride product on chlorination with Cl₂ in the presence of diffused sunlight is

1. 2,2-dimethylbutane
2. Neopentane
3. n-pentane
4. Isopentane.

Answer: 2. Neopentane

In the chlorination of alkanes, hydrogen is replaced by chlorine. In neo-pentane, only one type of hydrogen is present, hence the replacement of any hydrogen atom will give the same product.

Question 5. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is

1. CH ≡ CH
2. CH₂ = CH₂
3. CH₃ — CH₃
4. CH₄

Answer: 4. CH₄

Question 6. With respect to the conformers of ethane, which of the following statements is true?

1. The bond angle changes but the bond length remains the same.
2. Both bond angle and bond length change.
3. Both bond angle and bond length remain the same.
4. Bond angle remains the same but the bond length changes.

Answer: 3. Both bond angle and bond length remain the same.

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

Question 7. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is

1. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain
3. The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has torsional strain
4. The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.

Answer: 2. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain

Conformers of ethane have different dihedral angles but the same bond angle and bond lengths.

The magnitude of torsional strain depends upon the angle of rotation of the C-C bond. The staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. So, the staggered conformation of ethane is more stable than the eclipsed conformation.

Question 8. In the following the most stable conformation of n-butane is

Answer: 2

The anti-conformation is the most stable conformation of n-butane as in this, the bulky methyl groups are as far apart as possible thereby keeping steric repulsion at a minimum

Question 9. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by

1. Oxidation
2. Cracking
3. Distillation under reduced pressure
4. Hydrolysis.

Answer: 2. Cracking

The process of cracking converts higher alkanes into smaller alkanes and alkenes. This process can be used for the production of natural gas.

Question 10. Which of the following conformers for ethylene glycol is most stable?

Answer: 4

The conformation (4) is most stable because of intermolecular H-bonding.

Question 11. The dihedral angle in the staggered form of ethane is

1. 0°
2. 120°
3. 60°
4. 180°

Answer: 3. 60°

The staggered form of ethane has the following structure and the dihedral angle is 60°, which means ‘H’ atoms are at an angle of 60° to each other.

Question 12. Which of the following reactions is expected to readily give a hydrocarbon product in good yields?

Answer: 3.

When an aqueous solution of sodium or potassium salt of a carboxylic acid is electrolyzed, hydrocarbon is evolved at the anode.

At anode: \(2 \mathrm{RCOO}^{-}-2 e^{-} \rightarrow \underset{ Alkane }{R-R}+2 \mathrm{CO}_2\)

Question 13. In commercial gasoline, the type of hydrocarbons that are more desirable is

1. Linear unsaturated hydrocarbon
2. Toluene
3. Branched hydrocarbon
4. Straight-chain hydrocarbon.

Answer: 3. Branched hydrocarbon

The branching of the chain increases the octane number of a fuel. A high octane number means better fuel

Question 14. The most stable conformation of n-butane is

1. Gauche
2. Staggered
3. Skew boat
4. Eclipsed.

Answer: 2. Staggered

Newman projections for n-butane are

The staggered conformation has minimum repulsion between the hydrogen atoms attached tetrahedrally to the two carbon atoms. Thus, it is the most stable conformation.

Question 15. Which of the following is used as an antiknocking material?

1. Glyoxal
2. Freon
3. T.E.L.
4. Ethyl alcohol

Answer: 3. T.E.L.

Tetraethyllead(C₂H₅)₄Pb, is used as an antiknocking agent in gasoline used for running automobiles.

Question 16. The reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order

1. Tertiary > primary > secondary
2. Primary > secondary > tertiary
3. Both (1) and (2)
4. Tertiary > secondary > primary.

Answer: 4. Tertiary > secondary > primary.

The reactivity of the H-atom depends upon the stability of free radicals, therefore reactivity of the H-atom follows the order: 3°>2°>1°.

Question 17. Compound X on reaction with O₃ followed by Zn/H₂O gives formaldehyde and 2-methylpropanal as products. The compound X is

1. 3-methyl but-l-ene
2. 2-methyl but-l-ene
3. 2-methyl but-2-ene
4. pent-2-ene.

Answer: 1

So, X can be

Question 18. The major product of the following chemical reaction is

Answer: 2

In the presence of peroxide, an addition reaction takes place.

Question 19. An alkene on ozonolysis gives methanal as one of the products. Its structure is

Answer: 3

Question 20. An alkene A on reaction with O₃ and Zn—H₂O gives propanone and ethanal in equimolar ratio. The addition of HCl to alkene A gives B as the major product. The structure of product B is

Answer: 4

Addition of HCI to an alkene (1) will take place according to Markownikoff’s rule.

Question 21. The most suitable reagent for the following conversion is

1. Hg²⁺/H⁺,H₂O
2. Na/liquid NH₂
3. H₂, Pd/C, quinoline
4. Zn/HCl

Answer: 3. H₂, Pd/C, quinoline

Question 22. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?

Answer: 3

Question 23. The compound that will react most readily with gaseous bromine has the formula

1. C₃H₆
2. C₂H₂
3. C₄H₁₀
4. C₂H₄

Amswer: 1. C₃H₆

The rate of free radical substitution with Br_2(g) depends upon the stability of free radicals. Propenyl free radical is allylic free radical which is more stable.

Question 24. In the reaction with HCl, an alkene reacts in accordance with the Markovnikovs rule to give a product 1-chloro-1-methyl-cyclohexane. The possible alkene is

Answer: 3

Question 25. Which of the following is not the product of dehydration of isomerism?

Answer: 1

Question 26. In the following reaction

Answer: 1

Question 27. Which of the following compounds will exhibit cis-trans (geometrical) isomerism?

1. Butanol
2. 2-Butyne
3. 2-Butenol
4. 2-Butene

Answer: 4. 2-Butene

Cis-trans isomerism is exhibited by compounds having C – C, C – N, and N – N groups, due to restricted rotation around the double bond. Among the given options, only 2-butene exhibits geometrical isomerism.

Question 28.

Answer: 4

Question 29. Which of the compounds with molecular formula C₅H₁₀ yields acetone on ozonolysis?

1. 3-Methyl-1-butene
2. Cyclopentane
3. 2-Methyl-1-butene
4. 2-Methyl-2-butene

Answer: 4. 2-Methyl-2-butene

Question 30. Which one of the following alkenes will react faster with H₂ under catalytic hydrogenation conditions?

Answer: 1

The relative rates of hydrogenation decrease with the increase of steric hindrance. Most stable alkene, slowly it undergoes hydrogenation to give the product. The least substituted alkene is less stable and more reactive.

Order of stability is:

Hence, alkene which will react faster with H, is that which is most unstable

Question 31. The reaction of HBr with propene in the presence of peroxide gives

1. Isopropyl bromide
2. 3-bromopropane
3. Allyl bromide
4. n-propyl bromide.

Answer: 4. n-propyl bromide.

The formation of n-propyl bromide in the presence of peroxide can be explained as follows:

Step 1: Peroxide undergoes fission to give free radicals \(\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow 2 \mathrm{R}-\dot{\mathrm{O}}\)

Step 2: HBr combines with free radicals to form bromine free radicals. \(R-\dot{\mathrm{O}}+\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{Br}}\)

Step 3: Br attacks the double bond of the alkene to form a more stable free radical

Step 4: More stable free radical attacks on HBr. \(\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow \underset{n-\text { Propyl bromide }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{Br}}}\)

Step 5: \(\dot{\mathrm{Br}}+\dot{\mathrm{Br}} \longrightarrow \mathrm{Br}_2\)

Question 32. The compound,  on reaction with NaIO₄ in the presence of KMnO₂ gives

1. CH₃COCH₃
2. CH₃COCH₃ + CH₃COOH
3. CH₃COCH₃ + CH₃CHO
4. CH₃CHO + CO₃

Answer: 2. CH₃COCH₃ + CH₃COOH

Question 33. Geometrical isomers differ in

1. Position of functional group
2. Position of atoms
3. Spatial arrangement of atoms
4. Length of the carbon chain.

Answer: 3. Spatial arrangement of atoms

Geometrical isomers are those isomers that possess the same molecular and structural formula but differ in the arrangement of atoms or groups in space due to hindered rotation around the double-bonded atoms.

Question 34. In preparation of alkene from alcohol using Al₂O₃ which is the effective factor?

1. Porosity of Al₂O₃
2. Temperature
3. Concentration
4. Surface area of Al₂O₃

Answer: 2. Temperature

Temperature is an effective factor because at different temperatures it forms different products

This is intramolecular dehydration. At lower temperatures, intermolecular dehydration takes place between two molecules of alcohol and ether will be formed.

Question 35. Which reagent converts propane to 1-propanol?

1. \(\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{SO}_4\)
2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)
3. \(\mathrm{Hg}(\mathrm{OAc})_2, \mathrm{NaBH}_4 / \mathrm{H}_2 \mathrm{O}\)
4. Aq. KOH

Answer: 2. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}\)

Propene adds to diborane (B₂H₆) giving an additional product. The addition compound on oxidation gives l-propanol. Here the addition of water takes place according to anti-Markownikoff ‘s rule

Question 36. Which is maximum stable?

1. 1-Butene
2. cis-2-Butene
3. trans-2-Butene
4. All have the same stability.

Answer: 3. trans-2-Butene

This is the most stable as the repulsion between two methyl groups is the least.

Question 37. 2-Butene shows geometrical isomerism due to

1. Restricted rotation about the double bond
2. Free rotation about the double bond
3. Free rotation about single bond
4. Chiral carbon.

Answer: 1. Restricted rotation about the double bond

Due to restricted rotation about double bonds, 2-butene shows geometrical isomerism.

Question 38. 2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

1. trans-2-pentene
2. 1-pentene
3. 2-ethoxy pentane
4. 2-ds-pentene.

Answer: 1. trans-2-pentene

Question 39. In a reaction,    where, M = Molecule and R = Reagent. M and R are

1. CH₃CH₂OH and HCl
2. CH₂ = CH₂ and heat
3. CH₃CH₂Cl and NaOH
4. CH₂Cl – CH₂OH and aq. NaHCO₃

Answer: 4. CH₂Cl – CH₂OH and aq. NaHCO₃

Therefore, M = CH₂CI-CH₂OH and

R = ag.NaHCO₃

Question 40. The reaction, CH₂ = CH – CH₂ + HBr → CH₂CHBr – CH₃ is

1. Electrophilic substitution
2. Free radical addition
3. Nucleophilic addition
4. Electrophilic addition.

Answer: 4. Electrophilic addition.

In this reaction, HBr undergoes heterolytic fission as HBr → H⁺ + Br^–

This is an example of an electrophilic addition reaction

Question 41. Which of the following has zero dipole moment?

1. 1-Butene
2. 2-Methyl-1-propene
3. cis-2-Butene
4. trans-2-Butene

Answer: 4. trans-2-Butene

Question 42. One of the following which does not observe the anti-Markownikoff addition of HBr, is

1. Pent-2-ene
2. Propene
3. But-2-ene
4. But-1-ene.

Answer: 3. But-2-ene

In the case of but-2-ene (CH₃– CH-CH- CH₂)both double-bonded carbons are identical. Therefore, it does not observe the anti-Markownikoff addition of HBr

Question 43. Reduction of 2-butyne with sodium in liquid ammonia gives predominantly

1. cis-2-butene
2. No reaction
3. Trans-2-butene
4. n-butane.

Answer: 3. Trans-2-butene

Reduction of non-terminal alkynes with Na in Liq. NH₃ at 195 – 200 K gives trans-alkene.

Question 44. The restricted rotation about the carbon double bond in 2-butene is due to

1. Overlap of one s and sp²-hybridized orbitals
2. Overlap of two sp²-hybridized orbitals
3. Overlap of one p and one sp²-hybridized orbitals
4. Sideways overlap of two p-orbitals.

Answer: 4. Sideways overlap of two p-orbitals.

Restricted rotation is due to sideways overlap of two p-orbitals.

Question 45. Which one of the following can exhibit cis-trans isomerism?

1. CH₃ – CHCl – COOH
2. H – C ≡ C – Cl
3. ClCH =CHCl
4. ClCH₂ – CH₂Cl

Answer: 3. ClCH =CHCl

1, 2-Dichloroethene exhibits cis-trans (geometrical) isomerism

Question 46. In the following reaction, the number of sigma(σ) bonds present in the product A, is

1. 21
2. 9
3. 24
4. 18

Answer: 1. 21

There are 21 σ bonds.

Question 47. Which one is the correct order of acidity?

1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)
2. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{CH}_3\)
3. \(\mathrm{CH}_3-\mathrm{CH}_3>\mathrm{CH}_2=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)
4. \(\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\) \(>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\)

Answer: 1. \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}\) \(>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{CH}_3\)

Alkanes, alkenes, and alkynes follow the following trend in their acidic behavior:

⇒ \(\stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}>\stackrel{s p^2}{\mathrm{H}_2 \mathrm{C}}=\stackrel{s p^2}{\mathrm{CH}_2}>\stackrel{s p^3}{\mathrm{CH}_3}-\stackrel{s p^3}{\mathrm{CH}_3}\)

This is because sp-hybridised carbon is more electronegative than sp²-hybridized carbon which is further more electronegative than sp³-hybridised carbon.

Hence, in ethyne proton can be released more easily than ethene and ethane.

Among alkynes, the order of acidity is HC ≡ CH > CH₃ -C ≡ CH > CH₃ -C ≡ C- CH₃ This is due to the +1 effect of – CH₃ group.

Question 48. Predict the correct intermediate and product in the following reaction:

Answer: 3

In the case of unsymmetrical alkynes addition of H₂O occurs in accordance with Markownikoff’s rule.

Question 49. The pair of electrons in the given carbanion, CH₃C≡ C^–, is present in which of the following orbitals?

1. sp²
2. sp
3. 2p
4. sp³

Answer: 2. sp²

⇒ \(\mathrm{CH}_3-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}\)

Thus, a pair of electrons is present in the sp-hybridized orbital

Question 50. In the reaction H-C ≡ CH X and Y are

1. X = 2-butyne, Y = 2-hexyne
2. X = 1-butyne, Y = 2-hexyne
3. X = 1-butyne, Y = 3-hexyne
4. X = 2-butyne, Y = 3-hexyne.

Answer: 3. X = 1-butyne, Y = 3-hexyne

Question 51. Which of the following organic compounds has the same hybridization as its combustion product (CO₂)?

1. Ethane
2. Ethyne
3. Ethene
4. Ethanol

Answer: 2. Ethyne

⇒ \(\mathrm{C}_2 \mathrm{H}_2+\frac{5}{2} \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Both ethyne and CO₂ have sp-hybridization.

⇒ \(\mathrm{O}=\stackrel{s p}{\mathrm{C}}=\mathrm{O} \quad \stackrel{s p}{\mathrm{HC}} \equiv \stackrel{s p}{\mathrm{C}} \mathrm{H}\)

Question 52. In the following reaction: Product ‘P’ will not give

1. Tollens’ reagent test
2. Bradys reagent test
3. Victor Meyer test
4. Iodoform test.

Answer: 3. Victor Meyer test

Acetaldehyde does not give Victor Meyer a test.

Question 53. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?

1. NaNH₂
2. HCl
3. O₂
4. Br₂

Answer: 1. NaNH₂

Terminal alkynes (l-butyne) react with NaNH₂ to form sodium acetylide and evolve hydrogen but 2-butyne does not.

Question 54. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear.

1. CH₃ — CH = CH — CH₃
2. CH₃ — C ≡ C — CH₃
3. CH₃=CH —CH₃ —C ≡ CH
4. CH₃ — CH₂ — CH₃

Answer: 2. CH₃ — C ≡ C — CH₃

CH₃ – C ≡ C-CH₃

In the case of sp³ hybridized carbon, the bond angle is 109°28′; sp² hybridized carbon, the bond angle is 120° and sp hybridized carbon, the bond angle is 180°.

So, only   is linear.

Question 55. Base strength of

1. \(\mathrm{H}_3 \mathrm{C} \stackrel{-}{\mathrm{C}} \mathrm{H}_2\),
2. \(\mathrm{H}_2 \mathrm{C}=\stackrel{-}{\mathrm{C}} \mathrm{H}\)
3. and \(\mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}}\)

is in the order of

1. (A) > (C) > (B)
2. (A) > (B) > (C)
3. (B) > (A) > (C)
4. (C) > (B) > (A)

Answer: 2. (A) > (B) > (C)

⇒ \(\mathrm{H}-\underset{s p}{\mathrm{C}} \equiv \underset{s p}{\mathrm{C}}-\mathrm{H}>\underset{s p^2}{\mathrm{CH}_2}=\underset{s p^2}{\mathrm{CH}_2}>\underset{s p^3}{\mathrm{CH}_3} \mathrm{CH}_3\)

Conjugate base of the given acid: \(\stackrel{-}{\mathrm{C}} \equiv \mathrm{C}-\mathrm{H}<\stackrel{-}{\mathrm{C H}} =\mathrm{CH}_2<\stackrel{-}{\mathrm{C}} \mathrm{H}_2 \mathrm{CH}_3\)

The conjugate base of stronger acid is weaker and vice-versa.

Question 56. Predict the product C obtained in the following reaction of 1-butyne. 

Answer: 3

According to Markownikoff’s rule, during hydrohalogenation to an unsymmetrical alkene, the negative part of the addendum adds to a less hydrogenated (i.e. more substituted) carbon atom.

Question 57. Products of the following reaction: are

1. CH₃COOH + CO₂
2. CH₃COOH + HOOCCH₂CH₃
3. CH₃CHO + CH₃CH₂CHO
4. CH₃COOH + CH₃COCH₃

Answer: 2. CH₃COOH + HOOCCH₂CH₃

On ozonolysis, higher alkynes form diketones which are further oxidized to dicarboxylic acid.

Question 58. When CH₃CH₂CHCl₂ is treated with NaNH₂, the product formed is

Answer: 2

Question 59. When acetylene is passed through dil. H₂SO₄ in the presence of HgSO₄, the compound formed is

1. Acetic acid
2. Ketone
3. Ether
4. Acetaldehyde.

Answer: 4. Acetaldehyde.

Question 60. The cylindrical shape of an alkyne is due to

1. Two sigma C – C and one π C – C bonds
2. One sigma C – C and two π C-C bonds
3. Three sigma C – C bonds
4. Three π C – C bonds.

Answer: 2. One sigma C – C and two π C-C bonds

In alkyne, two carbon atoms constituting the triple bond are sp-hybridised’Carbon undergoes sp-hybridization to form two sp-hybrid orbitals. The two 2p-orbitals remain unhybridized. Hybrid orbitals form one sigma bond while two π-bonds are formed by unhybridised orbitals.

Question 61. The reagent is

1. Na
2. HCl in H₂O
3. KOH in C₂H₅OH
4. Zn in alcohol.

Answer: 3. KOH in C₂H₅OH

A powerful base is needed to carry out a second dehydrohalogenation reaction, for example, a hot alcoholic KOH solution or alkoxide ion.

Question 62. A compound is treated with NaNH₂ to give sodium salt. Identify the compound.

1. C₂H₂
2. C₆H₆
3. C₂H₆
4. C₂H₄

Answer: 1. C₂H₂

Alkynes react with strong bases like NaNH₂ to form sodium acetylide derivatives known as acrylics.

⇒ \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{NaNH}_2 \longrightarrow \mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}} \mathrm{Na}^{+}+1 / 2 \mathrm{H}_2\)

Question 63. The shortest C-C bond distance is found in

1. Diamond
2. Ethane
3. Benzene
4. Acetylene.

Answer: 4. Acetylene

The shortest C- C distance (1.20 Å) is in acetylene.

Question 64. Acetylenic hydrogens are acidic because

1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character
2. Acetylene has only open hydrogen in each carbon
3. Acetylene contains the least number of hydrogens among the possible hydrocarbons having two carbons
4. Acetylene belongs to the class of alkynes with the molecular formula, C_nH_2n-2.

Answer: 1. Sigma electron density of the C – H bond in acetylene is nearer to carbon, which has 50% s-character

The formation of a C-H bond in acetylene involves a sp-hybridised carbon atom. Since s-electrons are closer to the nucleus than p-electrons, the electrons present in a bond having more s-characters will be closer to the nucleus. In alkynes, the character is 50%, and the electrons constituting this bond are more strongly bonded by the carbon nucleus.

Thus acetylenic C-atom becomes more electronegative in comparison to sp², sp³ and hence the hydrogen atom Present on the carbon atom (≡C-H) can be easily removed

Question 65. Which is the most suitable reagent among the following to distinguish compound (3) from the rest of the compounds?

1. CH₃-C ≡ C-CH₃
2. CH₃ – CH₂ – CH₂ – CH₃
3. CH₃-CH₂C ≡ CH
4. CH₃-CH ≡ CH₂

1. Bromine in carbon tetrachloride
2. Bromine in acetic acid
3. Aik. KMnO₄
4. Ammoniacal silver nitrate

Answer: 4. Ammoniacal silver nitrate

Compound 3 possessing the terminal alkyne only reacts with ammoniacal AgNO₃ and thus can be distinguished from 1, 2, and 4 compounds.

Question 66. Consider the following compound/species:

The number of compounds/species which obey Huckel’s rule is __________

1. 2
2. 5
3. 4
4. 6

Answer: 3. 4

These compounds obey Huckel’s rule as these contain (4n + 2)π electrons.

Question 67. Which compound amongst the following is not an aromatic compound?

Answer: 4

Compound (4) is not an aromatic compound as reflected by the non-planarity of the methylene bridge (-CH₂-) with respect to other atoms. However, the tropylium cation is aromatic due to planarity.

Question 68. Among the following the reaction that proceeds through an electrophilic substitution is

Answer: 3

The attacking species in the reaction given in option (3) is an electrophile i.e., \(\begin{aligned}
& \delta+ \\
& \mathrm{Cl}
\end{aligned}\)

Therefore, it is an electrophilic substitution reaction.

Question 69. Which of the following can be used as the halide component for the Friedel-Crafts reaction?

1. Chlorobenzene
2. Bromobenzene
3. Chloroethene
4. Isopropyl chloride

Answer: 4. Isopropyl chloride

Friedel Crafts Reaction:

Chlorobenzene, bromobenzene, and chloroethene are not suitable halide components as the C-X bond acquires some double bond character due to the resonance of a lone pair of electrons with π-bond

Question 70. In which of the following molecules, all atoms are coplanar?

Answer: 1

Biphenyl is coplanar as all C-atoms are sp² hybridized.

Question 71. In pyrrole, the electron density is maximum on

1. 2 and 3
2. 3 and 4
3. 2 and 4
4. 2 and 5

Answer: 4. 2 and 5

Pyrrole has maximum electron density on 2 and 5. It generally interacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbonation that is a hybrid of structures (1) and (2). Attack at position 2 or 5  yields a carbonation that is a hybrid not only of structures (3) and (4) (analogous to 1 and 2) but also of structure (5).

The extra stabilization conferred by (5) makes this ion the more stable one. Also, attack at position 2 or 5 is laster because the developing positive charge is accommodated by three atoms of the ring instead of only two.

Question 72. In the given reaction, the product P is the product P is?

Answer: 3

Question 73. Consider the nitration of benzene using a mixed cone. H₂SO₄ and HNO₃. If a large amount of KHSO₄ is added to the mixture, the rate of nitration will be

1. Unchanged
2. Doubled
3. Faster
4. Slower.

Answer: 4. Slower.

Mechanism of nitration is: \(\mathrm{HNO}_3+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NO}_2^{+}+2 \mathrm{HSO}_4^{-}+\mathrm{H}_5 \mathrm{O}^{+}\)

If a large amount of KHSO₄ is added then conc. of HSO^–₄ ions increase and the reaction will be shifted in a backward direction hence, the rate of nilration will be slower.

Question 74. The oxidation of benzene by V2Os in the presence of air produces

1. Maleic anhydride
2. Benzoic acid
3. Benzaldehyde
4. Benzoic anhydride.

Answer: 1. Maleic anhydride

Question 75. What products are formed when the following compound is treated with Br₂ in the presence of FeBr₃?

Answer: 3

– CH₃ group is o,p-directing. Because of crowding, no substitution occurs at the carbon atom between the two – CH₃ groups in m-xylene, even though two – CH₃ groups activate that position.

Question 76. Some meta-directing substituents in aromatic substitution are given, Which one is most deactivating?

1. — COOH
2. — NO₂
3. — C ≡ N
4. — SO₃H

Answer: 2. — NO₂

-NO₂ is most deactivating due to – I and – M effect

Question 77. Which of the following compounds will not undergo Friedel-Crafts reaction easily?

1. Nitrobenzene
2. Toluene
3. Cumene
4. Xylene

Answer: 1. Nitrobenzene

Nitrobenzene is strongly deactivated, hence will not undergo Friedel-Crafts reaction.

Question 78. Which of the following chemical systems is nonaromatic?

Answer: 4

The molecules that do not satisfy the Huckel rule or (4n + 2)π-electron rule are said to be non-aromatic. The compound (4) has a total of 4πe^–. It does not follow (4n + 2)π tule. So, it is a non-aromatic compound.

Question 79. Among the following compounds, the one that is most reactive toward electrophilic nitration is

1. Benzoic acid
2. Nitrobenzene
3. Toluene
4. Benzene.

Answer: 3. Toluene

As the +I effect increases reactivity towards electrophilic reactions decreases and as -I or -M elect increases, reactivity toward electrophilic reactions decreases thus, the order is

Question 80. The reaction of toluene with Cl₂ in the presence of FeCl₃ gives X and the reaction in the presence of light gives Y. Thus, X and Y are

1. X = benzal chloride, Y = o-chlorotoluene
2. X = m-chlorotoluene, Y = p-chlorotoluene
3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene
4. X = benzyl chloride, Y= m-chlorotoluene.

Answer: 3. X = o- and p-chlorotoluene, Y = trichloromethyl benzene

The reaction of Cl₂ in the presence of FeCl₃, with toluene yields a ring substitution product.

In presence of sunlight, free radical reaction takes place.

Question 81. Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ to form

1. Chlorobenzene
2. Benzyl chloride
3. Xylene
4. Toluene.

Answer: 4. Toluene.

This is Friedel-Crafts alkylation.

Question 82. Nitrobenzene can be prepared from benzene by using a mixture of cones. HNO₃ and cone. H₂SO₄. In the mixture, nitric acid acts as an

1. Acid
2. Base
3. Catalyst
4. Reducing agent.

Answer: 2. Base

Question 83. Which one of the following is most reactive towards electrophilic attack?

Answer: 1

Groups like, -CI and -NO, show -I effet -I group attached to the benzene ring decrease the electron density and hence less; prone to electrophilic attack. -OH not only shows the -I effect but also the +M effect which predominates the -I character and electron density is increased in the benzene ring which facilitates electrophilic attack.

Question 84. The order of decreasing reactivity towards an electrophilic reagent, for the following, would be

1. Benzene
2. Toluene
3. Chlorobenzene
4. Phenol

1. (B) > (D) > (A) > (C)
2. (D) > (C) > (B) > (A)
3. (D) > (B) > (A) > (C)
4. (A) > (B) > (C) > (D)

Answer: 3. (D) > (B) > (A) > (C)

Benzene having any activaling group i.e., – OH, – R undergoes electrophilic substitution easily as compared to benzene itself. Thus, toluene and phenol undergo electrophilic substitution easily. Chlorine due to -I-reflect deactivates the ring. So, it is difficult to Larry about the electrophilic substitution in chlorobenzene.

Hence, the order is C₆H₅OH > C₆H₅CH₃ > C₆H₆ > C₆H₅Cl.

Question 85. Using anhydrous AlCl₃ as a catalyst, which one of the following reactions produces ethylbenzene (PhEt)?

1. H₃C – CH₂OH + C₆H₆
2. CH₃ – CH = CH₂ + C₆H₆
3. H₂C = CH₂ + C₆H₆
4. H₃C – CH₃ + C₆H₆

Answer: 3. H₂C = CH₂ + C₆H₆

Question 86. Which one of the following is a free-radical substitution reaction?

Answer: 1

Question 87. The correct order of reactivity towards the electrophilic substitution of the compounds aniline (1), benzene (2), and nitrobenzene (3) is

1. 3 > 2 > 1
2. 2 > 3 > 1
3. 1 < 2 > 3
4. 1 > 2 > 3

Answer: 4. 1 > 2 > 3

– NH₂ group is electron donating hence increasing electron density on ring. Benzene is also electron-rich due to the delocalization of electrons. -NO₂ the group is electron withdrawing hence, decreases electron density on the ring. Thus, the correct order for electrophilic substitution is 1>2>3.

Question 88. Increasing the order of electrophilic substitution for the following compounds

1. 4 < 1 < 2 < 3
2. 3 < 2 < 1 < 4
3. 1 < 4 < 3 < 2
4. 2 < 3 < 1 < 4

Answer: 1. 4 < 1 < 2 < 3

Due to -I effect of F atom, -CF₃ on the benzene ring, deactivates the ring and does not favor electrophilic substitution. While – CH₃ and – OCH₃ are an electron-donating group that favors electrophilic substitution in the benzene ring at the ‘ortho’ and ‘para’ positions. The +I effect of -OCH₃ is more than – CH₃, therefore the correct order for electrophilic substitution is

Question 89. In Friedel-Crafts reaction, toluene can be prepared by

1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{CH}_4\)
3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_2 \mathrm{Cl}_2\)
4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In the Friedel-Crafts reaction, toluene is obtained by the action of CH₃Cl on benzene in the presence of AlCl₃

Question 90. In Friedel-Crafts alkylation, besides AlCl₃ the other reactants are

1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)
2. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_4\)
3. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{NH}_3\)
4. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl}\)

Answer: 1. \(\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl}\)

In Friedel-Crafts reaction, Toluene an alkyl group is introduced into the benzene ring in the presence of a Lewis is acid (AlCl₃) catalyst. The reaction is

Question 91. Which of the following compounds will be most easily attacked by an electrophile?

Answer: 1

-OH, -Cl, and -CH₃ groups in benzene are ortho, para directing, groups. But among these – OH group is strongly activating while – CH₃ is weakly activating and – Cl is deactivating. Thus, phenol will be most easily attacked by an electrophile.

Question 92. Which one of these is not compatible with arenes?

1. Electrophilic additions
2. Delocalisation of π-electrons
3. Greater stability
4. Resonance

Answer: 1. Electrophilic additions

Arenes undergo electrophilic substitution reactions and are resistant to addition reactions, due to delocalization of m-electrons. These are also stabilized by resonance.

Question 93. Among the following compounds (1-3) the correct order of reaction with electrophiles is

1. 1>2>3
2. 1=2>3
3. 2>3>1
4. 3<1<2

Answer: 1. 1>2>3

In structure 3, the withdrawal of electrons by -NO₂ causes a decrease in reaction rate while in structure I, there is an electron-releasing effect by the – OCH₃ group which accelerates the reaction.

The order of reactivity towards electrophile is 1>2>3

Question 94. Electrophile in the case of chlorination of benzene in the presence of FeCl₃ is

1. Cl
2. FeCl₃
3. Cl⁺
4. Cl^–

Answer: 3. Cl⁺

⇒ \(\mathrm{Cl}_2+\mathrm{FeCl}_3 \longrightarrow \mathrm{FeCl}_4^{-}+\mathrm{Cl}^{+}\)

Question 95. The reactive species in the nitration of benzene is

1. \(\mathrm{NO}_3\)
2. \(\mathrm{HNO}_3\)
3. \(\mathrm{NO}_2^{+}\)
4. \(\mathrm{NO}_2^{-}\)

Answer: 3. \(\mathrm{NO}_2^{+}\)

Nitronium ion (NO⁺₂) is an electrophile that actually attacks the benzene ring.

Question 96. Which is the correct symbol relating to the two Kekule structures of benzene?

1. \(\rightleftharpoons\)
2. \(\longrightarrow\)
3. \(\equiv\)
4. \(\longleftrightarrow\)

Answer: 4. \(\longleftrightarrow\)

Benzene shows Kekule structures which are resonating structures and these structures are separated by a double-headed arrow (\(\leftrightarrow\)).

Question 97. Select the true statement about benzene amongst the following

1. Because of unsaturation benzene easily undergoes addition
2. There are two types of C – C bonds in the benzene molecule
3. There is cyclic delocalization of π-electrons in benzene
4. Monosubstitution of benzene gives three isomeric products.

Answer: 3. There is cyclic delocalization of π-electrons in benzene

Due to resonance all the C – C bonds in the benzene possess the same nature and the resonating structures are obtained because of the delocalization of π-electrons.

Electrochemistry

Question 1. The standard electrode potential (E°) values of Al³⁺/Al, Ag⁺/Ag, K⁺/K, and Cr³⁺/Cr are -1.66 V, 0.80 V, -2.93 V and -0.74 V, respectively. The correct decreasing order of reducing power of the metal is

1. Ag > Cr > Al > K
2. K > Al > Cr > Ag
3. K > Al > Ag > Cr
4. Al > K > Ag > Cr

Answer: 2. K > Al > Cr > Ag

The higher the value of E°_red, the stronger is the oxidizing power. Thus, the decreasing order of reducing the power of the metals K>AI>Cr>Ag.

Question 2. A button cell used in watches function as following: \(\mathrm{Zn}_{(s)}+\mathrm{Ag}_2 \mathrm{O}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons 2 \mathrm{Ag}_{(s)}+\mathrm{Zn}_{(a q)}^{2+}+2 \mathrm{OH}_{(a q)}^{-}\) If half cell potentials are \(\mathrm{Zn}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Zn}_{(s)} ; E^{\circ}=-0.76 \mathrm{~V}\); \(\mathrm{Ag}_2 \mathrm{O}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(l)}+2 e^{-} \rightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{OH}_{(a q)}^{-} ; E^{\circ}=0.34 \mathrm{~V}\)

The cell potential will be

1. 0.84 V
2. 1.34 V
3. 1.10 V
4. 0.42 V

Answer: 3. 1.10 V

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {O.P. }}^{\circ}+E_{\text {R.P. }}^{\circ}=0.76+0.34=1.10 \mathrm{~V}\)

Question 3. The standard reduction potentials of the half-reactions are given below:

• \(\mathrm{F}_{2(g)}+2 e^{-} \rightarrow 2 \mathrm{~F}_{(a q)}^{-} ; E^{\circ}=+2.85 \mathrm{~V}\)
• \(\mathrm{Cl}_{2(g)}+2 e^{-} \rightarrow 2 \mathrm{Cl}_{(a q)}^{-} ; E^{\circ}=+1.36 \mathrm{~V}\)
• \(\mathrm{Br}_{2(l)}+2 e^{-} \rightarrow 2 \mathrm{Br}_{(a q)}^{-} ; E^{\circ}=+1.06 \mathrm{~V}\)
• \(\mathrm{I}_{2(s)}+2 e^{-} \rightarrow 2 \mathrm{I}_{(a q)}^{-} ; E^{\circ}=+0.53 \mathrm{~V}\)

The strongest oxidizing and reducing agents respectively are

1. F₂ and I^–
2. Br₂ and Cl^–
3. Cl₂ and Br^–
4. Cl₂ and I₂

Answer: 1. F₂ and I^–

Less lower the value of reduction potential, the stronger the reducing agent thus, I am the strongest reducing agent. More positive, the value of reduction potential shows good oxidizing properties thus, the strongest oxidizing agent is F₂.

Read and Learn More NEET MCQs with Answers

Question 4. Standard electrode potentials of three metals X, Y, and Z are -1.2 V, + 0.5 V, and – 3.0 V respectively. The reducing power of these metals will be

1. Y>Z>X
2. Y> X> Z
3. Z>X>Y
4. X> Y> Z

Answer: 3. Z>X>Y

The more negative the value of reduction potential, the stronger will be the reducing agent.

So, Z (-3.0 V) > X (-1.2 V) > Y (+ 0.5 V)

Question 5. The standard electrode potential for Sn⁴⁺⁺/ Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is -0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be

1. + 1.19 V
2. + 0.89 V
3. + 0.18 V
4. + 1.83 V

Answer: 2. + 0.89 V

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\) = 0.15-(-0.74)=0.15+0.74=0.89 V

Question 6. A solution contains Fe²⁺, Fe³⁺, and I^– ions. This solution was treated with iodine at 35°C. E° for Fe³⁺/Fe²⁺ is + 0.77 V and E° for I₂/2I^– = 0.536 V. The favorable redox reaction is

1. I₂ will be reduced to I^–
2. There will be no redox reaction
3. I^– will be oxidized to I₂
4. Fe²⁺ will be oxidized to Fe³⁺.

Answer: 3. I will be oxidized to I₂

Since the reduction potential of \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) is greater than that of \(\mathrm{I}_{2} / \mathrm{I}^{-}\) will be reduced and I^– will be oxidised.

⇒ \(2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\)

Question 7. Consider the following relations for emf of an electrochemical cell

1. EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
2. EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
3. EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
4. EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

Which of the above relations is correct?

1. (3) and (1)
2. (1) and (2)
3. (3) and (4)
4. (2) and (4)

Answer: 4. (2) and (4)

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of the anode Oxidation potential of the cathode.

Question 8. On the basis of the following E° values, the strongest oxidizing agent is \({\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+e^{-} ; E^{\circ}=-0.35 \mathrm{~V}}\);  \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+e^{-} ; E^{\circ}=-0.77 \mathrm{~V}\)

1. \(\mathrm{Fe}^{3+}\)
2. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\)
3. \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}\)
4. \(\mathrm{Fe}^{2+}\)

Answer: 1. \(\mathrm{Fe}^{3+}\)

⇒ \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}, E^{\circ}\) = \(+0.35 \mathrm{~V}\); \(\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}^{2+} ; E^{\circ}=+0.77 \mathrm{~V}\)

The higher the +ve reduction potential stronger the oxidising agent. Oxidizing agent oxidizes other compounds and gets themselves reduced easily. Thus, Fe³⁺ is the strongest oxidizing agent.

Question 9. A hypothetical electrochemical cell is shown below: A/A⁺ (x M)||B⁺ (y M)| B The emf measured is + 0.20 V. The cell reaction is

1. \(A+B^{+} \rightarrow A^{+}+B\)
2. \(A^{+}+B \rightarrow A+B^{+}\)
3. \(A^{+}+e^{-} \rightarrow A ; B^{+}+e^{-} \rightarrow B\)
4. The cell reaction cannot be predicted.

Answer: 1. \(A+B^{+} \rightarrow A^{+}+B\)

From the given expression:

At anode A → A⁺ + e^– (oxidation)

At cathode B⁺ + e^– → B (reduction)

Overall reaction is A + B⁺ → A⁺ + B

Question 10. \(E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.441 \mathrm{~V} \text { and } E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}\), the standard EMF of the reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}\) will be

1. 0.111 V
2. 0.330 V
3. 1.653 V
4. 1.212 V

Answer: 4. 1.212 V

⇒ \(\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E^{\circ}=-0.441 \mathrm{~V}\)…..(1)

⇒ \(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; E^{\circ}=0.771 \mathrm{~V}\)….(2)

⇒ \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+} ; E^{\circ}=?\)

To get the above equation, (2) x 2 – (1)

Question 11. Standard electrode potentials are Fe²⁺/Fe; E° = -0.44 and Fe³⁺/Fe²⁺; E° = 0.77. Fe²⁺, Fe³⁺, and Fe blocks are kept together, then

1. Fe³⁺ increases
2. Fe³⁺ decreases
3. Fe²⁺/Fe³⁺ remains unchanged
4. Fe²⁺ decreases.

Answer: 2. Fe³⁺ decreases

⇒ \(E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V}\)

⇒ \(E_{\mathrm{Fe}^{3+}}^{\circ} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V}\)

If a cell is constructed by combining these two electrodes oxidation occurs at Fe²⁺ /Fe electrode.

If Fe²⁺, Fe^3+, and Fe blocks are kept together then Fe³⁺ reacts with Fe to yield Fe²⁺ i.e., the concentration of Fe³⁺ is decreased and that of Fe²⁺ is increased

Question 12. Electrode potential for the following half-cell reactions are

• \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}; E^{\circ}=+0.76 \mathrm{~V}\)
• \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 e^{-} ; E^{\circ}=+0.44 \mathrm{~V}\)

The EMF for the cell reaction \(\mathrm{Fe}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\) will be

1. -0.32 V
2. + 1.20 V
3. -1.20 V
4. + 0.32 V

Answer: 4. + 0.32 V

⇒ \(E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=+0.76 \mathrm{~V}\)

⇒ \(E_{\mathrm{Fe} / \mathrm{Fe}}{ }^{2+}=0.44 \mathrm{~V} \Rightarrow E_{\mathrm{Fe}}^{\circ}{ }^{2+} / \mathrm{Re}=-0.44 \mathrm{~V}\)

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {O.P. }}^{\circ}+E_{\text {R.P. }}^{\circ}=+0.76-0.44=+0.32 \mathrm{~V}\)

Question 13. An electrochemical cell is set up of: Pt;H₂ (1 atm) | HCl(0.1M) || CH₃COOH(0.1M) |H₂( 1 atm); Pt. The e.m.f. of this cell will not be zero, because

1. Acids used in two compartments are different
2. e.m.f. depends on the molarities of acids used
3. The temperature is constant
4. pH of 0.1 M HCl and 0.1 M CH₃COOH is not the same.

Answer: 4. pH of 0.1 M HCl and 0.1 M CH₃COOH is not the same

Since it is a concentration cell and the concentration of H⁺ ions in two electrolyte solutions (HCl and CH3₃COOH) are different i.e., pH of 0.1 M HCl and 0.1 M CH₃COOH is not the same, therefore e.m.f. of this cell will not be zero

Question 14. Standard reduction potentials at 25°C of Li⁺|Li, Ba²⁺|Ba, Na⁺|Na, and Mg²⁺|Mg are -3.05, -2.90, -2.71, and -2.37 volt respectively. Which one of the following is the strongest oxidizing agent?

1. Ba²⁺
2. Mg²⁺
3. Na⁺
4. Li⁺

Answer: 2. Mg²⁺

The more positive or less negative the reduction potential value, the stronger is the oxidizing agent.

Question 15. A solution of potassium bromide is treated with each of the following. Which one would liberate bromine?

1. Hydrogen iodide
2. Sulfur dioxide
3. Chlorine
4. Iodine

Answer: 3. Chlorine

A stronger oxidizing agent (Cl₂) displaces a weaker oxidizing agent (Br₂) from its salt solution.

⇒ \(2 \mathrm{KBr}+\mathrm{Cl}_2 \rightarrow 2 \mathrm{KCl}+\mathrm{Br}_2\)

Question 16. Given below are two statements: One is labelled as Assertion A and the other is labeled as Reason R;

Assertion A: In equation Δ_rG = -nFE_cell, the value of Δ_rG depends on n.

Reason R: cell is an intensive property and ArG is an extensive property.

In light of the above statements, choose the correct answer from the options given below.

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true and R is the correct explanation of A.
4. Both A and R are true and R is not the correct explanation of A.

Answer: 4. Both A and R are true and R is not the correct explanation of A.

Δ_rG =-nFE_cell

E_cell, is an intensive parameter but Δ_rG is an extensive thermodynamic property and the value of Δ_rG depends on n.

Question 17. Given below are half-cell reactions: \(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\); \(E_{\mathrm{Mn}^{2+} / \mathrm{MnO}_4^{-}}^{\circ}=-1.510 \mathrm{~V}\); \(\frac{1}{2} \mathrm{O}_2+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}; E_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}\) Will the permanganate ion, MnO₄ water in the presence of an acid?

1. Yes, because E°_cell= +0.287 V
2. No, because E°_cell = -0.287 v
3. Yes, because E°_cell = +2.733 V
4. No, because E°_cell = -2.733 V

Answer: 1. Yes, because E°_cell= +0.287 V

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=+1.510-1.223=+0.287 \mathrm{~V}\)

As \(E_{\text {cell }}^{\circ}\) is positive, hence the reaction is feasible.

Question 18. At 298 K the standard electrode potentials of Cu²⁺ / Cu, Zn²⁺/ Zn, Fe²⁺/ Fe, and Ag⁺/Ag are 0.34 V, -0.76 V, -0.44 V, and 0.80 V respectively. On the basis of standard electrode potential, predict which of the following reactions cannot occur.

1. \(\mathrm{CuSO}_{4(a q)}+\mathrm{Zn}_{(s)} \longrightarrow \mathrm{ZnSO}_{4(a q)}+\mathrm{Cu}_{(s)}\)
2. \(\mathrm{CuSO}_{4(a q)}+\mathrm{Fe}_{(s)} \longrightarrow \mathrm{FeSO}_{4(a q)}+\mathrm{Cu}_{(s)}\)
3. \(\mathrm{FeSO}_{4(a q)}^{4(a q)}+\mathrm{Zn}_{(s)} \longrightarrow \mathrm{ZnSO}_{4(a q)}+\mathrm{Fe}_{(s)}\)
4. \(2 \mathrm{CuSO}_{4(a q)}+2 \mathrm{Ag} g_{(s)} \longrightarrow 2 \mathrm{Cu}_{(s)}+\mathrm{Ag}_2 \mathrm{SO}_{4(a q)}\)

Answer: 4. \(2 \mathrm{CuSO}_{4(a q)}+2 \mathrm{Ag} g_{(s)} \longrightarrow 2 \mathrm{Cu}_{(s)}+\mathrm{Ag}_2 \mathrm{SO}_{4(a q)}\)

The values of the standard reduction potential of Cu and Ag suggest that Cu would undergo oxidation (lower reduction potential) and Ag would undergo reduction (higher reduction potential).

Hence, the cell reaction will be \(\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

Question 19. Find the emf of the cell in which the following reaction takes place at \(298 K \mathrm{Ni}_{(s)}+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \longrightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}_{(s)}\)

Given that \(E_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 R T}{F}=0.059 \text { at } 298 \mathrm{~K}\)

1. 1.0385 V
2. 1.385 V
3. 0.9615 V
4. 1.05 V

Answer: 3. 0.9615 V

According to the Nernst equation,

E = \(E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)

⇒ \(E_{\text {cell }}^{\circ}=1.05(\text { Given) }\)

E = \(1.05-\frac{0.059}{2} \log \frac{(0.001)}{(0.001)^2}\)

= \(1.05-\frac{0.059}{2} \log 10^3=1.05-\frac{0.059 \times 3}{2}\)

= \(1.05-0.0885=0.9615 \mathrm{~V}\)

Question 20. For the cell reaction: \(2 \mathrm{Fe}_{(a q)}^{3+}+2 \mathrm{I}_{(a q)}^{-} \rightarrow 2 \mathrm{Fe}_{(a q)}^{2+}+\mathrm{I}_{2(a q)}\) E°_cell = 0.24 V at 298 K. The standard Gibbs’ energy (Δ_rG°) of the cell reaction is [Given that Faraday constant, F = 96500 C mol^-1]

1. 23.16 kJ mol^-1
2. -46.32 kJ mol^-1
3. -23.16 kJ mol^-1
4. 46.32 kJ mol^-1

Answer: 2. -46.32 kJ mol^-1

The standard Gibbs energy, \(\left(\Delta G^{\circ}\right)=-n F E_{\text {cell }}^{\circ}\) Value of n=2

⇒ \(\Delta G^{\circ}=-2 \times 96500 \times 0.24=-46320 \mathrm{~J}\)

= \(-46.32 \mathrm{~kJ} / \mathrm{mol}\)

Question 21. For a cell involving one electron, E°_cell= 0.59 V at 298 K, the equilibrium constant for the cell reaction is [Given that \(\frac{2.303 R T}{F}\)= 0.059 V at T = 298 K]

1. 1.0 x 10³⁰
2. 1.0 x 10²
3. 1.0 x 10⁵
4. 1.0 x 10¹⁰

Answer: 4. 1.0 x 10¹⁰

According to Nernst equation, \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log Q_c\)

At equilibrium \(E_{\text {cell }}=0\), \(Q_c=K_c\)

⇒ \(E_{\text {cell }}^{\mathrm{a}}=\frac{0.059}{n} \log K_c \Rightarrow 0.59=\frac{0.059}{1} \log K_c\)

⇒ \(K_c=\text { antilog } 10 \Rightarrow K_c=1 \times 10^{10}\)

Question 22. In the electrochemical cell: Zn|ZnSO₄(0.01 M)||CuSO₄(1.0 M)|Cu, the emf of this Daniell cell is E₁ When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ changed to 0.01 M, the emf changes to E₁ From the followings, which one is the relationship between E₁ and E₂? (Given, RT/F = 0.059)

1. E₁ < E₂
2. E₁ > E₂
3. E₂ = 0 ≠ E₁
4. E₁ = E₂

Answer: 2. E₁ > E₂

⇒ \(E_{\text {cell }}=E_{\text {cell }}^o-\frac{0.059}{n} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)

⇒ \(E_1=E^{\circ}-\frac{0.059}{2} \log \frac{0.01}{1}\)

⇒ \(E_1=E^{\circ}-\frac{0.059}{2}(-2)=E^{\circ}+0.059\)

⇒ \(E_2=E^{\circ}-\frac{0.059}{2} \log \frac{1}{0.01}=E^{\circ}-0.059\)

Hence, \(E_1>E_2\).

Question 23. If the E°_cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and K_eq?

1. \(\Delta G^{\circ}>0 ; K_{\text {eq }}<1\)
2. \(\Delta G^{\circ}>0 ; K_{\text {eq }}>1\)
3. \(\Delta G^0<0 ; K_{\text {eq }}>1\)
4. \(\Delta G^{\circ}<0 ; K_{\text {eq }}<1\)

Answer: 1. \(\Delta G^{\circ}>0 ; K_{\text {eq }}<1\)

⇒ \(\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}\)

If \(E_{\text {cell }}^{\circ}=- ve then \Delta G^{\circ}=+ ve\)

i.e; \(\Delta G^{\circ}>0\).

⇒ \(\Delta G^{\mathrm{a}}=-n R T \log K_{\mathrm{eq}}\)

For \(\Delta G^{\circ}=+ \text { ve, } K_{\text {eq }}=- \text { ve i.e., } K_{e q}<1 \text {. }\)

Question 24. The pressure of H₂ required to make the potential of H₂ electrode zero in pure water at 298 K is

1. 10^-10 atm
2. 10^-4 atm
3. 10^-14 atm
4. 10^-12 atm.

Answer: 3. 10^-14 atm

pH = 7 for water.

⇒ \(-\log \left[\mathrm{H}^{+}\right]=7 \Rightarrow\left[\mathrm{H}^{+}\right]=10^{-7}\)

⇒ \(2 \mathrm{H}^{+}{ }_{(a q)}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2(g)}\)

⇒ \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{P_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\)

0 = \(0-\frac{0.0591}{2} \log \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}\)

⇒ \(\log \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=0 \Rightarrow \frac{P_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=1\) (because log 1=0)

⇒ \(p_{\mathrm{H}_2}=10^{-14} \mathrm{~atm}\)

Question 25. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH =10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of the electrode would be

1. 0.118 V
2. 1.18 V
3. 0.059 V
4. 0.59 V

Answer: 4. 0.59 V

⇒ \(\underset{1 \mathrm{~atm}}{\mathrm{H}_2} \longrightarrow \underset{10^{-10}}{2 \mathrm{H}^{+}}+2 e^{-}\)

⇒ \(E_{\mathrm{H}_2 / \mathrm{H}^{+}}=0-\frac{0.059}{2} \log \frac{\left(10^{-10}\right)^2}{1}\)

⇒ \(E_{\mathrm{H}_2 / \mathrm{H}^{+}}=+0.59 \mathrm{~V}\)

Question 26. Consider the half-cell reduction reaction \(\mathrm{Mn}^{2+}+2 e^{-} \rightarrow \mathrm{Mn}, E^{\circ}=-1.18 \mathrm{~V}\); \(\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+}+e^{-}, E^{\circ}=-1.51 \mathrm{~V}\) The \(E^{\circ}\) for the reaction, \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^0+2 \mathrm{Mn}^{3+}\) and possibility of the forward reaction are respectively

1. -4.18 V and yes
2. + 0.33 V and yes
3. + 2.69 V and no
4. – 2.69 V and no.

Answer: 4. – 2.69 V and no

⇒ \(\mathrm{Mn}^{2+}+2 e^{-} \longrightarrow \mathrm{Mn} ; E^{\circ}=-1.18 \mathrm{~V}\)

⇒ \(2 \mathrm{Mn}^{2+} \longrightarrow 2 \mathrm{Mn}^{3+}+2 e^{-} ; E^{\circ}=-1.51 \mathrm{~V}\)

By adding equations (1) and (2), we get an equation for the cell, \(3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}; E^{\circ}=-2.69 \mathrm{~V}\)

Since the E value is negative, so the process is non-spontaneous as ΔG° is positive.

Question 27. The Gibbs’ energy for the decomposition of Al₂O₃ at 500 °C is as follows \(\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} G=+960 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The potential difference needed for the electrolytic reduction of aluminum oxide (Al₂O₃) at 500 °C is at least

1. 4.5 V
2. 3.0 V
3. 2.5 V
4. 5.0 V

Answer: 3. 2.5 V

⇒ \(\Delta G^{\circ}=-n F E^{\circ}\)

F = \(96500, \Delta G^{\circ}=+960 \times 10^3 \mathrm{~J} / \mathrm{mol}\)

⇒ \(\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2\)

Total number of \(\mathrm{Al}\) atoms in \(\mathrm{Al}_2 \mathrm{O}_3=\frac{2}{3} \times 2=\frac{4}{3}\)

⇒ \(\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}\)

As \(3 e^{-}\) change occur for each Al-atom

total n = \(\frac{4}{3} \times 3=4\)

⇒ \(E^{\circ}=-\frac{\Delta G^{\circ}}{n F}=-\frac{960 \times 1000}{4 \times 96500} \Rightarrow E^{\circ}=-2.48 \approx-2.5 \mathrm{~V}\)

Question 28. The electrode potentials for, \(\mathrm{Cu}^{2+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(a q)}^{+} \text {and } \mathrm{Cu}^{+}{ }_{(a q)}+e^{-} \rightarrow \mathrm{Cu}_{(s)}\) are + 0.15 V and + 0.50 V respectively. The value of \(E^{\circ}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}\) will be

1. 0.500 V
2. 0.325 V
3. 0.650 V
4. 0.150 V

Answer: 2. 0.325 V

⇒ \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+e^{–} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{+} ; E_1^0=0.15 \mathrm{~V}\)

⇒ \(\mathrm{Cu}^{+}{ }_{(\mathrm{aq})}+e^{-} \longrightarrow \mathrm{Cu}_{(s)} ; E_2{ }^{\circ}=0.50 \mathrm{~V}\)

⇒ \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ}=?\)

Now, \(\Delta G^a=\Delta G_1{ }^{\circ}+\Delta G_2{ }^{\circ}\)

or, \(-n F E^{\circ}=-n_1 F E_1{ }^{\circ}-n_2 F E_2{ }^{\circ}\)

or, \(E^{\circ}=\frac{n_1 E_1^{\circ}+n_2 E_2^{\circ}}{n}=\frac{1 \times 0.15+1 \times 0.50}{2}=0.325 \mathrm{~V}\)

Question 29. For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25 °C. The value of standard Gibbs energy, ΔG° will be (F = 96500 C mol^-1)

1. – 89.0 kJ
2. – 89.0 J
3. -44.5 kJ
4. – 98.0 kJ

Answer: 1. – 89.0 kJ

The cell reaction can be written as \(\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

We know, \(\Delta G^{\circ}=-n F E^{\circ}{ }_{\text {cell }}\)

= – 2 x 96500 x 0.46 = – 88780 J

= – 88.78 kJ ≈ -89 kJ

Question 30. Given:

1. \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}, E^{\circ}=0.337 \mathrm{~V}\)
2. \(\mathrm{Cu}^{2+}+e^{-} \rightarrow \mathrm{Cu}^{+}, E^{\circ}=0.153 \mathrm{~V}\)

Electrode potential, E° for the reaction, \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}\), will be

1. 0.90 V
2. 0.30 V
3. 0.38 V
4. 0.52 V

Answer: 4. 0.52 V

Given, \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E_1^0=0.337 \mathrm{~V}\); \(\mathrm{Cu}^{2+}+e^{-} \longrightarrow \mathrm{Cu}^{+} ; E_2^{\circ}=0.153 \mathrm{~V}\)

The required reaction is \(\mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; E_3^{\circ}=?\)

Applying, \(\Delta G^{\circ}=-n F E^{\circ}, \Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)

⇒ \(-\left(n_3 F E_3^{\circ}\right)=-\left(n_1 F E_1^{\circ}\right)-\left(-n_2 F E_2^{\circ}\right)\)

or \(E_3^{\circ}=2 \times E_1^{\circ}-E_2^{\circ}\)

or \(E_3^{\circ}=(2 \times 0.337)-0.153=0.52 \mathrm{~V}\)

Question 31. Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4, and -8.2 for H₂O_(l), CO_2(g), and pentane_(g), respectively. The value of £°cell for the pentane-oxygen fuel cell is

1. 1.0968 V
2. 0.0968 V
3. 1.968 V
4. 2.0968 V

Answer: 1. 1.0968 V

⇒ \(\mathrm{C}_5 \mathrm{H}_{12(g)}+8 \mathrm{O}_{2(g)} \longrightarrow 5 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ \(\Delta G^{\circ}=[(-394.4 \times 5)+(-237.2 \times 6)]-[(-8.2)+(8 \times 0)]\) =-3387 kJ

Note that the standard free energy change of elementary substances is taken as zero.

For the fuel cell, the complete cell reaction is: \(\mathrm{C}_5 \mathrm{H}_{12(\mathrm{~g})}+8 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 5 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(0)}\)

which is the combination of the following two half-reactions: \(\mathrm{C}_5 \mathrm{H}_{12(g)}+10 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \longrightarrow 5 \mathrm{CO}_{2(g)}+32 \mathrm{H}^{+}+32 e^{-}\)

and \(8 \mathrm{O}_{2(\mathrm{~g})}+32 \mathrm{H}^{+}+32 e^{-} \longrightarrow 16 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Therefore, the number of electrons exchanged is 32 here, i.e., n=32.

⇒ \(\Delta G^{\circ}=-n F E^{\circ}=-3387 \times 10^3 \mathrm{~J}\)

= \(-32 \times 96500 \mathrm{~J} / \text { Volt } \times E^0\)

Thus, \(E^{\circ}=1.0968 \mathrm{~V}\)

Question 32. The equilibrium constant of the reaction: \(\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Cu}^{2+}{ }_{(a q)}+2 \mathrm{Ag}_{(s)}\); \(E^{\circ}=0.46 \mathrm{~V}\) at \(298 \mathrm{~K}\) is

1. \(2.0 \times 10^{10}\)
2. \(4.0 \times 10^{10}\)
3. \(4.0 \times 10^{15}\)
4. \(2.4 \times 10^{10}\)

Answer: 3. \(4.0 \times 10^{15}\)

For a cell reaction in equilibrium at 298 K, \(E_{\text {cell }}^{\circ}=\frac{0.0591}{n} \log K_c\)

where, \(K_c=\) equilibrium constant, n= number of electrons involved in the electrochemical cell reaction.

Given, \(E_{\text {cell }}^0=0.46 \mathrm{~V}, n=2\)

0.46 = \(\frac{0.0591}{2} \times \log K_c or, \log K_c=\frac{2 \times 0.46}{0.0591}=15.57\)

or, \(K_c=3.7 \times 10^{15} \approx 4 \times 10^{15}\)

Question 33. The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be

1. 2.0 x 10¹¹
2. 4.0 x 10¹²
3. 1.0 x 10²
4. 1.0 x 10¹⁰

(Given F = 96500 C mol^-1, R = 8.314 J K^-1 mol^-1)

Answer: 4. 1.0 x 10¹⁰

E = \(E^{\circ}-\frac{0,0591}{n} \log _{10} \mathrm{Q}\) at \(25^{\circ} \mathrm{C}\)

At equilibrium, E=0, Q=K

0 = \(E^{\circ}-\frac{0.0591}{n} \log _{10} K\)

or, \(K=\mathrm{antilog}\left[\frac{n E^{\circ}}{0.0591}\right]\)

K = \(\mathrm{antilog}\left[\frac{2 \times 0.295}{0.0591}\right]=\mathrm{antilog}\left[\frac{0.590}{0.0591}\right]\)

= \(\mathrm{antilog} 10=1 \times 10^{10}\)

Question 34. On the basis of the information available from the reaction, \(4 / 3 \mathrm{Al}+\mathrm{O}_2 \rightarrow 2 / 3 \mathrm{Al}_2 \mathrm{O}_3, \Delta G=-827 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { of } \mathrm{O}_2\), the minimum e.m.f. required to carry out an electrolysis of Al₂O₃ is (F = 96500 C mol^-1)

1. 2.14 V
2. 4.28 V
3. 6.42 V
4. 8.56 V

Answer: 1. 2.14 V

⇒ \(\Delta G^{\circ}=-n F E^{\circ}\)

⇒ \(E^{\circ}=\frac{\Delta G^{\circ}}{-n F}=\frac{-827000}{-4 \times 96500}=2.14 \mathrm{~V}\)

⇒ \(\left(1 \mathrm{Al} \equiv 3 e^{-}, \frac{4}{3} \mathrm{Al}=\frac{4}{3} \times 3 e^{-}=4 e^{-}\right)\)

Question 35. For the disproportionation of copper \(2 \mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}, E^{\circ}\) is (Given : \(E^{\circ}\) for \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) is \(0.34 \mathrm{~V}\) and \(E^{\circ}\) for \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\) is \(0.15 \mathrm{~V}\)

1. 0.49 V
2. -0.19 V
3. 0.38 V
4. -0.38 V

Answer: 3. 0.38 V

For the reaction, \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}\) the cathode is \(\mathrm{Cu}^{+} / \mathrm{Cu}\) and anode is \(\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}\).

Given, \(\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}_2 E_1^{\circ}=0.34 \mathrm{~V}\)…..(1)

\(\mathrm{Cu}^{2+}+e^{-} \longrightarrow \mathrm{Cu}^{+} ; E_2^{\circ}=0.15 \mathrm{~V}\)….(2)

\(\mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; E_3^{\circ}=?\)….(3)

Now, \(\Delta G_1^{\circ}=-n F E_1^{\circ}=-2 \times 0.34 \times F=-0.68 F\)

⇒ \(\Delta G_2^{\circ}=-1 \times 0.15 \times F, \Delta G_3^{\circ}=-1 \times E_3^{\mathrm{a}} \times F\)

Again \(\Delta G_1^{\circ}=\Delta G_2^{\circ}+\Delta G_3^{\circ}\)

-0.68 F = \(-0.15 F-E_3^0 \times F\)

⇒ \(E_3^{\circ}=0.68-0.15=0.53 \mathrm{~V}\)

⇒ \(E_{\text {cell }}^{\circ}=E_{\text {cathode }\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)}^{\circ}-E_{\text {anode }\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)}^{\circ}\)

= \(0.53-0.15=0.38 \mathrm{~V}\)

Question 36. E° for the cell, \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}{ }_{(a q)}\right|\left|\mathrm{Cu}^{2+}{ }_{(a q)}\right| \mathrm{Cu}\) is \(1.10 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Zn}+\mathrm{Cu}_{(\text {aq) }}^{2+} \rightarrow \mathrm{Cu}+\mathrm{Zn}_{(\text {aq })}^{2+}\) is of the order

1. \(10^{+18}\)
2. \(10^{+17}\)
3. \(10^{-28}\)
4. \(10^{+37}\)

Answer: 4. \(10^{+37}\)

Nernst equation is E = \(E^{\circ}-\frac{0.059}{2} \log K\)

⇒ \(E^{\circ}=\frac{0.059}{2} \log K\) (E=0 at equilibrium condition)

⇒ 1.1 = \(\frac{0.059}{2} \log K \Rightarrow K=1.9 \times 10^{+37}\)

Question 37. The conductivity of a centriolar solution of KCI at 25°C is 0.0210 ohm^-1 cm^-1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of the cell constant is

1. 1.26 cm^-1
2. 3.34 cm^-1
3. 1.34 cm^-1
4. 3.28 cm^-1

Answer: 1. 1.26 cm^-1

Given, \(\mathrm{K}=0.0210 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}, R=60 \mathrm{ohm}\)

k = \(\frac{1}{R} \times \frac{l}{a}\) (where \(\frac{l}{a}=\) cell constant) or \(0.0210=\frac{1}{60} \times \frac{l}{a}\)

or \(\frac{l}{a}=60 \times 0.0210=1.26 \mathrm{~cm}^{-1}\)

Question 38. The molar conductance of NaCl, HCl, and CH₃COONa at infinite dilution are 126.45, 426.16, and 91.0 S cm² mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is.

Choose the right option for your answer.

1. 540.48 S cm² mol^-1
2. 201.28 S cm² mol^-1
3. 390.71 S cm² mol^-1
4. 698.28 S cm² mol^-1

Answer: 3. 390.71 S cm² mol^-1

⇒ \(\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}\)

= \(91.0+426.16-126.45\)

= \(390.71 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)

Question 39. The molar conductivity of 0.007 M acetic acid is 20 S cm² mol^-1. What is the dissociation constant of acetic acid? Choose the correct option.

⇒ \(\left[\begin{array}{c}
\Lambda_{\mathrm{H}^{+}}^{\circ}=350 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
\Lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}=50 \mathrm{Scm}^2 \mathrm{~mol}^{-1}
\end{array}\right]\)

1. \(2.50 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
2. \(1.75 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
3. \(2.50 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
4. \(1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

Answer: 4. \(1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

⇒ \(\Lambda_{m\left(\mathrm{CH}_3 \mathrm{COOH}\right)}^{\circ}=\lambda_{\mathrm{H}^{+}}^{\circ}+\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}\)

= \(350+50=400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Degree of dissociation, \(\alpha=\frac{\Lambda_m^c}{\Lambda_m^{\circ}}=\frac{20}{400}=0.05\)

So, dissociation constant, \(K_a=c \alpha^2\) (for weak electrolytes)

= \(0.007(0.05)^2=1.75 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

 

Redox Reactions

Question 1. Which of the following is a redox reaction?

1. Evaporation of H₂O
2. Both oxidation and reduction
3. H₂SO₄ with NaOH
4. In the atmosphere O₃ from O₂ by lightning

Answer: 2. Both oxidation and reduction

Redox reactions are those chemical reactions which involve both oxidation and reduction simultaneously.

Question 2. Without losing its concentration, ZnCl₂ solution cannot be kept in contact with

1. Pb
2. Al
3. Au
4. Ag

Answer: 2. Al

Only ‘AI’ lies above ‘Zn’ in the electrochemical series, which can displace Zn from the ZnCl₂ solution. Therefore, conc. of ZnCl₂ will decrease when kept in an ‘Al’ container.

⇒ \(2 \mathrm{Al}+3 \mathrm{ZnCl}_2 \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{Zn}\)

Question 3. On balancing the given redox reaction, \(a \mathrm{Cr}_2 \mathrm{O}_7^{2-}+b \mathrm{SO}_{3(\mathrm{mq})}^{2-}+c \mathrm{H}_{(\mathrm{mq})}^{+} \rightarrow 2 a \mathrm{Cr}_{[(a q)}^{3+}+b \mathrm{SO}_{4(\mathrm{aq})}^{2-}+\frac{c}{2} \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\) the coefficient a, b and c are found to be respectively.

1. 1,8,3
2. 8,1,3
3. 1,3,8
4. 3,8, 1

Answer: 3. 1,3,8

⇒ \(\mathrm{Cr}_2 \mathrm{O}_{7(a q)}^{2-}+3 \mathrm{SO}_{3(a q)}^{2-}+8 \mathrm{H}_{(a q)}^{+} \longrightarrow 2 \mathrm{Cr}_{(a q)}^{3+}+3 \mathrm{SO}_{4(a q)}^{2-}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\)

Question 4. Which of the following reactions is the metal displacement reaction? Choose the right option.

1. \(2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow 2 \mathrm{PbO}+4 \mathrm{NO}_2+\mathrm{O}_2 \uparrow\)
2. \(2 \mathrm{KClO}_3\)\(\longrightarrow{\Delta}\)\(2 \mathrm{KCl}+3 \mathrm{O}_2\)
3. \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)
4. \(\mathrm{Fe}+2 \mathrm{HCl} \longrightarrow \mathrm{FeCl}_2+\mathrm{H}_2 \uparrow\)

Answer: 3. \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)

When a metal from the electrochemical series is mixed with the ions of a metal lower down in the electrochemical series, then more active metal displaces the less active one, this is known as metal displacement.

Hence, the correct reaction is \(\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al}\)\(\longrightarrow{\Delta}\)\(\mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}\)

Read and Learn More NEET MCQs with Answers

Question 5. What is the change in the oxidation number of carbon in the following reaction? \(\mathrm{CH}_{4(\mathrm{~g})}+4 \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{CCl}_{4(l)}+4 \mathrm{HCl}_{(\mathrm{g})}\)

1. +4 to+4
2. 0 to+4
3. -4 to+4
4. 0 to-4

Answer: 3. -4 to+4

In CH₄ the oxidation number of carbon is -4 while in CCl₄, the oxidation number of carbon is +4. Thus, the change in oxidation number of carbon in the given reaction is from -4 to +4.

Question 6. The correct structure of tribromooctaoxide is Which of the following reactions are disproportionation reactions?

Answer: 2

Question 7. Which of the following reactions are disproportionation reactions?

1. \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}^0\)
2. \(3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}\)
3. \(2 \mathrm{KMnO}_4\) \(\longrightarrow{\Delta}\)\(\mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2\)
4. \(2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}\)

Select the correct option from the following.

1. (1) and (4) only
2. (1) and (2) only
3. (1), (2) and (3)
4. (1), (3) and (4)

Answer: 2. (1) and (2) only

Disproportionation reactions are those in which the same element/compound gets oxidised and reduced simultaneously.

⇒ \(2 \mathrm{Cu}^{+}\)\(\longrightarrow\)\(\mathrm{Cu}^{2+}+\mathrm{Cu}^0\)

⇒ \(3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \stackrel{+7}{\mathrm{MnO}_4^{-}}+\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{O}\)

Question 8. The oxidation state of Cr in CrO₅ is

1. -6
2. +12
3. +6
4. +4

Answer: 3. +6

CrO₅ has a butterfly structure having two peroxo bonds.

Peroxo oxygen has -1 oxidation state.

Let the oxidation state of Cr be ‘x’

CrO₅: x+4(-1)+1(-2) =0 + x=+6

Question 9. The correct order of N-compounds in its decreasing order of oxidation states is

1. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{N}_2, \mathrm{NH}_4 \mathrm{Cl}\)
2. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{NH}_4 \mathrm{Cl}, \mathrm{N}_2\)
3. \(\mathrm{HNO}_3, \mathrm{NH}_4 \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_2\)
4. \(\mathrm{NH}_4 \mathrm{Cl}, \mathrm{N}_2, \mathrm{NO}, \mathrm{HNO}_3\)

Answer: 1. \(\mathrm{HNO}_3, \mathrm{NO}, \mathrm{N}_2, \mathrm{NH}_4 \mathrm{Cl}\)

⇒ \(\stackrel{+5}{\mathrm{HNO}_3,} \stackrel{+2}{\mathrm{NO}}, \stackrel{0}{\mathrm{~N}}, \stackrel{-3}{\mathrm{NH}_4 \mathrm{Cl}}\)

Question 10. For the redox reaction, \(\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\). The correct coefficients of the reactants for the balanced equation are

Answer: 2

The correct balanced equation is \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Question 11. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?

1. \(\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{S}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 3 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{C}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CO}_2+2 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}(\mathrm{NE}\)

Answer: 4. \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}(\mathrm{NE}\)

⇒ \(\mathrm{CaF}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{HF}\)

Here, the oxidation state of every atom remains the same so, it is not a redox reaction.

Question 12.

1. \(\mathrm{H}_2 \mathrm{O}_2+\mathrm{O}_3 \longrightarrow \mathrm{H}_2 \mathrm{O}+2 \mathrm{O}_2\)
2. \(\mathrm{H}_2 \mathrm{O}_2+\mathrm{Ag}_2 \mathrm{O} \longrightarrow 2 \mathrm{Ag}+\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2\)

The role of hydrogen peroxide in the above reactions is respectively

1. Oxidizing in (1) and reducing in (2)
2. Reducing in (1) and oxidizing in (2)
3. Reducing in (1) and (2)
4. Oxidizing in (1) and (2)

Answer: 3. Reducing in (1) and (2)

H₂O₂ acts as a reducing agent in both the reactions in which O₂ is evolved

Question 13. The pair of compounds that can exist together is

1. \(\mathrm{FeCl}_3, \mathrm{SnCl}_2\)
2. \(\mathrm{HgCl}_2, \mathrm{SnCl}_2\)
3. \(\mathrm{FeCl}_2, \mathrm{SnCl}_2\)
4. \(\mathrm{FeCl}_3, \mathrm{KI}\)

Answer: 3. \(\mathrm{FeCl}_2, \mathrm{SnCl}_2\)

Both FeCI₂ and SnCl₂ are reducing agents with low oxidation numbers.

Question 14. A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?

1. S
2. H
3. Cl
4. C

Answer: 3. Cl

⇒ \(\stackrel{+1+5-2}{\mathrm{KClO}_3}+(\mathrm{COOH})_2+\mathrm{H}_2 \stackrel{+6}{\mathrm{~S}} \mathrm{O}_4 \longrightarrow\)\(\mathrm{K}_2^{+6} \mathrm{SO}_4+\mathrm{KCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 15. Oxidation numbers of P in PO₄^3-, of S in SO₄^2- and of Cr in Cr₂O₇ are respectively

1. +3, +6 and +5
2. +5, +3 and +6
3. -3, +6 and +6
4. +5, +6 and +6

Answer: 4. +5, +6 and +6

Let the oxidation number of P in \(\mathrm{PO}_4^{3-}\) be r.

∴ x+4(-2)=-3 = x=+5

Let the oxidation number of S in \(\mathrm{SO}_4^{2-}\) be y.

∴ y+4(-2)=-2 + y=+6

Let oxidation number of Cr in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) be z.

∴ 2z+7(-2)=-2 + z=*6

Question 16. The number of moles of MnO₄ required to oxidize one mole of ferrous oxalate completely in an acidic medium will be

1. 7.5 moles
2. 0.2 moles
3. 0.6 moles
4. 0.4 moles.

Answer: 4. 0.4 moles.

⇒ \(\left[5 e^{-}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^* \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \ldots\right.(1)] \times 2\)

⇒ \(\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 e^{-}+2 \mathrm{CO}_2 \ldots \text { (2) }\right] \times 5\)

Question 17. Which is the best description of the behaviour of bromine in the reaction given below? \(\mathrm{H}_2 \mathrm{O}+\mathrm{Br}_2 \rightarrow \mathrm{HOBr}+\mathrm{HBr}\)

1. Proton acceptor only
2. Both oxidised and reduced
3. Oxidised only
4. Reduced only

Answer: 2. Both oxidised and reduced

Question 18. The oxidation states of sulphur in the anions \(\mathrm{SO}_3^2-\), \(\mathrm{S}_2 \mathrm{O}_4^{2-}\) and \(\mathrm{S}_2 \mathrm{O}_6^{2-}\) follow the order

1. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)
2. \(\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)
3. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}<\mathrm{SO}_3^{2-}\)
4. \(\mathrm{S}_2 \mathrm{O}_6^{2-}<\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}\)

Answer: 1. \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)

⇒ \(\mathrm{SO}_3^{2-}: x+(-2) 3=-2\) or \(x-6=-2\) or \(x=+4\)

⇒ \(\mathrm{S}_2 \mathrm{O}_4^{2-}: 2 x+(-2) 4=-2\)

or \(2 x-8=-2\) or \(2 x=+6\)

∴ x = +3

⇒ \(\mathrm{S}_2 \mathrm{O}_6^{2-}: 2 x+(-2) 6=-2\)

or \(2 x-12=-2\) or \(2 x=+10\)

∴ x = +5

Oxidation states follow the order : \(\mathrm{S}_2 \mathrm{O}_4^{2-}<\mathrm{SO}_3^{2-}<\mathrm{S}_2 \mathrm{O}_6^{2-}\)

Question 19. The oxidation state of Fe in Fe₂O₃ is

1. 5/4
2. 4/5
3. 3/2
4. 8/3

Answer: 4. 8/3

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4: 3 x+4(-2)=0 \Rightarrow x=+\frac{8}{3}\)

Question 20. Reaction of sodium thiosulphate with iodine gives

1. Tetrathionate ion
2. Sulphide ion
3. Sulphate ion
4. Sulphite ion.

Answer: 1. Tetrathionate ion

⇒ \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \underset{\mathrm{Sodium tetrathionate}}{\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6}+2 \mathrm{NaI}\)

Question 21. The oxide, which cannot act as a reducing agent is

1. CO₂
2. ClO₂
3. NO₂
4. SO₂

Answer: 1. CO₂

Since carbon is in its maximum oxidation state of +4, therefore, carbon dioxide (CO₂) cannot act as a reducing agent.

Question 22. Which substance is serving as a reducing agent in the following reaction? \(14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{Ni} \rightarrow 7 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Cr}^{3+}+3 \mathrm{Ni}^{2+}\)

1. H⁺
2. Cr₂O^2-₇
3. H₂O
4. Ni

Answer: 4. Ni

Since the oxidation number of Ni increases from 0 to 2, therefore it acts as a reducing agent.

Question 23. The oxidation state of I in H₄IO₆^– is

1. +1
2. – 1
3. + 7
4. + 5

Answer: 3. + 7

Let r = Oxidation state of l. Since oxidation state of H = +1 and oxidation state of O = – 2, therefore for H₄IO₆^–, we get (4 x 1) + x + (6x – 2) = -1 or x = +7

Question 24. Consider the change in the oxidation state of bromine corresponding to different emf values as shown in the given diagram:

Then the species undergoing disproportionation is

1. BrO^–₃
2. BrO^–₄
3. Br₂
4. HBrO

Answer: 4. HBrO

For a reaction to be spontaneous, E°_cell, should be positive as ΔG° = -nFE°_cell

HBrO → Br₂; E° = 1.595 V, SRP (cathode)

HBrO → BrO^–₃; E° = -1.5 V, SOP (anode)

2HBrO → Br₂ + BrO^–₃

E°_cell = SRP (cathode) – SRP (anode)

= 1.595 – 1.5 = 0.095 V

E°_cell> 0 ⇒ ΔG° < 0 (spontaneous)

Chemical Bonding And Molecular Structure

Question 1. Amongst the following, the total number of species not having eight electrons around the central atom in its outermost shell, is NH₃, AlCl₃, BeCl₂, CCl₄, PCl₅

1. 4
2. 1
3. 3
4. 2

Answer: 3. 3

AICl₃, BeCl_2, and PCl₅ do not have eight electrons around the central atom in their outermost shells.

Question 2. In \(\mathrm{PO}_4^{3-}\) ion, the formal charge on each oxygen atom and P—O bond order respectively are

1. -0.75, 1.25
2. -0.75, 1.0
3. -0.75,0.6
4. -3, 1.25

Answer: 1. -0.75, 1.25

The total charge = -3

So, the average formal charge on each ‘O’ atom is -3/4= -0.75

⇒ Average P-O bond order = \(\frac{\text { Total no. of bonds }}{\text { Total no. of resonating structures }}=\frac{5}{4}=1.25\)

Question 3. Among LiCl, BeCl₂, BCl₃ and CCl₄, the covalent bond character follows the order

1. \(\mathrm{BeCl}_2>\mathrm{BCl}_3>\mathrm{CCl}_4<\mathrm{LiCl}\)
2. \(\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4<\mathrm{LiCl}\)
3. \(\mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4\)
4. \(\mathrm{LiCl}>\mathrm{BeCl}_2>\mathrm{BCl}_3>\mathrm{CCl}_4\)

Answer: 3. \(\mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4\)

Along the period, as we move from Li>Be>B>C, the electronegativity increases, and hence the EN difference between the element and Cl decreases and accordingly, the covalent character increases. Thus LiCI < BeCl₂ < BCl₃ < CCl₄ is the correct order of covalent bond character.

Read and Learn More NEET MCQs with Answers

Question 4. Which one of the following formulae does not correctly represent the bonding capacities of the two atoms involved?

Answer: 4

The asterisk (*) marked carbon has a valency of 5 and hence, this formula is not correct because carbon has a maximum valency of 4.

Question 5. Among the following, which compound will show the highest lattice energy?

1. KF
2. NaF
3. CsF
4. RbF

Answer: 2. NaF

For compounds containing ions of the same charge, lattice energy increases as the size of ions decreases. Thus, NaF has the highest lattice energy

Question 6. Which of the following molecules is non-polar in nature?

1. NO₂
2. POCl₃
3. CH₂O
4. SbCl₅

Answer: 4. SbCl₅

Among the given molecules, SbCl₅ is non-polar in nature

Question 7. Which of the following, set of molecules will have zero dipole moment?

1. Ammonia, beryllium difluoride, water, 1, 4-dichlorobenzene
2. Boron trifluoride, hydrogen fluoride, carbon dioxide, 1, 3-dichlorobenzene
3. Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
4. Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene

Answer: 4. Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene

Question 8. Which of the following is the correct order of dipole moment?

1. \(\mathrm{NH}_3<\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{BF}_3<\mathrm{NH}_3<\mathrm{NF}_3<\mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{H}_2 \mathrm{O}<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{BF}_3\)

Answer: 2. \(\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3<\mathrm{H}_2 \mathrm{O}\)

Question 9. The species, having bond angles of 120° is

1. \(\mathrm{ClF}_3\)
2. \(\mathrm{NCl}_3\)
3. \(\mathrm{BCl}_3\)
4. \(\mathrm{PH}_3\)

Answer: 3. \(\mathrm{BCl}_3\)

BCl₃-Trigonal planar sp²-hybridised, 120° angle.

Question 10. Consider the molecules CH₄, NH₃, and H₂O. Which of the given statements is false?

1. The H — O — H bond angle in H₂O is smaller than the H — N — H bond angle in NH₃.
2. The H — C — H bond angle in CH₄ is larger than the H — N — H bond angle in NH₃.
3. The H — C — H bond angle in CH₄, the H — N — H bond angle in NH3, and the H — O — H bond angle in H₂O are all greater than 90°.
4. The H — O — H bond angle in H₂O is larger than the H — C — H bond angle in CH₄.

Answer: 4. The H — O — H bond angle in H₂O is larger than the H — C — H bond angle in CH₄.

Question 11. Which of the following molecules has the maximum dipole moment?

1. CO₂
2. CH₄
3. NH₃
4. NF₃

Answer: 3. NH₃

In NH₃, H is less electronegative than N and hence dipole moment of each N-H bond is towards N and creates a high net dipole moment whereas in NF₃, F is more electronegative than N, the dipole moment of each N-F bond is opposite to that of lone pair, hence reducing the net dipole moment.

Question 12. The correct order of increasing bond length of C – H, C – O, C – C and C = C is

1. C-H<C=C<C-O<C-C
2. C-C<C=C<C-O<C-H
3. C-O<C-H<C-C<C = C
4. C-H<C-O<C-C<C = C

Answer: 1. C-H<C=C<C-O<C-C

Increasing order of bond length is C-H<C-C<C-O<C-C

Question 13. Which of the following structures is the most preferred and hence of the lowest energy for SO₃?

Answer: 4

It has a maximum number of covalent bonds involving pπ – dπ bonding also.

Question 14. The correct order of increasing bond angles in the following triatomic species is

1. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2<\mathrm{NO}_2^{-}\)
2. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}<\mathrm{NO}_2\)
3. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}<\mathrm{NO}_2\)
4. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Answer: 4. \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Structures of \(\mathrm{NO}_2^{-}, \mathrm{NO}_2\) and \(\mathrm{NO}_2^{+}\) is given as

The correct order of increasing bond angles in the following triatomic species is \(\mathrm{NO}_2^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}\)

Question 15. The correct order of C – O bond length among CO, CO^2-₃, CO₂ is

1. \(\mathrm{CO}<\mathrm{CO}_3^{2-}<\mathrm{CO}_2\)
2. \(\mathrm{CO}_3^{2-}<\mathrm{CO}_2<\mathrm{CO}\)
3. \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)
4. \(\mathrm{CO}_2<\mathrm{CO}_3^{2-}<\mathrm{CO}\)

Answer: 3. \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)

The more single bond character in resonance hybrid more is the bond length. Hence, the increasing bond length is \(\mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-}\)

Question 16. The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH₃ (1.5 D) is larger than that of NF₃ (0.2 D). This is because

1. In NH₃ the atomic dipole and bond dipole are in the opposite directions whereas in NF₃ these are in the same direction
2. In NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction
3. In NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions
4. In NH₃ as well as in NF₃ the atomic dipole and bond dipole are in opposite directions.

Answer: 3. In NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions

The dipole moment of NF₃ is 0.24 D, and of NH₃ is 1.48 D. The difference is due to the fact that the dipole moment due to N – F bonds in NF₃ are in opposite directions to the direction of the dipole moment of the lone pair on N atom which partly cancels out.

The dipole moment of N – H bonds in NH₃ are in the same direction as the dipole moment of the lone pair on the N atom which adds up as shown

Question 17. The correct order in which the O-O bond length increases in the following is

1. \(\mathrm{O}_2<\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_3\)
2. \(\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_2\)
3. \(\mathrm{H}_2 \mathrm{O}_2<\mathrm{O}_2<\mathrm{O}_3\)
4. \(\mathrm{O}_2<\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2\)

Answer: 4. \(\mathrm{O}_2<\mathrm{O}_3<\mathrm{H}_2 \mathrm{O}_2\)

Bond lengths of \(\mathrm{O}-\mathrm{O}\) in \(\mathrm{O}_2\) is 1.21Å, in \(\mathrm{H}_2 \mathrm{O}_2\) is 1.48 Å and in \(\mathrm{O}_3\) is 1.28 Å. Therefore, correct order of the \(\mathrm{O}-\mathrm{O}\) bond length is \(\mathrm{H}_2 \mathrm{O}_2>\mathrm{O}_3>\mathrm{O}_2\).

Question 18. The correct sequence of increasing covalent character is represented by

1. LiCl < NaCl < BeCl₂
2. BeCl₂ < LiCl < NaCl
3. NaCl < LiCl < BeCl₂
4. BeCl₂ < NaCl < LiCl

Answer: 3. NaCl < LiCl < BeCl₂

The covalent character in a compound is found by Fajan’s rule.

Fajan’s Rule: The smaller the size of the cation and the larger the size of the anion, the greater is the covalent character of an ionic bond. The greater the charge on the cation, the greater is the covalent character of the ionic bond.

Question 19. Which of the following would have a permanent dipole moment?

1. SiF₄
2. SF₄
3. XeF₄
4. BF₃

Answer: 2. SF₄

For dipole moment, we have to know the hybridization and shape.

Question 20. H₂O is dipolar, whereas BeF₂ is not. It is because

1. The electronegativity of F is greater than that of O
2. H₂O involves hydrogen bonding whereas BeF₂ is a discrete molecule
3. H₂O is linear and BeF₂ is angular
4. H₂O is angular and BeF₂ is linear.

Answer: 4. H₂O is angular and BeF₂ is linear.

The overall value of the dipole moment of a polar molecule depends on its geometry and shape, i.e., vectorial addition of the dipole moment of the constituent bonds. Water has an angular structure with a bond angle of 105°, it has a dipole moment. However, BeF₂ is a linear molecule thus, dipole moment summation of all the bonds present in the molecule cancels each other.

Question 21. Which of the following molecules does not possess a permanent dipole moment?

1. \(\mathrm{CS}_2\)
2. \(\mathrm{SO}_3\)
3. \(\mathrm{H}_2 \mathrm{~S}\)
4. \(\mathrm{SO}_2\)

Answer: 1. \(\mathrm{CS}_2\)

The structure of CS₂ is linear and therefore it does not have a permanent dipole moment. It is represented as S = C = S.

Question 22. The table shown below gives the bond dissociation energies (E_diss) for single covalent bonds of carbon (C) atoms with elements A, B, C, and D. Which element has the smallest atoms?

1. C
2. D
3. A
4. B

Answer: 2. D

The smaller the atom, the stronger the bond, and the greater the bond dissociation energy. Therefore, the bond C-D has the greatest energy or D has the smallest atoms.

Question 23. The strongest bond is in between

1. CsF
2. NaCl
3. Both (1) and (2)
4. None of the above.

Answer: 1. CsF

According to the Fajans rule, ionic character increases with an increase in the size of the cation, (Cs > Rb > K > Na) and with a decrease in the size of the anion (F > CI > Br > I). Thus, CsF has a higher ionic character than NaCl and hence, the bond in CsF is stronger than in NaCl.

Question 24. Which of the following bonds will be most polar?

1. N – Cl
2. O – F
3. N – F
4. N – N

Answer: 3. N – F

The polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. The greater the electronegativity difference, the more is the polarity of the bond.

⇒ \(\begin{array}{cccc}
\mathrm{N}-\mathrm{Cl} & \mathrm{O}-\mathrm{F} & \mathrm{N}-\mathrm{F} & \mathrm{N}-\mathrm{N} \\
3.04-3.16 & 3.5-4.0 & 3.04-4.0 & 3.04-3.04
\end{array}\)

Question 25. Amongst the following which one will have the maximum lone pair lone pair electron

1. \(\mathrm{ClF}_3\)
2. \(\mathrm{IF}_5\)
3. \(\mathrm{SF}_4\)
4. \(\mathrm{XeF}_2\)

Answer: 4. \(\mathrm{XeF}_2\)

ClF₃, IF₃, SF₄, and Xe₂, contain 2, 1, 1, and 3 lone pairs of electrons on the central atom respectively. Hence, XeF₂ has maximum lone pair – lone pair repulsions.

Question 26. Match List-A with List-B

Choose the correct answer from the options given below.

1. (1) – (D), (2) – (C), (3) – (B), (4) – (A)
2. (1) – (D), (2) – (C), (3) – (A), (4) – (B)
3. (1) – (B), (2) – (C), (3) – (D), (4) – (A)
4. (1) – (C), (2) – (A), (3) – (A), (4) – (B)

Answer: 2. (1) – (D), (2) – (C), (3) – (A), (4) – (B)

PCl₅: Trigonal bipyramidal

SF₆: Octahedral

BrF₅: Square pyramidal

BF₃: Trigonal planar

Question 27. In the structure of ClF₃, the number of lone pairs of electrons on the central atom ‘Cl’ is

1. One
2. Two
3. Four
4. Three

Answer: 2. Two

The structure of CIF₃ is

Hence, Cl has 2 lone pairs of electrons

Question 28. Predict the correct order among the following:

1. Bond pair – bond pair > lone pair-bond pair > lone pair – lone pair
2. Lone pair – bond pair > bond pair-bond pair > lone pair – lone pair
3. Lone pair – lone pair > lone pair-bond pair > bond pair-bond pair
4. Lone pair – lone pair > bond pair-bond pair > lone pair-bond pair

Answer: 3. Lone pair – lone pair > lone pair-bond pair > bond pair-bond pair

According to VSEPR theory, the repulsive forces between lone pair and lone pair are greater than between lone pair and bond pair which are further greater than bond pair and bond pair.

Question 29. Which of the following species contains three bond pairs and one lone pair around the central atom?

1. \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{BF}_3\)
3. \(\mathrm{NH}_2^{-}\)
4. \(\mathrm{PCl}_3\)

Answer: 4. \(\mathrm{PCl}_3\)

Question 30. Which of the following is not a correct statement?

1. Multiple bonds are always shorter than corresponding single bonds.
2. The electron-deficient molecules can act as Lewis acids.
3. The canonical structures have no real existence.
4. Every AB₅ molecule has have square pyramid structure.

Answer: 4. Every AB₅ molecule does in fact have a square pyramid structure

For AB₅ molecules, there are three possible geometries i.e. plantar pentagonal, square pyramidal, and trigonal bipyramidal.

Out of these three geometries, it is only a trigonal bipyramidal shape in which bond pair-bond pair repulsions are minimal and hence, this geometry is the most probable geometry of AB₅ molecule.

Question 31. Which of the following is not isostructural with SiCl₄?

1. \(\mathrm{NH}_4^{+}\)
2. \(\mathrm{SCl}_4\)
3. \(\mathrm{SO}_4^{2-}\)
4. \(\mathrm{PO}_4^{3-}\)

Answer: 2. \(\mathrm{SCl}_4\)

∴ \(\mathrm{SiCl}_4, \mathrm{NH}_4^{+}, \mathrm{SO}_4^{2-}\) and \(\mathrm{PO}_4^{3-}\) ions are the examples of molecules/ions which are of AB, type and have tetrahedral structures. SCl₄ is AB₄ (lone pair) type species.

Although the arrangement of five sp³d hybrid orbitals in space is trigonal bipyramidal, due to the presence of one lone pair of electrons in the basal hybrid orbital, the shape of AB₅ (lone pair) species gets distorted and trichomes distorted tetrahedral or see-saw.

Question 32. In which of the following molecules all the bonds are not equal?

1. \(\mathrm{NF}_3\)
2. \(\mathrm{ClF}_3\)
3. \(\mathrm{BF}_3\)
4. \(\mathrm{AlF}_3\)

Answer: 2. \(\mathrm{ClF}_3\)

The Cl – F (Cl – F_eq) bond length is equal to 1.60 Å while each of the two axial CI – F (Cl – F_a) bond lengths is equal to 1.70 Å.

Question 33. Which of the following molecules has trigonal planar geometry?

1. \(\mathrm{BF}_3\)
2. \(\mathrm{NH}_3\)
3. \(\mathrm{PCl}_3\)
4. \(\mathrm{IF}_3\)

Answer: 1

Question 34. In a regular octahedral molecule, MX₆ the number of X- M -A bonds at 180° is

1. Three
2. Two
3. Six
4. Four.

Answer: 1. Three

In an octahedral molecule, six hybrid orbitals direct towards the corners of a regular octahedron with a bond angle of 90°.

According to this geometry., the X number of X – M – X bonds at 180° must be three

Question 35. In BrF₃ molecule, the lone pairs occupy equatorial positions to minimize

1. Lone pair – bond pair repulsion only
2. Bond pair – bond pair repulsion only
3. Lone pair-lone pair repulsion and lone pair-bond pair repulsion
4. Lone pair – lone pair repulsion only

Answer: 4. Lone pair – lone pair repulsion only

Bent T-shaped geometry in which both lone pairs occupy the equatorial positions of the trigonal bipyramid. Here (lp – lp) repulsion = 0, (lp – bp) repulsion = 4, and (bp – bp) repulsion = 2.

Question 36. In \(\mathrm{NO}_3^{-}\) ion, the number of bond pair and lone pair of electrons on the nitrogen atom are

1. 2,2
2. 3, 1
3. 1,3
4. 4,0

Answer: 4. 4,0

In NO₃^– ion, nitrogen has 4 bond pairs of electrons and no lone pair of electrons.

Question 37. In which of the following bond angles is maximum?

1. \(\mathrm{NH}_3\)
2. \(\mathrm{NH}_4^{+}\)
3. \(\mathrm{PCl}_3\)
4. \(\mathrm{SCl}_2\)

Answer: 2. \(\mathrm{NH}_4^{+}\)

The bond angle is maximum in NH₄⁺ tetrahedral molecules with a bond angle of 109º.

Question 38. BCl₃ is a planar molecule whereas NCl₃ is pyramidal because

1. The nitrogen atom is smaller than the boron atom
2. BCl₃ has no lone pair but NCl₃ has a lone pair of electrons
3. B—Cl bond is more polar than N—Cl bond
4. The N-Cl bond is more covalent than the B—Cl bond.

Answer: 2. BCl₃ has no lone pair but NCl₃ has a lone pair of electrons

There is no lone pair on boron in BCl₃. hence, no repulsion takes place. There is a lone pair of nitrogen in NCl₃, hence, repulsion takes place. Therefore, BCl₃ is a planar molecule but NCl₃ is a pyramidal molecule.

Question 39. In compound X, all the bond angles are exactly 109°28′, and X is

1. Chloromethane
2. Carbon tetrachloride
3. Iodoform
4. Chloroform.

Answer: 2. Carbon tetrachloride

As C – Cl bonds are directed towards the corner of a regular tetrahedron

Question 40. Which of the following species contains an equal number of σ and π bonds?

1. \((\mathrm{CN})_2\)
2. \(\mathrm{CH}_2(\mathrm{CN})_2\)
3. \(\mathrm{HCO}_3^{-}\)
4. \(\mathrm{XeO}_4\)

Answer: 4. \(\mathrm{XeO}_4\)

Question 41. Which one of the following molecules contains no π bond?

1. \(\mathrm{SO}_2\)
2. \(\mathrm{NO}_2\)
3. \(\mathrm{CO}_2\)
4. \(\mathrm{H}_2 \mathrm{O}\)

Answer: 4. \(\mathrm{H}_2 \mathrm{O}\)

Question 42. Which one of the following statements is not correct for sigma- and pi- bonds formed between two carbon atoms?

1. A sigma-bond is stronger than a pi-bond.
2. Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively.
3. Free rotation of atoms about a sigma bond is allowed but not in the case of a pi bond.
4. Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard.

Answer: 2. Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively.

Sigma bond dissociation energy. = 347 kJ/mol

Pi-bond dissociation energy = 264kJ/mol

Question 43. The main axis of a diatomic molecule is z, molecular orbital p_x, and p_y overlap to form which of the following orbitals?

1. π molecular orbital
2. σ molecular orbital
3. δ molecular orbital
4. No bond will form

Answer: 1. π molecular orbital

For π overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

Hence, only sidewise overlapping takes place

Question 44. Which statement is not correct?

1. A sigma bond is weaker than a pi bond.
2. A sigma bond is stronger than a pi bond.
3. A double bond is stronger than a single bond.
4. A double bond is shorter than a single bond.

Answer: 1. A sigma bond is weaker than a π bond.

A σ bond is stronger than a t-bond

Question 45. Linear combination of two hybridized orbitals belonging to two atoms each having one electron leads to the formation of

1. Sigma bond
2. Double bond
3. Co-ordinate covalent bond
4. Pi bond.

Answer: 1. Sigma bond

Question 46. Which of the following does not apply to metallic bonds?

1. Overlapping valence orbitals
2. Mobile valence electrons
3. Delocalized electrons
4. Highly directed bonds

Answer: 4. Highly directed bonds

Metallic bonds have electrostatic attractions on all sides and hence, do not have directional characteristics.

Question 47. The angle between the overlapping of one s-orbital and one p-orbital is

1. 180°
2. 120°
3. 109°28′
4. 120°, 60°

Answer: 1. 180°

The type of overlapping between s- and p-orbitals occurs along the internuclear axis and hence, the angle is 180°.

Question 48. Which of the following pairs of compounds is isoelectronic and isostructural?

1. \(\mathrm{TeI}_2, \mathrm{XeF}_2\)
2. \(\mathrm{IBr}_2^{-}, \mathrm{XeF}_2\)
3. \(\mathrm{IF}_3, \mathrm{XeF}_2\)
4. \(\mathrm{BeCl}_2, \mathrm{XeF}_2\)

Answer: None

In this question, in place of isoelectronic, there should be the same number of valence electrons.

Question 49. The hybridizations of atomic orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-} \text {and } \mathrm{NH}_4^{+}\) respectively are

1. \(s p, s p^3\) and \(s p^2\)
2. \(s p^2, s p^3\) and \(s p\)
3. \(s p, s p^2\) and \(s p^3\)
4. \(s p^2, s p\) and \(s p^3\)

Answer: 3. \(s p, s p^2\) and \(s p^3\)

X = \(\frac{1}{2}(V E+M A-c+a)\)

For \(\mathrm{NO}_2^{+}, X=\frac{1}{2}(5+0-1)=2\) i.e., sp hybridisation

For \(\mathrm{NO}_3^{-}, X=\frac{1}{2}(5+0+1)=3\) i.e., \(s p^2\) hybridisation

For \(\mathrm{NH}_4^{+}, X=\frac{1}{2}(5+4-1)=4\) i.e, \(s p^3\) hybridisation

Question 50. Which of the following pairs of ions is isoelectronic and isostructural?

1. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
2. \(\mathrm{ClO}_3^{-}, \mathrm{CO}_3^{2-}\)
3. \(\mathrm{SO}_3^{2-}, \mathrm{NO}_3^{-}\)
4. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)

Answer: 1. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\) and 4. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)

1. \(\mathrm{CO}_3^{2-}: 6+24+2=32 ; s p^2\); trigonal planar \(\mathrm{NO}_3^{-}: 7+24+1=32 ; s p^2\); trigonal planar

Hence, these are isoelectronic as well as isostructural.

2. \(\mathrm{ClO}_3^{-}: 17+24+1=42 ; s p^3\), trigonal pyramidal \(\mathrm{CO}_3^{2-}: 6+24+2=32 ; s p^2\), trigonal planar

Hence, these are neither isoelectronic nor isostructural.

3. \(\mathrm{SO}_3^{2-}: 16+24+2=42 ; s p^3\), trigonal pyramidal \(\mathrm{NO}_3^{-}: 7+24+1=32 ; s p^2\), trigonal planar These are neither isoelectronic nor isostructural.

4. \(\mathrm{ClO}_3^{-}: 17+24+1=42; s p^3\), trigonal pyramidal \(\mathrm{SO}_3^{2-}: 16+24+2=42; s p^3\), trigonal pyramidal Hence, these are isoelectronic as well as isostructural.

Question 51. The correct geometry and hybridization for XeF₄ are

1. Octahedral, sp³d²
2. Trigonal bipyramidal, sp³d
3. Planar triangle, sp³d³
4. Square planar, sp³d2.

Answer: 1. Octahedral, sp³d²

Question 52. In which of the following pairs, both the species are not isostructural?

1. Diamond, Silicon carbide
2. \(\mathrm{NH}_3, \mathrm{PH}_3\)
3. \(\mathrm{XeF}_4, \mathrm{XeO}_4\)
4. \(\mathrm{SiCl}_4, \mathrm{PCl}_4^{+}\)

Answer: 3. \(\mathrm{XeF}_4, \mathrm{XeO}_4\)

In diamond and silicon carbide, the central atom is sp³ hybridized and hence, both are isostructural.

NH₃ and PH₃, both are pyramidal and the central atom in both cases is sp³ hybridized. SiCl₄ and PCl⁺₄, both are tetrahedral and the central atom in both cases is sp³ hybridized.

In XeF₄, Xe is sp³d² hybridized and the structure is square planar while in XeO₄, Xe is sp³ hybridized and the structure is tetrahedra.

 

Question 53. Maximum bond angle at nitrogen is present in which of the following?

1. \(\mathrm{NO}_2^{+}\)
2. \(\mathrm{NO}_3^{-}\)
3. \(\mathrm{NO}_2\)
4. \(\mathrm{NO}_2^{-}\)

Answer: 1. \(\mathrm{NO}_2^{+}\)

So, \(\mathrm{NO}_2^{+}\) has maximum bond angle.

Question 54. Which one of the following species has a planar triangular shape?

1. \(\mathrm{N}_3\)
2. \(\mathrm{NO}_3^{-}\)
3. \(\mathrm{NO}_2^{-}\)
4. \(\mathrm{CO}_2\)

Answer: 2. \(\mathrm{NO}_3^{-}\)

Question 55. XeF₂ is isostructural with

1. \(\mathrm{SbCl}_3\)
2. \(\mathrm{BaCl}_2\)
3. \(\mathrm{TeF}_2\)
4. \(\mathrm{ICl}_2^{-}\)

Answer: 4. \(\mathrm{ICl}_2^{-}\)

Question 56. Which of the following is a polar molecule?

1. \(\mathrm{SiF}_4\)
2. \(\mathrm{XeF}_4\)
3. \(\mathrm{BF}_3\)
4. \(\mathrm{SF}_4\)

Answer: 4. \(\mathrm{SF}_4\)

SF₄ has sp³d-hybridisation and see-saw shape with (4 bp + 1 lp)

 

Question 57. In which of the following pairs, the two species are isostructural?

1. \(\mathrm{SO}_3^{2-}\) and \(\mathrm{NO}_3^{-}\)
2. \(\mathrm{BF}_3\) and \(\mathrm{NF}_3\)
3. \(\mathrm{BrO}_3^{-}\) and \(\mathrm{XeO}_3\)
4. \(\mathrm{SF}_4\) and \(\mathrm{XeF}_4\)

Answer: 3. \(\mathrm{BrO}_3^{-}\) and \(\mathrm{XeO}_3\)

Hybridisation of Brin \(\mathrm{BrO}_3^{-}\): H = 1/2(7 + 0 – 0 + 1) = 4 i.e.sp³ hybridisation

Hybridisation of Xe in XeO₃ : H = 1/2(8 +0 -0+0)= 4 i.e. sp³ hybridisation

In both BrO^–₃ and XeO₃, the central atom is sp³ hybridized and contains one lone pair of electrons, hence in both cases, the structure is trigonal pyramidal.

 

 

Question 58. Which of the following species has a linear shape?

1. \(\mathrm{O}_3\)
2. \(\mathrm{NO}_2^{-}\)
3. \(\mathrm{SO}_2\)
4. \(\mathrm{NO}_2^{+}\)

Answer: 4. \(\mathrm{NO}_2^{+}\)

∴ \(\mathrm{NO}_2^{-}\): Due to sp² hybridization of N-atom and the presence of one lone pair on it, \(\mathrm{NO}_2^{-}\) has an angular shape.

SO₂: Due to the presence of one lone pair of electrons in one of the three sp²-hybrid orbitals, the SO₂ molecule has an angular (V-shaped) structure.

∴ \(\mathrm{NO}_2^{+}\): Due to sp hybridisation of \(\mathrm{N}^{+}, \mathrm{NO}_2^{+}\) ion has linear shape.

Question 59. The correct order regarding the electronegativity of hybrid orbitals of carbon is

1. \(s p<s p^2<s p^3\)
2. \(s p>s p^2<s p^3\)
3. \(s p>s p^2>s p^3\)
4. \(s p<s p^2>s p^3\)

Answer: 3. \(s p>s p^2>s p^3\)

The electronegativity of carbon atom is not fixed. It varies with the state of hybridization. The electronegativity of carbon increases as the s-character of the hybrid orbital increases.

C(sp)>C(sp²)>c(sp³)

Question 60. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

1. \(\mathrm{NO}<\mathrm{O}_2^{-}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
2. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
3. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}\)
4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Answer: 4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Thus, bond order increases as: \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Question 61. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH₃OH to a gas?

1. Dipole-dipole interaction
2. Covalent bonds
3. London dispersion force
4. Hydrogen bonding

Answer: 4. Hydrogen bonding

Methanol can undergo intermolecular association through H-bonding as the -OH group in alcohols is highly polarised.

As a result, in order to convert Iiquid CH₃OH to a gaseous state, the strong hydrogen bonds must be broken.

Question 62. Which of the following has pπ – dπ bonding?

1. \(\mathrm{NO}_3^{-}\)
2. \(\mathrm{SO}_3^{2-}\)
3. \(\mathrm{BO}_3^{3-}\)
4. \(\mathrm{CO}_3^{2-}\)

Answer: 2. \(\mathrm{SO}_3^{2-}\)

In sulfite ion, the central atom sulfur is sp³ hybridisedElectronic structure of S atom in an excited state

The three p electrons form o bonds with three oxygen atoms with one position (of the tetrahedron) being occupied by a lone pair. The d electron (excluded from hybridization) forms π bond with one oxygen atom. i.e. pπ – dπ bonding occurs.

Thermodynamics

Question 1. Which of the following are not state functions?

1. q+w
2. q
3. w
4. H-TS

1. (1), (2) and (3)
2. (2) and (3)
3. (1) and (4)
4. (2),(3) and (4)

Answer: 2. (2) and (3)

State functions or state variables are those which depend only on the state of the system and not on how the state was reached.

q + w = ΔE (internal energy)

H – Ts = G (free energy)

Path function depends on the path followed during a process. Work and heat are the path functions.

Question 2. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true?

1. ΔE= W≠0, q=0
2. ΔE= W=q≠0
3. ΔE =0, W = q≠0
4. W=0, ΔE=q≠0

Answer: 1. ΔE= W≠0, q=0

The mathematical form of the first law of thermodynamics q = ΔE + w

Since the system is closed and insulated, q = 0

Paddlework is done on the system.

∴ w ≠ 0.

Temperature and hence internal energy of the system increases.

∴ ΔE ≠ 0.

Read and Learn More NEET MCQs with Answers

Question 3. Which of the following is the correct equation?

1. ΔU= ΔW+ ΔQ
2. ΔU= ΔQ – W
3. ΔW = ΔU+ ΔQ
4. None of these

Answer: 2. ΔU= ΔQ – W

This is the mathematical relation of the first law of thermodynamics. Here ΔU = change in internal energy; ΔQ = heat absorbed by the system and W = work done by the system.

Question 4. Which of the following options is the correct relation between change in enthalpy and change in internal energy?

1. ΔH – ΔU= -ΔnRT
2. ΔH + ΔU = ΔnR
3. ΔH = ΔU – Δn_gRT
4. ΔH= ΔU+ Δn_gRT

Answer: 4. ΔH= ΔU+ Δn_gRT

Question 5. Which of the following P-V curve represents maximum work done?

Answer: 2

Question 6. Which one among the following is the correct option for the right relationship between C_P and C_V for one mole of an ideal gas?

1. C_V=RC_P
2. C_P+C_V=R
3. C_P– C_V=R
4. C_P = RC_V

Answer: 3. C_P – C_v=R

C_P = C_V+ nR

For one mole of an ideal gas, C_P= C_V + R or C_P – C_V = R

Question 7. The correct option for free expansion of ideal gas under adiabatic conditions is

1. q = 0, ΔT = 0 and w = 0
2. q = 0, ΔT < 0 and w > 0
3. q < 0, ΔT = 0 and w = 0
4. q > 0, ΔT > 0 and w > 0

Answer: 1. q = 0, ΔT = 0 and w = 0

For the free expansion of an ideal gas, P_ex = 0,

w=-P_ex ΔV=0

For adiabatic process, q = 0

According to the first law of thermodynamics, ΔU=q+w=0

As the internal energy of an ideal gas is a function of temperature, ΔU=0,

∴ T =0

Question 8. Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1 L bar = 100 J]

1. 30 J
2. -30 J
3. 5 kJ
4. 25J

Answer: 2. -30 J

Expansion of a gas against a constant external pressure is an irreversible process. The work done in an irreversible process

= \(-P_{\text {ext }} \Delta \mathrm{V}=-P_{\text {ext }}\left(V_2-V_1\right)=-2(0.25-0.1)\)

= \(-2 \times 0.15 \mathrm{~L} \text { bar }=-0.30 \times 100 \mathrm{~J}=-30 \mathrm{~J}\)

Question 9. Reversible expansion of an ideal gas under isothermal and adiabatic conditions as shown in the

AB → Isothermal expansion

AC → Adiabatic expansion

Which of the following options is not correct?

1. \(\Delta S_{\text {isothermal }}>\Delta S_{\text {adiabatic }}\)
2. \(T_A=T_B\)
3. \(W_{\text {isothermal }}>W_{\text {adiabatic }}\)
4. \(T_C>T_A\)

Answer: 4. \(T_C>T_A\)

For an ideal gas, internal energy is a function of temperature. The final temperature i.e., T_C for adiabatic process is less than its initial temperature i.e., T_A

∴ T_C < T_A

Question 10. An ideal gas expands isothermally from 10^-3 m³ to 10^-2 m³ at 300 K against a constant pressure of 10⁵ N m^-2. The work done on the gas is

1. +270 kJ
2. -900 J
3. +900 kJ
4. -900 kJ

Answer: 2. -900 J

w = \(-P d V=-P\left(V_2-V_1\right)\)

= \(-10^5 \mathrm{~N} \mathrm{~m}^{-2}\left(10^{-2}-10^{-3}\right) \mathrm{m}^3=-10^5 \mathrm{~N} \mathrm{~m}^{-2}\left(9 \times 10^{-3}\right) \mathrm{m}^3\)

= \(-9 \times 10^2 \mathrm{~N} \mathrm{~m}=-900 \mathrm{~J}\)

Question 11. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

1. -500 J
2. -505 J
3. +505 J
4. 1136.25 J

Answer: 2. -505 J

w = \(-P_{{ext}} \Delta V=-2.5(4.50-2.50)\)

= \(-5 \mathrm{~L} \text { atm }=-5 \times 101.325 \mathrm{~J}=-506.625 \mathrm{~J}\)

ΔU=q+w

As, the container is insulated, thus q=0

Hence, \(\Delta U=w=-506.625 \mathrm{~J}\)

Question 12. Equal volumes of two monatomic gases, A and B at the same temperature and pressure are mixed. The ratio of specific heat (C_P/C_V) of the mixture will be

1. 0.83
2. 1.50
3. 3.3
4. 1.67

Answer: 4. 1.67

C_P for a monoatomic gas mixture of the same volume

= \(\frac{5}{2} R, C_V=\frac{3}{2} R\)

∴ \(\frac{C_P}{C_V}=\frac{\frac{5}{2} R}{\frac{3}{2} R}=\frac{5}{3}=1.67\)

Question 13. Which of the following is the correct option for free expansion of an ideal gas under adiabatic conditions?

1. q=0, ΔT≠0, w = 0
2. q≠0, ΔT= 0, w= 0
3. q=0, ΔT=0, w=0
4. q=0, ΔT<0, w≠0

Answer: 3. q=0, ΔT=0, w=0

For free expansion of an ideal gas under adiabatic condition q = 0, ΔT = 0, w = 0.

For free expansion, w = 0, adiabatic process, q = 0

ΔU = q + w = 0

Internal energy remains constant means ΔT = 0.

Question 14. Three moles of an ideal gas expanded spontaneously into the vacuum. The work done will be

1. Infinite
2. 3 Joules
3. 9 Joules
4. Zero.

Answer: 4. Zero.

Since the ideal gas expands spontaneously into the vacuum, P_ext = 0, hence work done is also zero.

Question 15. Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE?

1. \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)
2. \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)
3. \(\mathrm{C}_{(s)}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightarrow 2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}\)
4. \(\mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(\mathrm{~g})}\)

Answer: 2. \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)

∴ \(\Delta H=\Delta E+\Delta n_g R T\)

For \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}\)

∴ \(\Delta n_g=2-(1+1)=0 \text {. i.e. } \Delta H=\Delta E\)

Question 16. The work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm is (1 L atm = 101.32 J)

1. -6J
2. -608J
3. +304J
4. -304J

Answer: 2. -608J

Work = -P_ext x volume change = -3 x (6 – 4) x 101.32 = 6 x 101.32 = – 607.92 J = – 608 J

Question 17. For the reaction, \(\mathrm{C}_3 \mathrm{H}_{8(\mathrm{~g})}+5 \mathrm{O}_{2(g)} \rightarrow 3 \mathrm{CO}_{2(\mathrm{~g})}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\) at constant temperature, ΔH – ΔE is

1. +RT
2. -3RT
3. +3RT
4. -RT

Answer: 2. -3RT

⇒ \(\mathrm{C}_3 \mathrm{H}_{g(g)}+5 \mathrm{O}_{2(g)} \rightarrow 3 \mathrm{CO}_{2(g)}+4 \mathrm{H}_2 \mathrm{O}_{(l)}\)

⇒ \(\Delta n_g=3-6=-3\)

∴ \(\Delta H=\Delta E+P \Delta V\) or \(\Delta H-\Delta E=P \Delta V\)

∴ \(\Delta H-\Delta E=\Delta n_g R T=-3 R T\)

Question 18. The molar heat capacity of water at constant pressure, C, is 75 J K^-1 mol^-1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

1. 1.2 K
2. 2.4 K
3. 4.8 K
4. 6.6 K

Answer: 2. 2.4 K

Molar heat capacity = 75J K^-1 mol^-1

18 g of water = 1 mole = 75J K^-1 mol^-1

1 g of water = \(\frac{75}{18}\) J K^-1

100 g ofwater= \(\frac{75}{18}\) x 100 J K^-1

Q = \(m \cdot C \cdot \Delta T \text { or } 1000=100 \times \frac{75}{18} \times \Delta T\)

⇒ \(\Delta T=\frac{10 \times 18}{75}=2.4 \mathrm{~K}\)

Question 19. When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?

1. q = w = 500 J, ΔE = 0
2. q = ΔE = 500 J, w = 0
3. q = w = 500 J, ΔE = 0
4. ΔE = 0, q = w = -500 J

Answer: 2. q = ΔE = 500 J, w = 0

ΔH = ΔE + PΔV

When ΔV = 0; w = 0.

ΔH = ΔE + 0 or ΔH = ΔE

As ΔE = q + w, ΔE = q

In the present problem, ΔH = 500 J

ΔH= ΔE = 500 J, q = 500 J, w = 0

Question 20. For the reaction, \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)}+3 \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}+3 \mathrm{H}_2 \mathrm{O}_{(l)}\) which one is true?

1. ΔH = ΔE – RT
2. ΔH = ΔE + RT
3. ΔH = ΔE + 2RT
4. ΔH = ΔE – 2RT

Answer: 1. ΔH = ΔE – RT

ΔH = ΔE + PΔV also PV = nRT (ideal gas equation)

or PΔV = Δn_gRT

Δn_g = Change in the number of gaseous moles

∴ ΔH= ΔE+ Δn_gRT

⇒ Δn_g = 2 – 3 = -1

⇒ ΔH = ΔE – RT

Question 21. In an endothermic reaction, the value of ΔH is

1. Negative
2. Positive
3. Zero
4. Constant.

Answer: 2. Positive

In endothermic reactions, the energy of reactants is less than the energy of products. Thus, E_R < E_P.

ΔH = E_P – E_R = +ve

Question 22. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The ΔE for this process is (R = 2cal mol^-1 K^-1)

1. 1381.1 cal
2. Zero
3. 163.7 cal
4. 9 L atm

Answer: 2. Zero

Change in internal energy depends upon temperature. At constant temperature, the internal energy of the gas remains constant, so ΔE = 0.

Question 23. During isothermal expansion of an ideal gas, its

1. Internal energy increases
2. Enthalpy decreases
3. Enthalpy remains unaffected
4. Enthalpy reduces to zero.

Answer: 3. Enthalpy remains unaffected

During isothermal expansion of an ideal gas, ΔT = 0, ΔE = 0

∴ ΔH = 0

H = E + PV

ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT)

∴ ΔH= ΔE + nRΔT = 0 + 0 = 0

Change in enthalpy is zero, which means its enthalpy remains the same or unaffected.

Question 24. For the reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3, \Delta H\)=?

1. ΔE + 2RT
2. ΔE – 2RT
3. ΔH = RT
4. ΔE – RT

Answer: 2. ΔE – 2RT

Δn_g =2 – 4= -2, ΔH = ΔE -2RT

Question 25. If ΔH is the change in enthalpy and ΔT, the change in internal energy accompanying a gaseous reaction, then

1. ΔH is always greater than ΔT
2. ΔH < ΔE only if the number of moles of the products is greater than the number of moles of the reactants
3. ΔH is always less than ΔE
4. ΔH < ΔE only if the number of moles of products is less than the number of moles of the reactant

Answer: 4. ΔH < ΔE only if the number of moles of products is less than the number of moles of the reactant

lf n_p < n_r; Δn_g= n_p – n_r= -ve.

Hence, ΔH < ΔE.

Question 26. Three thermochemical equations are given below:

1. \(\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(g)} ; \Delta_{,} H^{\circ}=x \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)} ; \Delta_r H^{\circ}=y \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta_{\mathrm{r}} H^{\circ}=z \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Based on the above equations, find out which of the relationships given below is correct.

1. z = x + y
2. x = y + z
3. y = 2z – x
4. x = y – z

Answer: 2. x = y + z

According to Hesst’s law, equation (1) is equal to equations (2) + (3) i.e., x = y + z

Question 27. The standard enthalpy of vaporisation Δ_vapH° for water at 100°C is 40.66 kJ mol^-1. The internal energy of the vaporisation of water at 100°C (in kJ mol^-1) is

1. +37.56
2. -43.76
3. +43.76
4. +40.66

(Assume water vapour to behave like an ideal gas)

Answer: 1. +37.56

∴ \(\Delta_{\text {vap }} H^{\circ}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

T = \(100+273=373 \mathrm{~K}, \Delta E=?\)

∴ \(\Delta H=\Delta E+\Delta n_g R T \Rightarrow \Delta E=\Delta H-\Delta n_g R T\)

⇒ \(\Delta n_g\) = number of gaseous moles of products – number of gaseous moles of reactants

⇒ \(\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(g)}\)

⇒ \(\Delta n_g=1-0=1\)

∴ \(\Delta E=\Delta H-R T\)

ΔE = (40.66 x 10³) – (8314 x 373)

= 37559 J/mol or 37.56 kJ/mol

Question 28. Consider the following processes:

⇒ \(\begin{array}{lc}
& \Delta \boldsymbol{H}(\mathbf{k J} / \mathbf{m o l}) \\
1 / 2 A \rightarrow B & +150 \\
3 B \rightarrow 2 C+D & -125 \\
E+A \rightarrow 2 D & +350
\end{array}\)

For \(B+D \rightarrow E+2 C, \Delta H\) will be

1. \(525 \mathrm{~kJ} / \mathrm{mol}\)
2. \(-175 \mathrm{~kJ} / \mathrm{mol}\)
3. \(-325 \mathrm{~kJ} / \mathrm{mol}\)
4. \(325 \mathrm{~kJ} / \mathrm{mol}\)

Answer: 2. \(-175 \mathrm{~kJ} / \mathrm{mol}\)

Adding all the equations, we get

⇒ \(\begin{array}{lc}
& \Delta H \\
A \rightarrow 2 B & 300 \mathrm{~kJ} / \mathrm{mol} \\
3 B \rightarrow 2 C+D & -125 \mathrm{~kJ} / \mathrm{mol} \\
2 D \rightarrow A+E & -350 \mathrm{~kJ} / \mathrm{mol} \\
\hline B+D \rightarrow E+2 C ; \Delta H=(300-125-350)=-175 \mathrm{~kJ} / \mathrm{mol}
\end{array}\)

Question 29. The following two reactions are known

• \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)} ; \Delta H=-26.8 \mathrm{~kJ}\)
• \(\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=-16.5 \mathrm{~kJ}\)

The value of \(\Delta H\) for the following reaction \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+\mathrm{CO}_{(g)} \rightarrow 2 \mathrm{FeO}_{(s)}+\mathrm{CO}_{2(g)}\) is

1. +10.3 kJ
2. -43.3 kJ
3. -10.3 kJ
4. +6.2kJ

Answer: 4. +6.2kJ

⇒ \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)}\); \(\Delta H=-26.8 \mathrm{~kJ}\)…..(1)

⇒ \(\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=-16.5 \mathrm{~kJ}\)….(2)

⇒ \(\mathrm{Fe}_2 \mathrm{O}_{3(s)}+\mathrm{CO}_{(g)} \rightarrow 2 \mathrm{FeO}_{(s)}+\mathrm{CO}_{2(g)} ; \Delta H=?
\)…(3)

Equation (3) can be obtained as: (1) -2(2) = \(-26.8-2(-16.5)=-26.8+33.0=+6.2 \mathrm{~kJ}\)

Question 30. For which one of the following equations is \(\Delta H^{\circ} \text { reaction }\) equal to \(\Delta H^{\circ}{ }_f\) Affreaction equal to ïH°_f, for the product?

1. \(\mathrm{N}_{2(g)}+\mathrm{O}_{3(g)} \rightarrow \mathrm{N}_2 \mathrm{O}_{3(g)}\)
2. \(\mathrm{CH}_{4(g)}+2 \mathrm{Cl}_{2(g)} \rightarrow \mathrm{CH}_2 \mathrm{Cl}_{2(b)}+2 \mathrm{HCl}_{(g)}\)
3. \(\mathrm{Xe}_{(g)}+2 \mathrm{~F}_{2(g)} \rightarrow \mathrm{XeF}_{4(g)}\)
4. \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}\)

Answer: 3. \(\mathrm{Xe}_{(g)}+2 \mathrm{~F}_{2(g)} \rightarrow \mathrm{XeF}_{4(g)}\)

For (3), \(\Delta H_{\text {reaction }}^{\circ}\)

= \(\Delta H^{\circ}{ }_f\left(\mathrm{XeF}_4\right)-\left[\Delta H_f^{\circ}(\mathrm{Xe})+2 \Delta H^{\circ}{ }_f\left(\mathrm{~F}_2\right)\right]\)

Enthalpies of formation of elementary substances \(\mathrm{Xe}\) and \(\mathrm{F}_2\) are taken as zero.

Thus, \(\Delta H_{\text {reaction }}^{\circ}=\Delta H_f^{\circ}\left(\mathrm{XeF}_4\right)\)

Question 31. Heat of combustion \(\Delta H\) for \(\mathrm{C}_{(s)}, \mathrm{H}_{2(g)}\) and \(\mathrm{CH}_{4(\mathrm{~g})}\) are -94,-68 and -213 \(\mathrm{kcal} / \mathrm{mol}\), then \(\Delta H\) for \(\mathrm{C}_{(s)}+2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)}\) is

1. -17 k cal
2. -111 k cal
3. -170 k cal
4. -85 k cal

Answer: 1. -17 k cal

1. \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H_{\mathrm{i}}=-94 \mathrm{kcal} / \mathrm{mole}\)
2. \(2 \mathrm{H}_{2(\mathrm{gg})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{f})} ; \Delta H_{\mathrm{ii}}=-68 \times 2 \mathrm{kcal} / \mathrm{mole}\)
3. \(\mathrm{CH}_{4(g)}+2 \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(l)} ; \Delta H_{\mathrm{iii}}=-213 \mathrm{kcal} / \mathrm{mole}\)
4. \(\mathrm{C}_{(s)}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{4(\mathrm{~g}} ; \Delta H_{\mathrm{iv}}=\)?

By applying Hess’s law, we can compute \(\Delta H_{\mathrm{iv}}\).

∴ \(\Delta H_{\mathrm{iv}}=\Delta H_{\mathrm{1}}+\Delta H_{\mathrm{1}}-\Delta H_{\mathrm{3}}\)

= \((-94-68 \times 2+213) \mathrm{kcal}=-17 \mathrm{kcal}\)

Question 32. Change in enthalpy for reaction, \(2 \mathrm{H}_2 \mathrm{O}_{2(l)} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{O}_{2(\mathrm{~g})}\) if the heat of formation of H₂O_2(l) and H₂O_(l) are -188 and -286 kJ/mol respectively, is

1. -196 kJ/mol
2. +196 kJ/mol
3. +948 kJ/mol
4. -948 kJ/mol

Answer: 1. -196 kJ/mol

∴ \(\Delta H_f^{\circ}=\Sigma H_{f \text { (products) }}^{\circ}-\Sigma H_{f \text { (reactants) }}^{\circ}\)

For the given reaction, \(2 \mathrm{H}_2 \mathrm{O}_{2(\mathrm{~J})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{J})}+\mathrm{O}_{2(\mathrm{~g})}\)

∴ \(\Delta H_f^{\circ}=2 \times \Delta H_{f\left(\mathrm{H}_2 \mathrm{O}\right)}^{\mathrm{o})}-2 \times \Delta H_{f\left(\mathrm{H}_2 \mathrm{O}_2\right)}^{\mathrm{o}}\)

= \(2 \times-286 \mathrm{~kJ} \mathrm{~mol}^{-1}-2 \times(-188) \mathrm{kJ} \mathrm{mol}^{-1}\)

= \(-196 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 33. Enthalpy of \(\mathrm{CH}_4+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}\) is negative. If the enthalpy of combustion of CH₄ and CH₃OH are x and y respectively, then which relation is correct?

1. x>y
2. x<y
3. x=y
4. x≥y

Answer: 1. x>y

⇒ \(\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}, \Delta H_1=-x\)…..(1)

⇒ \(\mathrm{CH}_3 \mathrm{OH}+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}, \Delta H_2=-y\)….(2)

Subtracting (2) from (1), we get \(\mathrm{CH}_4+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}, \Delta H_3=- \text { ve }\)

i.e., \(-x-(-y)=-\mathrm{ve}\)

y – x = -ve. Hence, x>y.

Question 34. In the reaction \(\mathrm{S}+3 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3+2 x\) and \(\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3+y\) kcal the heat of formation of SO₂ is

1. (2x + y)
2. (x – y)
3. (x + y)
4. (2x – y)

Answer: 4. (2x – y)

⇒ \(\mathrm{S}+\frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3+2 x \) kcal….(1)

⇒ \(\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3+y \) kcal….(2)

By subtracting equation (2) from (1) we get, \(\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2+(2 x-y)\) kcal

The heat of formation of \(\mathrm{SO}_2\) is \((2 x-y)\) kcal/mole.

Question 35. Given that \(\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2, \Delta H^{\circ}=-x \mathrm{~kJ}\); \(2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2, \Delta H^{\circ}=-y \mathrm{~kJ}\) The enthalpy of formation of carbon monoxide will be

1. \(\frac{y-2 x}{2}\)
2. \(2 x-y\)
3. \(y-2 x\)
4. \(\frac{2 x-y}{2}\)

Answer: 1. \(\frac{y-2 x}{2}\)

⇒ \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H^{\circ}=-x \mathrm{~kJ}\)….(1)

⇒ \(\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} ; \Delta H^{\circ}=\frac{-y}{2} \mathrm{~kJ}\)….(2)

By subtracting equation (2) from (1) we get, \(\mathrm{C}_{(s)}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(g)} ;\)

∴ \(\Delta H^{\circ}=-x-\left(-\frac{y}{2}\right)=\frac{y-2 x}{2} \mathrm{~kJ}\)

Question 36. If enthalpies of formation for C₂H_4(g), CO_2(g) and H₂O_(l) at 25°C and 1 atm pressure are 52, -394 and -286 kJ/mol respectively, then enthalpy of combustion of C₂H_4(g) will be

1. +141.2 kJ/mol
2. +1412 kJ/mol
3. -141.2 kJ/mol
4. -1412 kJ/mol

Answer: 4. -1412 kJ/mol

⇒ \(\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

∴ \(\Delta H^{\circ}=\Delta H_{\text {products }}^o-\Delta H_{\text {reactants }}^{\circ}\)

= \(2 \times(-394)+2 \times(-286)-(52+0)=-1412 \mathrm{~kJ} / \mathrm{mol}\)

Question 37. The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1: 0.5: 1. ΔH for the formation of XY is -200 kJ mol^-1. The bond dissociation energy of X₂ will be

1. 200 kJ mol^-1
2. 100 kJ mol^-1
3. 800 kJ mol^-1
4. 400 kJ mol^-1

Answer: 3. 800 kJ mol^-1

Let B.E. of \(X_2, Y_2\) and \(X Y\) are \(x \mathrm{~kJ} \mathrm{~mol}^{-1}, 0.5 x \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(x \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively.

⇒ \(\frac{1}{2} X_2+\frac{1}{2} Y_2 \rightarrow X Y ; \Delta H=-200 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\Delta H=\Sigma(B . E .)_{\text {Reactants }}-\Sigma(B . E .)_{\text {Products }}\)

∴ \(-200=\left[\frac{1}{2} \times(x)+\frac{1}{2} \times(0.5 x)\right]-[1 \times(x)]\)

B.E. of \(X_2=x=800 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 38. The heat of combustion of carbon to CO₂ is -393.5 kJ/mol. The heat released upon the formation of 35.2 g of CO₂ from carbon and oxygen gas is

1. +315 kJ
2. -630 kJ
3. -3.15 kJ
4. 315 kJ

Answer: None

Given: \(\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \longrightarrow \mathrm{CO}_{2(g)}, \Delta H=-393.5 \mathrm{~kJ} / \mathrm{mol}\)

Amount of heat released on formation of 44 g CO₂ = 393.5kJ

∴ Amount of heat released on formation of 35.2 g of CO₂

= \(\frac{393.5}{44} \times 35.2=314.8 \approx 315 \mathrm{~kJ}\)

-ve or +ve sign considering the reaction is exothermic or endothermic.

Question 39. When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0ºC and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in \(\mathrm{kJ}\left(\Delta H_{\text {comb }}\left(\mathrm{CH}_4\right)=890 \mathrm{~kJ} \mathrm{~mol}^{-1}\right.\), \(\left.\Delta H_{\text {comb }}\left(\mathrm{C}_3 \mathrm{H}_8\right)=2220 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) is\)

1. 38
2. 317
3. 477
4. 32

Answer: 2. 317

⇒ \(\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}\)

Number of moles in gaseous mixture \(\mathrm{CH}_4+\mathrm{C}_3 \mathrm{H}_8=\frac{5}{22.4}=0.22 \text { moles }\)

Number of moles of \(\mathrm{O}_2=\frac{16}{22.4}=0.71 \mathrm{moles}\)

Let x moles of CH₄ be there in a gaseous mixture so, a number of moles of C₃H₈ would be 0.22 – x.

Then moles of O₂ consumed, 2x + (0.22 – x)5 = 0.71 or x = 0.13

The total amount of heat liberated = 0.13 x 890 + 0.09 x 22210 = 315.5 J

Question 40. Enthalpy change for the reaction, 4 \(\mathrm{H}_{(g)} \rightarrow 2 \mathrm{H}_{2(g)} \text { is }-869.6 \mathrm{~kJ}\) The-dissociation energy of H – H bond is

1. -434.8 kJ
2. -869.6 kJ
3. +434.8 kJ
4. +217.4 kJ

Answer: 3. +434.8 kJ

The dissociation energy of H-H bond is \(\frac{869.6}{2}=434.8 \mathrm{~kJ}\)

Question 41. From the following bond energies:

1. H – H bond energy: 431.37 k) mol^-1
2. C: C bond energy: 606.10 kJ mol^-1
3. C – C bond energy: 336.49 kJ mol^-1
4. C – H bond energy: 410.50 kJ mol^-1

Enthalpy for the reaction, will be

1. -243.6 kJ mol^-1
2. -120.0 kJ mol^-1
3. 553.0 kJ mol^-1
4. 1523.6 kJ mol^-1

Answer: 2. -120.0 kJ mol^-1

For the given reaction, the enthalpy of the reaction can be calculated as

= \(\Sigma B \cdot E_{.}(\text {reactants })-\Sigma B \cdot E .(\text { products) }\)

= \(\left[B \cdot E_{(\mathrm{C}=\mathrm{C})}+B \cdot E_{(\mathrm{H}-\mathrm{H})}+4 \times B_{\cdot} E_{(\mathrm{C}-\mathrm{H})}\right]\)

∴ \(-\left[B \cdot E_{\cdot(\mathrm{C}-\mathrm{C})}+6 \times B \cdot E_{\cdot(\mathrm{C}-\mathrm{H})}\right]\)

= \([606.10+431.37+4 \times 410.50]-[336.49+6 \times 410.50]\)

= \(2679.47-2799.49=-120.02 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 42. Bond dissociation enthalpy of H₂, Cl₂, and HCI are 434,242 and 431 kJ mol^-1 respectively. Enthalpy of the formation of HCl is

1. -93 kJ mol^-1
2. 245 kJ mol^-1
3. 93 kJ mol^-1
4. -245 kJ mol^-1

Answer: 1. -93 kJ mol^-1

⇒ \(\mathrm{H}_2+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}\)

⇒ \(\Delta H_{\text {reaction }}=\Sigma(B \cdot E)_{\text {reactants }}-\Sigma(B \cdot E)_{\text {products }}\)

= \(\left[(B \cdot E)_{\mathrm{H}-\mathrm{H}}+(B \cdot E)_{\mathrm{Cl}-\mathrm{Cl}}\right]-\left[2 B \cdot E_{(\mathrm{H}-\mathrm{Cl})}\right]\)

= \(434+242-(431) \times 2\)

∴ \(\Delta H_{\text {reaction }}=-186 \mathrm{~kJ}\)

Question 43. Consider the following reactions:

1. \(\mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}=\mathrm{H}_2 \mathrm{O}_{(b)}, \Delta H=-X_1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(\mathrm{H}_{2(g)}+1 / 2 \mathrm{O}_{2(g)}=\mathrm{H}_2 \mathrm{O}_{(l)}, \Delta H=-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(\mathrm{CO}_{2(g)}+\mathrm{H}_{2(\mathrm{~g})}=\mathrm{CO}_{(g)}+\mathrm{H}_2 \mathrm{O}_{(l)}, \Delta H=-X_3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(\mathrm{C}_2 \mathrm{H}_{2(g)}+5 / 2 \mathrm{O}_{2(g)}=2 \mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(j)},\Delta H=+X_4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Enthalpy of the formation of H₂O_(l) is

1. \(+X_3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(-X_4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(+X_1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 4. \(-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent elements is called the heat of formation or enthalpy of formation.

Equation (1) represents the neutralisation reaction, (3) represents the hydrogenation reaction and (4) represents the combustion reaction.

Thus, enthalpy of formation of \(\mathrm{H}_2 \mathrm{O}_{(J)} \text {is }-X_2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 44. Given that the bond energy of H – H and Cl – Cl are 430 kJ mol^-1 and 240 kJ mol^-1 respectively and ΔH_f for HCl is -90 kJ mol^-1, the bond enthalpy of HCl is

1. 380 kJ mol^-1
2. 425 kJ mol^-1
3. 245 kJ mol^-1
4. 290 kJ mol^-1

Answer: 2. 425 kJ mol^-1

∴ \(\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \rightarrow \mathrm{HCl}\)

ΔH = \(=\Sigma B \cdot E_{\text {(reactants) }}-\Sigma B \cdot E_{\text {(products) }}\)

= \(\frac{1}{2}\left[B \cdot E_{\left(\mathrm{H}_2\right)}+B \cdot E_{\cdot\left(\mathrm{Cl}_2\right)}\right]-B \cdot E_{(\mathrm{HCl})}=-90\)

∴ \(\frac{1}{2}(430+240)-B \cdot E_{\cdot(\mathrm{HCl})}=-90\)

∴ \(B \cdot E_{(\mathrm{HCl})}=\frac{1}{2}(430+240)+90=425 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 45. The absolute enthalpy of neutralisation of the reaction: \(]\mathrm{MgO}_{(s)}+2 \mathrm{HCl}_{(a q)} \rightarrow \mathrm{MgCl}_{2(a q)}+\mathrm{H}_2 \mathrm{O}_{(f)}\) will be

1. -57.33 kJ mol^-1
2. Greater than -57.33 kJ mol^-1
3. Less than -57.33 kJ mol^-1
4. 57.33 kJ mol^-1

Answer: 3. Less than -57.33 kJ mol^-1

MgO is the oxide of a weak base and we know that the heat of neutralisation of 1 eq. of a strong acid with a strong base is -57.33 kJ/mol.

⇒ With a weak base, some heat is absorbed in the dissociation of the weak base.

⇒ The heat of neutralisation of weak base with strong acid will be less than -57.33kJ/mol

 

 

States Of Matter Multiple Choice Question and Answers

Question 1. Intermolecular forces are forces of attraction and repulsion between interacting particles that will include

1. Dipole-dipole forces
2. Dipole-induced dipole forces
3. Hydrogen bonding
4. Covalent bonding
5. Dispersion forces.

Choose the most appropriate answer from the options given below:

1. 1, 2, 3, 5 are correct.
2. 1, 3, 4, 5 are correct.
3. 2, 3, 4, and 5 are correct.
4. 1, 2, 3, 4 are correct.

Answer: 1. 1, 2, 3, 5 are correct.

Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms and molecules).

This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together, i.e., covalent bonds.

Question 2. Dipole-induced dipole interactions are present in which of the following pairs?

1. HCl and He atoms
2. SiF₄ and He atoms
3. H₂O and alcohol
4. Cl₂ and CCl₄

Answer: 1. HCl and He atoms

HCI is polar (μ ≠ 0) and He is non-polar (μ = 0) giving dipole-induced dipole interactions.

Question 3. Which one of the following is the correct order of interactions?

1. Covalent < hydrogen bonding < van der Waals < dipole-dipole
2. van der Waals < hydrogen bonding < dipole- dipole < covalent
3. van der Waals < dipole-dipole < hydrogen bonding < covalent
4. Dipole-dipole < van der Waals’ < hydrogen bonding < covalent.

Answer: 2. van der Waals < hydrogen bonding < dipole- dipole < covalent

The strength of interaction follows the order: van der Waals < hydrogen-bonding < dipole-dipole < covalent. It is so because the bond length of the H-bond is larger than that of a covalent bond.

Also covaient bond is strongest because, tire greater the extent of overlapping, the stronger the bond formed.

Read and Learn More NEET MCQs with Answers

Question 4. Which of the following statements is wrong for gases?

1. Confined gas exerts uniform pressure on the walls of its container in all directions.
2. The volume of the gas is equal to the volume of the container confining the gas.
3. Gases do not have a definite shape and volume.
4. The mass of a gas cannot be determined by weighing a container in which it is enclosed.

Answer: 4. Mass of a gas cannot be determined by weighing a container in which it is enclosed.

Mass of the gas = Mass of the cylinder including gas – Mass of empty cylinder.

So, the mass of a gas can be determined by weighing the container in which it is enclosed.

Thus, the statement (4) is wrong or gases

Question 5. Which of the following options is the correct graphical representation of Boyle Law?

Answer: 4

The graph in option (4) correctly the Boyle’s law.

Question 6. Choose the correct option for a graphical representation of Boyle’s law, which shows a graph of pressure vs volume of a gas at different temperatures.

Answer: 1

According to Boyle’s law, \(P \propto \frac{1}{V}\) (at constan T)

The graph between P and V is a hyperbola

Question 7. At 25°C and 730 mm pressure, 380 mL of dry oxygen was collected. If the temperature is constant, what volume will the oxygen occupy at 760 mm pressure?

1. 569 mL
2. 365 mL
3. 265 mL
4. 621 mL

Answer: 2. 365 mL

V₁ = 380 mL, P₁ =730 mm, V₂ = ? P₂= 760 mm.

From Boyle’s Iaw, P₁V₁ = P₂V₂

⇒ \(V_2=\frac{P_1 V_1}{P_2}=\frac{730 \times 380}{760}=365 \mathrm{~mL}\)

Question 8. Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by a definite fraction of its volume at

1. 0°C
2. Its critical temperature
3. Absolute zero
4. It’s Boyle temperature.

Answer: 1. 0°C

According to Charles’ law which states that the volume of the given mass of a gas increases or decreases by 1/27.3 of its volume at 0°C for a degree rise or fall of temperature at constant pressure.

∴ \(V_t=V_0\left(1+\frac{t}{273}\right) \text { at constant } P \text { and } n \text {. }\)

Question 9. Which one is not the correct mathematical equation for Daltons’ Law of partial pressure? Here p = total pressure of the gaseous mixture

1. \(p=p_1+p_2+p_3\)
2. \(p=n_1 \frac{R T}{V}+n_2 \frac{R T}{V}+n_3 \frac{R T}{V}\)
3. \(p_i=x_i p\), where \(p_i=\) partial pressure of \(i^{\text {th }}\) gas \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture
4. \(p_i=x_i p_i^{\circ}\), where \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture, \(p_i^{\circ}=\) pressure of \(i^{\text {th }}\) gas in pure state

Answer: 4. \(p_i=x_i p_i^{\circ}\), where \(x_i=\) mole fraction of \(i^{\text {th }}\) gas in gaseous mixture, \(p_i^{\circ}=\) pressure of \(i^{\text {th }}\) gas in pure state

Question 10. A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O₁ gas is behaving ideally). The pressure inside the flask in the bar is (Given R = 0.0831 L bar K^-1 mol^-1)

1. 2.5
2. 498.6
3. 49.8
4. 4.9

Answer: 4. 4.9

V = 10 L, Mass of O₁ = 64 g, T = 300 K

According to the ideal gas equation, PV = nRT

P = \(\frac{n R T}{V}=\frac{64}{32} \times \frac{0.0831 \times 300}{10} \Rightarrow P=4.986 \mathrm{bar}\)

Question 11. Choose the correct option for the total pressure (in atm) in a mixture of 4 g O₂ and 2 g H₂ confined in a total volume of one litre at 0° C is [Given R = 0.082 L atm mol^-1 K^-1, T = 273 K]

1. 26.02
2. 2.518
3. 2.602
4. 25.18

Answer: 4. 25.18

Applying PV = \(\frac{w}{M} R T\)

∴ \(P_{\mathrm{O}_2}=\frac{4}{32} \times \frac{0.0821 \times 273}{1}=2.80 \mathrm{~atm}\)

∴ \(P_{\mathrm{H}_2}=\frac{2}{2} \times \frac{0.0821 \times 273}{1}=22.4 \mathrm{~atm}\)

Now, according to Dalton’s law, \(P_{\text {total }}=P_{\mathrm{O}_2}+P_{\mathrm{H}_2}=2.80+22.4=25.2 \mathrm{~atm}\)

Question 12. A mixture of N₂ and Ar gases in a cylinder contains 7 g of N₂ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N₂ is [Use atomic masses (in g mol^-1): N₂= 14, Ar = 40]

1. 9 bar
2. 12 bar
3. 15 bar
4. 18 bar.

Answer: 3. 15 bar

Number of moles of \(\mathrm{N}_2=\frac{7}{28}\) = 0.25 mol

Number of moles of Ar = \(\frac{8}{40}\) = 0.2 mol

Mole fraction of \(\mathrm{N}_2=\frac{0.25}{0.25+0.2}=\frac{0.25}{0.45}=0.55\)

Partial pressure of N₂ gas = mole fraction x total pressure = 0.55 x 27 = 14.85 ≈ 15 bar

Question 13. The volume occupied by 1.8 g of water vapour at 374°C and 1 bar pressure will be [Use R = 0.083 bar L K^-1 mol^-1]

1. 96.66 L
2. 55.87 L
3. 3.10 L
4. 5.37 L

Answer: 4. 5.37 L

m = \(1.8 \mathrm{~g} \Rightarrow n=\frac{m}{M}=\frac{1.8}{18}=0.1 \mathrm{~mol}\)

T = \(374^{\circ} \mathrm{C}=647 \mathrm{~K}, P=1\) bar, R=\(0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)

V = \(\frac{n R T}{P}=\frac{0.1 \times 0.083 \times 647}{1}=5.37 \mathrm{~L}\)

Question 14. Equal moles of hydrogen and oxygen gases are placed in a container with a pinhole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

1. 3/8
2. 1/2
3. 1/8
4. 1/4

Answer: 3. 1/8

Let the number of moles of each gas = x

Fraction of hydrogen escaped \(=\frac{1}{2} x\)

⇒ \(\frac{r_{\mathrm{O}_2}}{r_{\mathrm{H}_2}}=\sqrt{\frac{M_{\mathrm{H}_2}}{M_{\mathrm{O}_2}}} \Rightarrow \frac{n_{\mathrm{O}_2} / t}{\frac{x}{2} / t}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}\)

⇒ \(\frac{n_{\mathrm{O}_2} / t}{\frac{x}{2} / t}=\frac{1}{4} \Rightarrow n_{\mathrm{O}_2}=\frac{1}{8} x\)

Hence, the fraction of oxygen escaped = 1/8

Question 15. What is the density of N₂ gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)

1. 1.40 g/mL
2. 2.81 g/mL
3. 3.41 g/mL
4. 0.29 g/mL

Answer: 3. 3.41 g/mL

PV=nRT

PV = \(\frac{w}{M} R T\left[n=\frac{\text { Weight of the gas taken }(W)}{\text { Mol. mass of gas }(M)}\right]\)

P = \(\frac{w}{M} \times \frac{R T}{V} ; P=\frac{d R T}{M} \quad\left[\text { Density }=\frac{\text { Mass }}{\text { Volume }}\right]\)

d = \(\frac{P M}{R T}=\frac{5 \times 28}{0.0821 \times 500}=3.41 \mathrm{~g} / \mathrm{mL}\)

Question 16. 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively to effuse through a pinhole under similar conditions. If the molecular mass of gas B is 36, the molecular mass of gas A will be

1. 96
2. 128
3. 32
4. 64

Answer: None

According to Graham’s law of diffusion, \(\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{M_2}{M_1}}, r_A=\frac{V_A}{T_A}, \quad r_B=\frac{V_B}{T_B}\)

⇒ \(\frac{V_A / T_A}{V_B / T_B}=\sqrt{\frac{M_B}{M_A}}\)

⇒ \(V_A=V_B, T_A=150 \mathrm{sec}, T_B=200 \mathrm{sec}, M_B=36, M_A=?\)

⇒ \(\frac{T_B}{T_A}=\sqrt{\frac{M_B}{M_A}} \Rightarrow \frac{200}{150}=\sqrt{\frac{36}{M_A}}\)

⇒ \(\frac{4}{3}=\sqrt{\frac{36}{M_A}} \text { or } \frac{4 \times 4}{3 \times 3}=\frac{36}{M_A} \text { or } M_A=\frac{36}{4 \times 4} \times 3 \times 3=20.25\)

Question 17. A certain gas takes three times as long to effuse out as helium. Its molecular mass will be

1. 27 u
2. 36 u
3. 64 u
4. 9u

Answer: 2. 36 u

According to Graham’s law of diffusion, \(r \propto \frac{1}{\sqrt{d}} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)

Rate of diffusion = \(\frac{\text { Volume of gas diffused }(V)}{\text { Time taken }(t)}\)

∴ \(\frac{V_1 / t_1}{V_2 / t_2}=\sqrt{\frac{M_2}{M_1}}\)

If the same volume of two gases diffuses, then \(V_1=V_2\)

∴ \(\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}\)

Here \(t_2=3 t_1, M_1=4 \mathrm{u}, M_2=\)?

∴ \(\frac{3 t_1}{t_1}=\sqrt{\frac{M_2}{4}} \Rightarrow 3=\sqrt{\frac{M_2}{4}} \Rightarrow 9=\frac{M_2}{4} \Rightarrow M_2=36 \mathrm{u}\)

Question 18. Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. The molecular mass of B will be

1. 50.00 u
2. 12.25 u
3. 6.50 u
4. 25.00 u

Answer: 2. 12.25 u

We know that \(\frac{r_A}{r_B}=\frac{V / t_A}{V / t_B}=\sqrt{\frac{M_B}{M_A}}\)

⇒ \(\frac{t_B}{t_A}=\sqrt{\frac{M_B}{M_A}} \Rightarrow \frac{10}{20}=\sqrt{\frac{M_B}{49}}\)

⇒ \(\left(\frac{10}{20}\right)^2=\frac{M_B}{49} \Rightarrow \frac{100}{400}=\frac{M_B}{49}\)

⇒ \(M_B=\frac{49 \times 100}{400}=12.25 \mathrm{u}\)

Question 19. A gaseous mixture was prepared by taking equal moles of CO and N₂. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N₂) in the mixture is

1. 0.5 atm
2. 0.8 atm
3. 0.9 atm
4. 1 atm

Answer: 1. 0.5 atm

⇒ \(p_{\mathrm{CO}}+p_{\mathrm{N}_2}=1 \mathrm{~atm}\)

⇒ \(2 p_{\mathrm{N}_2}=1\)

⇒ \(p_{\mathrm{N}_2}=\frac{1}{2}=0.5 \mathrm{~atm}\)

Question 20. A bubble of air is underwater at a temperature 15°C and a pressure of 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

1. The volume will become greater by a factor of 1.6.
2. The volume will become greater by a factor of 1.1.
3. The volume will become smaller by a factor of 0.70.
4. The volume will become greater by a factor of 2.5

Answer: 1. Volume will become greater by a factor of 1.6.

From ideal gas equation, \(V \propto \frac{T}{P}\)

Given \(T_1=15+273=288 \mathrm{~K}, P_1=1.5\) bar \(T_2=25+273=298 \mathrm{~K}, P_2=1\) bar \(V_1 \propto \frac{288}{1.5}\)

i.e., \(V_1 \propto 192\) and \(V_2 \propto \frac{298}{1}\)

∴ \(\frac{V_2}{V_1}=\frac{298}{192}=1.55 \approx 1.6\)

Question 21. The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J K^-1 mol^-1)

1. 275216 Pa
2. 13409 Pa
3. 41648 Pa
4. 31684 Pa

Answer: 3. 41648 Pa

Given, the mass of CH₄, w = 6g

Volume of CH₄, V= 0.03 m³

T = 129°C = 129 +273 = 402 K, R = 8.314 J K^-1 mol^-1

Molecular mass of CHr, M = 12.01 + 4 x 1.01 = 16.05

PV = nRT = \(\frac{w}{M} R T\)

∴ P = \(\frac{w}{M} \frac{R T}{V}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}\)

= \(41647.7 \mathrm{~Pa} \approx 41648 \mathrm{~Pa}\)

Question 22. Which of the following mixtures of gases does not obey the Daltorx law of partial pressure?

1. Cl₂ and SO₂
2. CO₂ and He
3. O₂ and CO₂
4. N₂ and O₂

Asnwer: 1. Cl₂ and SO₂

∴ \(\mathrm{Cl}_2+\mathrm{SO}_2\) \(\underrightarrow{\text { Sunlight }}\) \(\underset{\mathrm{Sulphuryl chloride}}{\mathrm{SO}_2 \mathrm{Cl}_2}\)

Dalton’s law of partial pressure is applicable only in those cases where gases are non-reacting. As Cl₂ and SO₂ react to form SO₂Cl₂ this law is not obeyed in a given case.

Question 23. At what temperature, the rate of effusion of N₂ would be 1.625 times the rate of SO₂ at 50°C?

1. 373°C
2. 620°C
3. 100°C
4. 173°C

Answer: 3. 10°C

∴ \(r_1=1.625 r_2\) and \(T_2=50^{\circ} \mathrm{C}=323 \mathrm{~K}\)

We know that \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1} \times \frac{T_1}{T_2}}\)

or \(1.625=\sqrt{\frac{64}{28} \times \frac{T_1}{323}}\) or \(T_1=\frac{(1.625)^2 \times 28 \times 323}{64}=373.15 \mathrm{~K}=100.15^{\circ} \mathrm{C}\)

Question 24. 50 mL of hydrogen diffuses out through a small hole of a vessel, in 20 minutes. The time taken by 40 mL of oxygen to diffuse out is

1. 32 minutes
2. 64 minutes
3. 8 minutes
4. 12 minutes

Answer: 2. 64 minutes

Volume of hydrogen = 50 mI; Time for diffusion (t) = 20 min and volume of oxygen = 40 mI.

Rate of diffusion of hydrogen (r₁) = 50/20 =2.5 mL/min

Rate of diffusion of oxygen (r₂) = 40/t mL/min

Since the molecular mass of hydrogen (M₁) = 2 and that of oxygen (M₂) = 32

Therefore \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \Rightarrow \frac{2.5}{40 / t}=\sqrt{\frac{32}{2}} \Rightarrow \frac{t}{16}=4 \Rightarrow t=64 \text { minutes }\)

Question 25. Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of I atm but also a concentration of 1-mole litre^-1? (R = 0.082 litre atm mol^-1 deg^-1)

1. At STP
2. When V = 22.4 litres
3. When T=12K
4. Impossible under any conditions

Answer: 3. When T= 12K

PV = nRT or \(P=\frac{n}{V} R T=C R T\)

Hence, \(1=1 \times 0.082 \times T \Rightarrow T=\frac{1}{0.082}=12 \mathrm{~K}\)

Question 26. The correct value of the gas constant ‘R’ is close to

1. 0.082 litre-atmosphere K
2. 0.082 litre-atmosphere K^-1 mol^-1
3. 0.082 litre-atmosphere^-1 K mol^-1
4. 0.082 litre^-1 atmosphere^-1 K mol.

Answer: 2. 0.082 litre-atmosphere K^-1 mol^-1

Question 27. Select one correct statement. In the gas equation, PV = nRT

1. n is the number of molecules of a gas
2. V denotes the volume of one mole of the gas
3. n moles of the gas have a volume of V
4. P is the pressure of the gas when only one mole of gas is present.

Answer: 3. n moles of the gas have a volume of V

In the ideal gas equation, PV=nRT

n moles of the gas have volume V

Question 28. At constant temperature, in a given mass of an ideal gas

1. The ratio of pressure and volume always remains constant
2. Volume always remains constant
3. Pressure always remains constant
4. The product of pressure and volume always remains constant.

Answer: 4. The product of pressure and volume always remains constant.

According to Boyle’s law at a constant temperature, \(P \propto \frac{1}{V} \text { or } P V=\text { constant }\)

Question 29. If P, V M, T and R are pressure, volume, molar mass, temperature and gas constant respectively, then for an ideal gas, the density is given by

1. \(\frac{R T}{P M}\)
2. \(\frac{P}{R T}\)
3. \(\frac{M}{V}\)
4. \(\frac{P M}{R T}\)

Answer: 4. \(\frac{P M}{R T}\)

Ideal gas equation is, \(P V=n R T=\frac{m}{M} R T\)

or \(P M=\frac{m}{V} R T=d R T\) [here d= density]

⇒ d = \(\frac{P M}{R T}\)

Question 30. The correct gas equation is

1. \(\frac{V_1 T_2}{P_1}=\frac{V_2 T_1}{P_2}\)
2. \(\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)
3. \(\frac{P_1 T_1}{V_1}=\frac{P_2 V_2}{T_2}\)
4. \(\frac{V_1 V_2}{T_1 T_2}=P_1 P_2\)

Answer: 2. \(\frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)

∴ \(\frac{P V}{T}=\text { constant or } \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \Rightarrow \frac{P_1 V_1}{P_2 V_2}=\frac{T_1}{T_2}\)

Question 31. By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?

1. 2.0
2. 2.8
3. 4.0
4. 1.4

Answer: 4. 1.4

Average velocity = \(\sqrt{\frac{8 R T}{\pi M}}\)

When T becomes 2T then average velocity = \(\sqrt{\frac{8 R(2 T)}{\pi M}}\)

i.e., \(\sqrt{2}\) or 1.41 times increase.

Question 32. The temperature of a gas is raised from 27°C to 927°C. The root mean square speed of the gas

1. Remains same
2. \(\text { gets } \sqrt{\frac{927}{27}} \text { times }\)
3. Gets halved
4. Gets doubled.

Answer: 4. Gets doubled.

T₁ =27°C = 300 K and T₂ = 927°C= 1200 K

We know that root means square speed (v) ∝ √T.

Therefore root mean square speed of the gas, when its temperature is raised = \(\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=2 \text { times }\)

Question 33. The ratio among most probable velocity, mean velocity and root mean square velocity is given by

1. \(1: 2: 3\)
2. \(1: \sqrt{2}: \sqrt{3}\)
3. \(\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}\)
4. \(\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}\)

Answer: 4. \(\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}\)

Most probable velocity, \(\left(u_{m p}\right)=\sqrt{\frac{2 R T}{M}}\)

Mean velocity, \((\bar{v})=\sqrt{\frac{8 R T}{\pi M}}\)

Root mean square velocity, \(\left(u_{\text {r.m.s }}\right)=\sqrt{\frac{3 R T}{M}}\)

∴ \(u_{m p}: \bar{v}: u_{r, m, s}=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}}=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}\)

Question 34. The root mean square velocity at STP for the gases H₂, N₂, O₂ and HBr are in the order

1. \(\mathrm{H}_2<\mathrm{N}_2<\mathrm{O}_2<\mathrm{HBr}\)
2. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{N}_2<\mathrm{H}_2\)
3. \(\mathrm{H}_2<\mathrm{N}_2=\mathrm{O}_2<\mathrm{HBr}\)
4. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{H}_2<\mathrm{N}_2\)

Answer: 2. \(\mathrm{HBr}<\mathrm{O}_2<\mathrm{N}_2<\mathrm{H}_2\)

PV = \(\frac{1}{3} m n u^2=\frac{1}{3} M u^2\) or \(u=\sqrt{3 P V / M}\),

At STP, \(u \propto \sqrt{\frac{1}{M}}\) and molecular masses of \(\mathrm{H}_2, \mathrm{~N}_2, \mathrm{O}_2\) and \(\mathrm{HBr}\) are 2, 28, 32 and 81 .

Question 35. The root mean square velocity of a gas molecule is proportional to

1. \(m^{1 / 2}\)
2. \(m^0\)
3. \(m^{-1 / 2}\)
4. m

Answer: 3. \(m^{-1 / 2}\)

PV = \(\frac{1}{3} m N u^2\), here u= root mean square velocity.

Now \(u^2=\frac{3 P V}{m N}\) or \(u \propto \frac{1}{\sqrt{m}}\)

Question 36. The energy absorbed by each molecule (A₂) of a substance is 4.4 x 10^-19 J and bond energy per molecule is 4.0 x 10^-19 J. The kinetic energy of the molecule per atom will be

1. \(2.2 \times 10^{-19} \mathrm{~J}\)
2. \(2.0 \times 10^{-19} \mathrm{~J}\)
3. \(4.0 \times 10^{-20} \mathrm{~J}\)
4. \(2.0 \times 10^{-20} \mathrm{~J}\)

Answer: 4. \(2.0 \times 10^{-20} \mathrm{~J}\)

Energy absorbed by each molecule = 4.4 x 10^-19 J

The energy required to break the bond = 4.0 x 10^-19 J

The remaining energy is converted to kinetic energy

= (4.4 x 10^-19 – 4.0 x 10^-19 )J = 0.4 x 10^-19 J  per molecule

∴ Kinetic energy per atom = 0.2 x 10^-19 J or 2 x 10^-20 J

Question 37. If a gas expands at a constant temperature, it indicates that

1. The kinetic energy of molecules remains the same
2. Number of the molecules of gas increases
3. The kinetic energy of molecules decreases
4. The pressure of the gas increases.

Answer: 1. Kinetic energy of molecules remains the same

The average translationalK’E. of one molecule of an ideal gas will be given by \(E_t=\frac{K. E .}{N_A}=\frac{3 / 2 R T}{N_A}=\frac{3}{2} k T\)

where \(R / N_A=\) Boltzmann constant

i.e. \(E_t \propto T\)

Question 38. The average molar kinetic energy of CO and N₂ at the same temperature is

1. KE₁ = KE₂
2. KE₁ > KE₂
3. KE₁ < KE₂
4. Can’t say anything. Both volumes are not given.

Answer: 1. KE₁ = KE₂

K.E = \(\frac{3}{2}\)RT (for one mole of a gas)

As temperatures are the same and KE is independent of molecular mass KE₁ = KE₂.

Question 39. The average kinetic energy of an ideal gas, per molecule in S.I. units, at 25°C will be

1. \(6.17 \times 10^{-20} \mathrm{~J}\)
2. \(7.16 \times 10^{-20} \mathrm{~J}\)
3. \(61.7 \times 10^{-20} \mathrm{~J}\)
4. \(6.17 \times 10^{-21} \mathrm{~J}\)

Answer: 4. \(6.17 \times 10^{-21} \mathrm{~J}\)

Temperature (T) = 25°C=298K.

Therefore, K.E. per molecule = \(\frac{3 R T}{2 N_A}=\frac{3 \times 8.314 \times 298}{2 \times\left(6.02 \times 10^{23}\right)}=6.17 \times 10^{-21} \mathrm{~J}\)

Question 40. At STP, 0.50 mol H₂ gas and 1.0 mol He gas

1. Have equal average kinetic energies
2. Have equal molecular speeds
3. Occupy equal volumes
4. Have equal effusion rates.

Answer: 1. Have equal average kinetic energies

Because average kinetic energy depends only on temperature K.E = \(\frac{3}{2}\) K T

Question 41. Internal energy and pressure of a gas per unit volume are related as

1. \(P=\frac{2}{3} E\)
2. \(P=\frac{3}{2} E\)
3. \(P=\frac{1}{2} E\)
4. \(P=2 E\)

Answer: 1. \(P=\frac{2}{3} E\)

PV = \(\frac{1}{3} m n u^2=\frac{1}{3} M u^2\)

= \(\frac{2}{3} \cdot \frac{1}{2} M u^2=\frac{2}{3} E\) (because \(\frac{1}{2} M u^2=E\))

or P = \(\frac{2}{2} E\) per unit volume.

Question 42. A closed flask contains water in all its three states solid, liquid and vapour at 0°C. In this situation, the average kinetic energy of water molecules will be

1. The greatest in all the three states
2. The greatest in vapour state
3. The greatest in the liquid state
4. The greatest in the solid state.

Answer: 2. The greatest vapour state

Velocity and hence average KE. of water molecules is maximum in the gaseous state.

Question 43. Which is not true in the case of an ideal gas?

1. It cannot be converted into a liquid.
2. There is no interaction between the molecules.
3. All molecules of the gas move at the same speed.
4. At a given temperature, PV is proportional to the amount of the gas.

Answer: 3. All molecules of the gas move at the same speed.

Molecules in an ideal gas range with different speeds. Due to collision between the particles their speed changes.

Question 44. A gas at 350 K and 15 bar has a molar volume 20 per cent smaller than that of an ideal gas under the same conditions. Tire correct option about the gas and its compressibility factor (Z) is

1. Z < 1 and repulsive forces are dominant
2. Z > 1 and attractive forces are dominant
3. Z > 1 and repulsive forces are dominant
4. Z < 1 and attractive forces are dominant.

Answer: 4. Z < 1 and attractive forces are dominant.

∴ \(V_{\text {ideal }}=V, V_{\text {real }}=V-0.2 \mathrm{~V}=0.8 \mathrm{~V}\)

Z = \(\frac{V_{\text {real }}}{V_{\text {ideal }}}=0.8\)

Question 45. A gas such as carbon monoxide would be most likely to obey the ideal gas law at

1. Low temperatures and high pressures
2. High temperatures and high pressures
3. Low temperatures and low pressures
4. High temperatures and low pressures.

Answer: 4. High temperatures and low pressures

Real gases show ideal gas behaviour at high temperatures and low pressures.

Question 46. Maximum deviation from ideal gas is expected from

1. \(\mathrm{CH}_{4(g)}\)
2. \(\mathrm{NH}_{3(g)}\)
3. \(\mathrm{H}_{2(g)}\)
4. \(\mathrm{N}_{2(g)}\)

Answer: 2. \(\mathrm{NH}_{3(g)}\)

NH₃ is a polar molecule, thus more attractive forces between NH₃ molecules.

Question 47. For real gases, van der Waals’ equation is written as \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T \) where a and b are van der Waals’ constants. Two sets of gases are

1. \(\mathrm{O}_2, \mathrm{CO}_2, \mathrm{H}_2\) and \(\mathrm{He}\)
2. \(\mathrm{CH}_4, \mathrm{O}_2\) and \(\mathrm{H}_2\)

The gases given in set-1 in increasing order of 2 and gases given in set-2 in decreasing order of 1 are arranged below. Select the correct order from the following:

1. 1. \(\mathrm{He}<\mathrm{H}_2<\mathrm{CO}_2<\mathrm{O}_2\)
   2. \(\mathrm{CH}_4>\mathrm{H}_2>\mathrm{O}_2\)
2. 1.  \(\mathrm{O}_2<\mathrm{He}<\mathrm{H}_2<\mathrm{CO}_2\)
   2. \(\mathrm{H}_2>\mathrm{O}_2>\mathrm{CH}_4\)
3. 1. \(\mathrm{H}_2<\mathrm{He}<\mathrm{O}_2<\mathrm{CO}_2\)
   2. \(\mathrm{CH}_4>\mathrm{O}_2>\mathrm{H}_2\)
4. 1. \(\mathrm{H}_2<\mathrm{O}_2<\mathrm{He}<\mathrm{CO}_2\)
   2. \(\mathrm{O}_2>\mathrm{CH}_4>\mathrm{H}_2\)

Answer: 3. \(\mathrm{H}_2<\mathrm{He}<\mathrm{O}_2<\mathrm{CO}_2\) and \(\mathrm{CH}_4>\mathrm{O}_2>\mathrm{H}_2\)

Question 48. van der Waals’ real gas, acts as an ideal gas, at which conditions?

1. High temperature, low pressure
2. Low temperature, high pressure
3. High temperature, high pressure
4. Low temperature, low pressure

Answer: 1. High temperature, low pressure

At low-pressure and high-temperature van der Waals, real gas acts as an ideal gas and is observed to obey PV = nRT relation.

At very low pressure when the gas volume is quite large the space occupied by the molecules themselves becomes negligible comparatively and because the molecules are then far apart, the force of mutual attraction becomes too feeble, the real gas would satisfy the postulates of kinetic theory.

As the temperature is raised, the volume of the gas increases and

we can consider \(\left(P+\frac{n^2 a}{V^2}\right)\) term as P and at low pressure (V – nb) term as V. \(\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T\) (van der Waals equation)

This equation becomes PV = nRT

This is an ideal gas equation

Question 49. When is deviation more in the behaviour of a gas from the ideal gas equation PV = nRT.

1. At high temperatures and low pressure
2. At low temperatures and high pressure
3. At high temperatures and high pressure
4. At low temperatures and low pressure

Answer: 2. At low temperatures and high pressure

At low temperatures and high pressure, there is a deviation from the ideal behaviour in gases.

Question 50. A gas is said to behave like an ideal gas when the relation PV/T = constant. When do you expect a real gas to behave like an ideal gas?

1. When the temperature is low.
2. When both the temperature and pressure are low.
3. When both the temperature and pressure are high
4. When the temperature is high and pressure is low.

Answer: 4. When the temperature is high and pressure is low.

At high temperatures and low pressure, the effect of a/V² and b is negligible.

As we know, PV = nRT (Ideal gas equation)

PV = \(R T \text { or } \frac{P V}{R T}=1\)

∴ Z=1 [Z is compressibility factor]

Hence gas shows ideal behaviour.

Question 51. In van der Waals’ equation of state for a non-ideal gas, the term that accounts for intermolecular forces is

1. \((V-b)\)
2. \((R T)^{-1}\)
3. \(\left(P+\frac{a}{V^2}\right)\)
4. R T

Answer: 3. \(\left(P+\frac{a}{V^2}\right)\)

Vander Waals’ equation for 1 mole is \(\left(P+\frac{a}{V^2}\right)\)(V-b)=RT

Here, \(\left(P+\frac{a}{V^2}\right)\) represents the intermolecular forces and (V-b) is the correct volume.

Question 52. Given van der Waals’ constant for NH₃, H₂, O₂ and CO₂ are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquified?

1. NH₃
2. H₂
3. O₂
4. HO₂

Answer: 1. NH₃

van der Waals constant signifies the intermolecular forces of attraction between the particles of gas.

So, the higher the value of ‘a’, the easier will be the liquefaction of gas.

Question 53. An ideal gas, obeying the kinetic theory of gases cannot be liquefied, because

1. It solidifies before becoming a liquid
2. Forces acting between its molecules are negligible
3. Its critical temperature is above 0°c
4. Its molecules are relatively small in size.

Answer: 2. Forces acting between its molecules are negligible

A gas can only be liquefied, if some forces of attraction are acting in its molecules. According to kinetic theory, an ideal gas is devoid of the force of attraction in its molecules, therefore it cannot be liquefied.

Question 54. The beans are cooked earlier in a pressure cooker because

1. The boiling point increases with increasing pressure
2. The boiling point decreases with increasing pressure
3. The extra pressure of the cooker softens the beans
4. Internal energy is not lost while cooking in a pressure cooker

Answer: 1. Boiling point increases with increasing pressure

The more the pressure, the greater the boiling point.

 

Classification Of Elements And Periodicity In Properties

Question 1. The IUPAC name of an element with atomic number 119 is

1. Ununennium
2. Unnilennium
3. Unununnium
4. Ununoctium.

Answer: 1. Ununennium

The IUPAC name of an element with atomic number 119 is ununennium.

Question 2. Identify the incorrect match.

1. 1-A
2. 2-B
3. 3-C
4. 4-D

Answer: 4. 4-D

Unnilunium – Mendelevium ⇒ (1)-(A)

Unniltrium – Lawrencium ⇒ (2)-(B)

Unnilhexium – Seaborgium ⇒ (3)-(C)

Unununnium – Roentgenium ⇒ (4)-(D)

Question 3. The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?

1. Carbon family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^2\)
2. Oxygen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^4\)
3. Nitrogen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^6\)
4. Halogen family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^5\)

Answer: 1. Carbon family, \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^2\)

The electronic configuration of the element with Z = 114 (Flerovium) is \([\mathrm{Rn}] 5 f^4 6 d^{10} 7 s^2 7 p^2\)

Hence, it belongs to the carbon family which has the same outer electronic configuration.

Read and Learn More NEET MCQs with Answers

Question 4. An atom has electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s² you will place it in

1. Fifth group
2. Fifteenth group
3. Second group
4. Third group.

Answer: 1. Fifth Group

The electronic configuration of an atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

In the configuration, the last electron of the atom is filled in d-subshell as 3d³. Thus, this element belongs to the d-block of the periodic table with group number VB or 5.

Question 5. The electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p³ What is the atomic number of the element, which is just below the above element in the periodic table?

1. 36
2. 49
3. 33
4. 34

Answer: 3. 33

The atomic number of the given element is 15 and it belongs to group 15. Therefore the atomic number of the element below the above element = 15 + 18 = 33

Question 6. If the atomic number of an element is 33, it will be placed in the periodic table in the

1. First group
2. Third group
3. Fifth group
4. Seventh group.

Answer: 3. Fifth Group

The electronic configuration of an element with Z = 33 is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p³

Hence, it lies in the VA or the 15th group

Question 7. The electronic configuration of four elements is given below. Which elements do not belong to the same family as others?

1. \([\mathrm{Xe}] 4 f^{14} 5 d^{10} 4 s^2\)
2. \([\mathrm{Kr}] 4 d^{10} 5 s^2\)
3. \([\mathrm{Ne}] 3 s^2 3 p^5\)
4. \([\mathrm{Ar}] 3 d^{10} 4 s^2\)

Answer: 3. \([\mathrm{Ne}] 3 s^2 3 p^5\)

Elements (1), (2) and (4) belong to the same group since each one of them has two electrons in the valence shell. In contrast, element (3) has seven electrons in the valence shell, and hence it lies in another group.

Question 8. The element expected to form the largest ion to achieve the nearest noble gas configuration is

1. N
2. Na
3. O
4. F

Answer: 1. N

N^3-, O^2-, F^– and Nat have 10 electrons each, hence these are isoelectronic. For isoelectronic species, the size of the species decreases as the nuclear charge increases. Hence, the size decreases as N^3- > O^2- > F^– > Na⁺

Hence, among the given elements, nitrogen is expected to form the largest ion to achieve the nearest noble gas configuration.

Question 9. For the second-period elements, the correct increasing order of first ionization enthalpy is

1. Li < Be < B < C < O < N < F < Ne
2. Li < Be < B < C < N < O < F < Ne
3. Li < B < Be < C < O < N < F < Ne
4. Li < B < Be < C < N < O < F < Ne

Answer: 3. Li < B < Be < C < O < N < F < Ne

As we move across a period, ionisation enthalpy increases, because of increased nuclear charge and decrease in atomic radii. However, abnormal values are observed for Be, N and Ne due to the extra stability of half-filled and fully-filled orbitals.

Thus, the actual order is, Li<B<Be<C<O<N<F<Ne.

Question 10. Match the oxide given in column A with its property given in column B

Which of the following options has all the correct pairs?

1. (1)-B, (2)-A, (3)-D, (4)-C
2. (1)-C, (2)-B, (3)-A, (4)-D
3. (1)-A, (2)-D, (3)-B, (4)-C
4. (1)-B, (2)-D, (3)-A, (4)-C

Answer: 4. (1)-B, (2)-D, (3)-A, (4)-C

Na₂O – Basic oxide, AI₂O₃ – Amphoteric oxide, N₂O – Neutrai oxide, Cl₂O₇ – Acidic oxide.

Question 11. Which of the following oxides is most acidic in nature?

1. MgO
2. BeO
3. BaO
4. CaO

Answer: 2. BeO

In metals, on moving down the group, metallic character increases, so basic nature increases hence most acidic will be BeO

Question 12. In which of the following options the order of arrangement does not agree with the variation of the property indicated against it?

1. I < Br < Cl < F (increasing electron gain enthalpy)
2. Li < Na < K < Rb (increasing metallic radius)
3. Al³⁺ < Mg²⁺ < Na⁺ < F^– (increasing ionic size)
4. B < C < N < O (increasing first ionisation enthalpy)

Answer: 1. I < Br < Cl < F (increasing electron gain enthalpy) and 4. B < C < N < O (increasing first ionisation enthalpy)

The correct order of increasing negative electron gain enthalpy is: I < Br < F < CI due to electron-electron repulsion in small-sized F atom and the correct order of increasing first ionisation enthalpy is B < C < O < N due to extra stability of half-filled orbitals in N-atom.

Question 13. The formation of the oxide ion, O^2-_(g) from the oxygen atom requires first an exothermic and then an endothermic step as shown below:

• \(\mathrm{O}_{(g)}+e^{-} \rightarrow \mathrm{O}_{(g)}^{-} ; \Delta_f H^{\circ}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
• \(\mathrm{O}_{(g)}^{-}+e^{-} \rightarrow \mathrm{O}_{(g)}^{2-} ; \Delta_f H^{\circ}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Thus, the process of formation of O^2- in the gas phase is unfavourable even though O^2- is isoelectronic with neon. It is due to the fact that,

1. O^– ion has a comparatively smaller size than the oxygen atom
2. Oxygen is more electronegative
3. The addition of an electron in oxygen results in a larger size of ion
4. Electron repulsion outweighs the stability gained by achieving noble gas configuration.

Answer: 4. Electron repulsion outweighs the stability gained by achieving noble gas configuration.

Question 14. Which of the following orders of ionic radii is correctly represented?

1. \(\mathrm{H}^{-}>\mathrm{H}^{+}>\mathrm{H}\)
2. \(\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}\)
3. \(\mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+}\)
4. \(\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{N}^{3-}\)

Answer: None

Cations lose electrons and are smaller in size than the parent atom, whereas anions add electrons and are larger in size than the parent atom. Hence, the order is H^–>H>H⁺.

For isoelectronic species, the ionic radii decreases with increase in atomic number i.e., orclear charge. Hence, the correct orders are \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+} \text {and } \mathrm{N}^{3-}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\)

Question 15. Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O?

1. Al < Ca < O < C <F
2. Al < O < C < Ca < F
3. C < F < O < Al < Ca
4. Ca < Al < C < O < F

Answer: 4. Ca < Al < C < O < F

Electron gain enthalpy becomes less negative from top to bottom in a group while it becomes more negative from left to right within a period.

Question 16. In which of the following arrangements the given sequence is not strictly according to the property indicated against it?

1. HF < HCl < HBr < HI: increasing acidic strength
2. H₂O < H₂S < H₂Se < H₂Te: increasing pK_a values
3. NH₃ < PH₃ < AsH₃ < SbH₃: increasing acidic character
4. CO₂ < SiO₂ < SnO₂ < PbO₂: increasing oxidising power

Answer: 2. H₂O < H₂S < H₂Se < H₂Te: increasing pKa values

The acidic strength of hydrides increases with an increase in molecular mass.

Thus, the order of acidic strength is

HF<HCl<HBr<HI

H₂O < H₂S < H₂Se < H₂Te

NH₃ < PH₃ < AsH₃ < SbH₃

and as acidic strength increases, pK_a decreases. Thus order of \(\mathrm{p} K_p, \mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{Te}\)

Question 17. Identify the wrong statement in the following.

1. Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.
2. Amongst isoelectronic species, the greater the negative charge on the anion, the larger the ionic radius.
3. The atomic radius of the elements increases as one moves down the first group of the periodic table.
4. Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.

Answer: 1. Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.

As positive charge on the cation increases, effective nuclear charge increases. Thus, atomic size decreases.

Question 18. What is the value of electron gain enthalpy of Na⁺ if IE₁ of Na = 5.1 eV?

1. -5.1 eV
2. -10.2 eV
3. +2.55 eV
4. +10.2 eV

Answer: 1. -5.1 eV

⇒ \(\mathrm{Na} \rightarrow \mathrm{Na}^{+}+e^{-} ; \Delta H=5.1 \mathrm{eV}\)

⇒ \(\mathrm{Na}^{+}+e^{-} \rightarrow \mathrm{Na} ; \Delta H=-5.1 \mathrm{eV}\)

Question 19. Which of the following oxides is amphoteric?

1. SnO₂
2. CaO
3. SiO₂
4. CO₂

Answer: 1. SnO₂

SnO₂ reacts with acid as well as base. So, SnO₂ is an amphoteric oxide.

⇒ \(\mathrm{SnO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{SnO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_3+\mathrm{H}_2 \mathrm{O}\)

Question 20. The correct order of the decreasing ionic radii among the following isoelectronic species is

1. \(\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}\)
2. \(\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}\)
3. \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)
4. \(\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Cl}^{-}>\mathrm{S}^{2-}\)

Answer: 3 \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)

S^2- > Cl^– > K⁺ > Ca²⁺

Among isoelectronic species, ionic radii increase with the increase in negative charge. This happens because the effective nuclear charge (Z_eff) decreases.

Similarly, ionic radii decrease with an increase in positive charge as Z_eff increases.

Question 21. Which of the following represents the correct order of increasing electron gain enthalpy with a negative sign for the elements O, S, F and Cl?

1. Cl < F < O < S
2. O < S < F < Cl
3. F < S < O < Cl
4. S < O < Cl < F

Answer: 2. O < S < F < Cl

Cl atom has the highest electron affinity in the periodic table. F being a member of group 17 has higher electron gain enthalpy than S which belongs to group 16. This in turn is higher than the electron affinity of O atom. Thus, Cl >F>S>O

It is worth noting that the electron gain enthalpy of oxygen and fluorine, the members of the second period, have less negative values of electron gain enthalpy than the corresponding elements sulphur and chlorine of the third period.

This is due to small size of the atoms of oxygen and fluorine. As a result, there is a strong inter-electronic repulsion when an extra electron is added to these atoms, i.e., the electron density is high and the addition of an extra electron is not easy.

Question 22. Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is

1. Mg < Ca < Cl < P
2. Cl < P < Mg < Ca
3. P < Cl < Ca < Mg
4. Ca < Mg < P < Cl

Answer: 2. Cl < P < Mg < Ca

The atomic radii decrease on moving from left to right in a period, the order of sizes for Cl, P and Mg is Cl < P < Mg. Down the group size increases. thus, overall order is Cl<P<Mg<Ca.

Question 23. Among the following which one has the highest cation to anion size ratio?

1. CsI
2. CsF
3. LiF
4. NaF

Answer: 2. CsF

The cation-to-anion size ratio will be maximum when the cation is of the largest size and the anion is of the smallest size.

Among the given species, Cs⁺ has the maximum size among given cations and F^– has the smallest size among given anions, thus CsF has the highest r_c/r_a ratio.

Question 24. Amongst the elements with the following electronic configurations, which one of them may have the highest ionisation energy?

1. \(\mathrm{Ne}\left[3 s^2 3 p^2\right]\)
2. \({Ar}\left[3 d^{10} 4 s^2 4 p^3\right]\)
3. \(\mathrm{Ne}\left[3 s^2 3 p^1\right]\)
4. \(\mathrm{Ne}\left[3 s^2 3 p^3\right]\)

Answer: 4. \(\mathrm{Ne}\left[3 s^2 3 p^3\right]\)

Among options (1), (3) and (4), option (4) has the highest ionisation energy because of extra stability associated with, a half-filled 3p-orbital.

In option (2), the presence of 3d¹⁰ electrons offers a shielding effect, as a result, the 4p³ electrons do not experience much nuclear charge and hence, the electrons can be removed easily.

Question 25. Identify the correct order of the size of the following.

1. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
2. \(\mathrm{Ar}<\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
3. \(\mathrm{Ca}^{2+}<\mathrm{Ar}<\mathrm{K}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)
4. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{S}^{2-}<\mathrm{Cl}^{-}\)

Answer: 1. \(\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Ar}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}\)

Among isoelectronic ions, ionic radii of anions is more than those of cations. Further size of the anion increases with an increase in negative charge and size of the cation decreases with an increase in positive charge.

Question 26. With which of the following electronic configuration an atom has the lowest ionisation enthalpy?

1. \(1 s^2 2 s^2 2 p^3\)
2. \(1 s^2 2 s^2 2 p^5 3 s^1\)
3. \(1 s^2 2 s^2 2 p^6\)
4. \(1 s^2 2 s^2 2 p^5\)

Answer: 2. \(1 s^2 2 s^2 2 p^5 3 s^1\)

The larger the atomic size, the smaller the value of the ionisation enthalpy. Again higher the screening effect, the lesser the value of ionisation potential. Hence, option (2) has the lowest ionisation enthalpy.

Question 27. Which one of the following ionic species has the greatest proton affinity to form a stable compound?

1. \(\mathrm{NH}_2^{-}\)
2. \(\mathrm{F}^{-}\)
3. \(\mathrm{I}^{-}\)
4. \(\mathrm{HS}^{-}\)

Answer: 1. \(\mathrm{NH}_2^{-}\)

In going from Ieft to right across a period in the periodic table, the basicity (l.e., proton affinity) decreases as the electronegativity of the atom possessing the lone pair of electrons increases.

Hence, the basicity of NH₂^– is higher than F^–. On moving down a group, as the atomic size increases, basicity decreases.

Hence, F is more basic than I^– and HO^– is more basic than HS^–. Hence, among the given ionic species, NH₂^– has maximum proton affinity.

Question 28. Which of the following is the most basic oxide?

1. SeO₂
2. Al₂O₃
3. Sb₂O₃
4. Bi₂O₃

Answer: 4. Bi₂O₃

⇒ \(\mathrm{SeO}_2 \longrightarrow\) acidic oxide,

⇒ \(\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Sb}_2 \mathrm{O}_3 \longrightarrow\) amphoteric, \(\mathrm{Bi}_2 \mathrm{O}_3 \longrightarrow\) basic oxide.

Question 29. What is the correct relationship between the pH of isomolar solutions of sodium oxide, Na₂O (pH₁), sodium sulphide, Na₂S (pH₂), sodium selenide, Na₂Se (pH₃) and sodium telluride Na₂Te (pH₄)?

1. \(\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4\)
2. \(\mathrm{pH}_1>\mathrm{pH}_2=\mathrm{pH}_3>\mathrm{pH}_4\)
3. \(\mathrm{pH}_1<\mathrm{pH}_2<\mathrm{pH}_3<\mathrm{pH}_4\)
4. \(\mathrm{pH}_1<\mathrm{pH}_2<\mathrm{pH}_3=\mathrm{pH}_4\)

Answer: 1. \(\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4\)

⇒ \(\begin{array}{l|l}
\mathrm{Na}_2 \mathrm{O} & \text { Basic character } \\
\mathrm{Na}_2 \mathrm{~S} & \text { decreases down the group } \\
\mathrm{Na}_2 \mathrm{Se} & \\
\mathrm{Na}_2 \mathrm{Te} &
\end{array}\)

pH ∝ basic character

Hence, pH₁ > pH₂ > PH₃ > pH₄

Question 30. Ionic radii are

1. Inversely proportional to the effective nuclear charge
2. Inversely proportional to the square of the effective nuclear charge
3. Directly proportional to the effective nuclear charge
4. Directly proportional to the square of effective nuclear charge.

Answer: 2. Inversely proportional to the square of effective nuclear charge

Question 31. The ions \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) are isoelectronic. Their ionic radii show

1. A significant increase from O^2- to Al³⁺
2. A significant decrease from O^2- to Al³⁺
3. An increase from O^2- to F^– and then decrease from Na⁺ to Al³⁺
4. A decrease from O^2- to F^– and then increase from Na⁺ to Al³⁺

Answer: 2. A significant decrease from O^2- to Al³⁺

Amongst isoelectronic ions, the ionic radii of anions are more than that of cations. Further size of the anion increases with an increase in the +ve charge and the size of the cation decreases with an increase in the +ve charge.

Hence, correct order is \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} .\)

Question 32. Which of the following orders is wrong?

1. \(\mathrm{NH}_3<\mathrm{PH}_3<\mathrm{AsH}_3\)-acidic
2. \(\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}-1^{\text {st }}\) IP
3. \(\mathrm{Al}_2 \mathrm{O}_3<\mathrm{MgO}<\mathrm{Na}_2 \mathrm{O}<\mathrm{K}_2 \mathrm{O}-\) basic
4. \(\mathrm{Li}^{+}<\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Cs}^{+}-\) ionic radius.

Answer: 2. \(\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}-1^{\text {st }}\) IP

Li, Be, B, C – these elements belong to the same period. Generaliy the value of 1’t ionisation potential increases on moving from left to right in a period, since the nuclear charge of the elements also increase in the same direction.

But the ionisation potential of boron (B → 2s² 2p¹) is lower than that of beryllium (Be → 2s²) since in the case of boron, 2p¹ electron has to be removed to get B⁺ while in the case of Be (2s²), s-electron has to be removed to get Be⁺ (2s¹). p-electron can be removed more easily than s-electron so the energy required to remove electron will be less in case of boron.

The order will be Li < B < Be < C.

Question 33. The correct order of 1st ionisation potential following elements Be, B, C, N, O is

1. B<Be<C<O<N
2. B<Be<C<N<O
3. Be<B<C<N<O
4. Be<B<C<O<N

Answer: 1. B<Be<C<O<N

The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionisation energy. The ionisation potential increases as the size of the atom decreases. Atoms with fully or partly filled orbitals have high ionisation potential.

Question 34. Which of the following elements has the maximum electron affinity?

1. I
2. Br
3. Cl
4. F

Answer: 3. Cl

Among the halogens the electron affinity value of ‘F’ should be maximum. But due to small size, there is interelectronic repulsion thus, there is difficulty in the entry of new electrons. Thus, the E.A. value is slightly lower than chlorine andthe order is I < Br < F < Cl.

Question 35. The first ionization potentials (eV) of Be and B respectively are

1. 8.29,8.29
2. 9.32,9.32
3. 8.29,9.32
4. 9.32,8.29

Answer: 4. 9.32,8.29

⇒ \({ }_4 \mathrm{Be} \rightarrow 1 s^2 2 s^2,{ }_5 \mathrm{~B} \rightarrow 1 s^2 2 s^2 2 p^1\)

Due to the stable fully-fiIled ‘s’-orbital arrangement of electrons in the ‘Be’ atom, more energy is required to remove an electron from the valence shell than the ‘ B’-atom. Therefore ‘Be’ has a higher ionisation potential than ‘B’.

Question 36. Which one of the following is the correct order of the size of iodine species?

1. \(I^{+}>I^{-}>\mathrm{I}\)
2. \(I^{-}>I^{\text {I }^{+}}\)
3. \(I>I^{-}>I^{+}\)
4. \(I^{\text {I }^{+}}>I^{-}\)

Answer: 2. \(I^{-}>I^{\text {I }^{+}}\)

Positive ion is always smaller and negative ion is always larger than the parent atom.

Question 37. Which of the following ions is the largest in size?

1. K⁺
2. Ca²⁺
3. Cl^–
4. S^2-

Answer: 4. S^2-

Since all of these ions contain 18 electrons each, these are isoelectronic. For isoelectronic ions, the anion having a large negative charge is the largest in size i.e., S^2-.

Question 38. Which of the following has the smallest size?

1. \(\mathrm{Al}^{3+}\)
2. \(\mathrm{F}\)
3. \(\mathrm{Na}^{+}\)
4. \(\mathrm{Mg}^{2+}\)

Answer: 1. \(\mathrm{Al}^{3+}\)

These are isoelectronic ions (ions with the same number of electrons) and for isoelectronic ions, the greater the positive charge, the greater the force of attraction on the electrons by the nucleus and the smaller the size of the ion. Thus, Al³⁺ has the smallest size.

Question 39. Among the following oxides, the one which is most basic is

1. ZnO
2. MgO
3. Al₂O₃
4. N₂O₅

Answer: 2. MgO

Al₂O₃ and ZnO are amphoteric. N₂O₅ is strongly acidic. MgO is the rnost basic.

Question 40. Which of the following has the largest size?

1. Na
2. Na⁺
3. Na^–
4. Can’t be predicted.

Answer: 3. Na^–

The cations are always smaller than the neutral atom and anions are always larger in size, Na^– > Na > Na⁺

Question 41. \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+} \text { and } \mathrm{Si}^{4+}\) are isoelectronic. The order of their ionic size is

1. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}<\mathrm{Si}^{4+}\)
2. \(\mathrm{Na}^{+}<\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)
3. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)
4. \(\mathrm{Na}^{+}<\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}<\mathrm{Si}^{4+}\)

Answer: 3. \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{Si}^{4+}\)

In isoelectronic ions, the size of the cation decreases as the magnitude of the positive charge increases

Question 42. In the periodic table from left to right in a period, the atomic volume

1. Decreases
2. Increases
3. Remains same
4. First decreases then increases.

Answer: 4. First decreases then increases.

Within a period from left to right, atomic volume first decreases and then increases.

Question 43. Which electronic configuration of an element has an abnormally high difference between second and third ionization energy?

1. \(1 s^2, 2 s^2, 2 p^6, 3 s^1\)
2. \(1 s^2, 2 s^2, 2 p^6, 3 s^1, 3 p^1\)
3. \(1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^2\)
4. \(1 s^2, 2 s^2, 2 p^6, 3 s^2\)

Answer: 4. \(1 s^2, 2 s^2, 2 p^6, 3 s^2\)

An abnormally high difference between 2nd and 3rd ionisation energy means that the element has two valence electrons, which is the case in configuration (4)

Question 44. One of the characteristic properties of non-metals is that they

1. Are reducing agents
2. Form basic oxides
3. Form cations by electron gain
4. Are electronegative.

Answer: 4. Are electronegative.

Question 45. Which one of the following has the minimum value of the cation/anion ratio?

1. NaCl
2. KCl
3. MgCl₂
4. CaF₂

Answer: 3. MgCl₂

The order of ionic size for given ions will be K²⁺ > Ca²⁺ > Mg²⁺ and that of CI^– > F^–. Therefore, MgCl₂ has a minimum value of cation/anion (Mg²⁺/Cl^–) ratio.

Question 46. Which of the following sets has the strongest tendency to form anions?

1. Ga, Ni, Tl
2. Na, Mg, Al
3. N, O, F
4. V, Cr, Mn

Answer: 3. N, O, F

N, O and F are highly electronegative non-metals and will have the strongest tendency to form anions by gaining electrons from metal atoms.

Question 47. Elements of which of the following groups will form anions most readily?

1. Oxygen family
2. Nitrogen family
3. Halogens
4. Alkali metals

Answer: 3. Halogens

As halogens have seven electrons (ns²np⁵) in the valence shell, they have a strong tendency to acquire the nearest inert gas configuration by gaining an electron from the metallic atom and forming halide ions easily.

Question 48. In the periodic table, with the increase in atomic number, the metallic character of an element

1. Decreases in a period and increases in a group
2. Increases in a period and decreases in a group
3. Increases both in a period and the group
4. Decreases in a period and the group.

Answer: 1. Decreases in a period and increases in a group

Metallic character decreases in a period and increases in a group

Question 49. Which of the following atoms will have the smallest size?

1. Mg
2. Na
3. Be
4. Li

Answer: 3. Be

The atomic size decreases within a period from left to right, therefore Li > Be and Na > Mg. The size increases in a group from top to bottom. Hence, the size of Na is greater than Li. Overall order Na > Mg > Li > Be. Thus, Be has the smallest size.

 

Structure Of Atom Multiple Choice Question and Answers

Question 1. Select the correct statements from the following

1. Atoms of all elements are composed of two fundamental particles.
2. The mass of an electron is 9.10939 x 10^-31 kg.
3. All the isotopes of a given element show the same chemical properties.
4. Protons and electrons are collectively known as nucleons.
5. Dalton’s atomic theory regarded the atom as an ultimate particle of matter.

Choose the correct answer from the options given below:

1. 1 and 5 only
2. 2, 3 and 5 only
3. 1, 2 and 3 only
4. 3, 4 and 5 only

Answer: 2. 2, 3 and 5 only

Atoms of all elements are composed of three fundamental particles: electrons, Protons, and neutrons.

Protons and neutrons are collectively known as nucleons

Hence, 2, 3, and 5 are correct statements

Question 2. From the following pairs of ions which one is not an iso-electronic pair?

1. \(\mathrm{Fe}^{2-}, \mathrm{Mn}^{2-}\)
2. \(\mathrm{O}^{2-}, \mathrm{F}\)
3. \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}\)
4. \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{3-}\)

Answer: 1. \(\mathrm{Fe}^{2-}, \mathrm{Mn}^{2-}\)

Among the given pairs of ions, Fe²⁺ and Mn²⁺ is not an iso-electronic pair.

Question 3. The number of protons, neutrons and electrons in \({ }_{71}^{175} \mathrm{Lu}_3\) respectively, are:

1. 71, 104 and 71
2. 104, 71 and 71
3. 71, 71 and 104
4. 175,104 and 71

Answer: 1. 71, 104 and 71

⇒ \({ }_{71}^{175} \mathrm{Lu}_3\) Number of protons = Number of electrons =

Atomic number = 71

Number of neutrons = Mass number – Atomic number

= 175 – 71 = 104

Read and Learn More NEET MCQs with Answers

Question 4. Be²⁺ is isoelectronic with which of the following ions?

1. \(\mathrm{H}^{+}\)
2. \(\mathrm{Li}^{+}\)
3. \(\mathrm{Na}^{+}\)
4. \(\mathrm{Mg}^{2+}\)

Answer: 2. \(\mathrm{Li}^{+}\)

⇒ \(\begin{array}{cc}
\text { Species } & \text { No. of electrons } \\
\mathrm{Be}^{2+} & 2 \\
\mathrm{H}^{+} & 0 \\
\mathrm{Li}^{+} & 2 \\
\mathrm{Na}^{+} & 10 \\
\mathrm{Mg}^{2+} & 10
\end{array}\)

 

Question 5. Isoelectronic species are

1. \(\mathrm{CO}, \mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{C}_2^{2-}\)
2. \(\mathrm{CO}^{-}, \mathrm{CN}^{-} \mathrm{NO}, \mathrm{C}_2^{-}\)
3. \(\mathrm{CO}^{+}, \mathrm{CN}^{+}, \mathrm{NO}^{-}, \mathrm{C}_2\)
4. \(\mathrm{CO}, \mathrm{CN}, \mathrm{NO}, \mathrm{C}_2\)

Answer: 1. \(\mathrm{CO}, \mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{C}_2^{2-}\)

Species having the same number of electrons are called isoelectronic species.

The no. of electrons in \(\mathrm{CO}=\mathrm{CN}^{-}=\mathrm{NO}^{+}=\mathrm{C}_2^{2-}=14\). So, these are isoelectronic species

Question 6. The ion that is isoelectronic with CO is

1. \(\mathrm{CN}^{-}\)
2. \(\mathrm{N}_2^{+}\)
3. \(\mathrm{O}^{2-}\)
4. \(\mathrm{N}_2^{-}\)

Answer: 1. \(\mathrm{CN}^{-}\)

Since both CO and CN^– have 14 electrons, therefore these are isoelectronic species (i.e. having the same number of electrons)

Question 7. Which one of the following is not isoelectronic with O^2-?

1. \(\mathrm{Tl}^{+}\)
2. \(\mathrm{Na}^{+}\)
3. \(\mathrm{N}^{3-}\)
4. \(\mathrm{F}^{-}\)

Answer: 1. \(\mathrm{Tl}^{+}\)

The number of electrons in O^2- N⁺₂, F^-, and Nan is 10 each, but a number of electrons in Tl⁺ is 80.

Question 8. A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1, 368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is [speed of light. c = 3.0 x 10⁸ ms^-1]

1. 21.92 cm
2. 219.3 m
3. 219.2 m
4. 2192 m

Answer: 2. 219.3 m

v = \(\frac{c}{\lambda} ; \lambda=\frac{c}{v}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{1368 \times 10^3 \mathrm{~s}^{-1}}=219.3 \mathrm{~m}\)

Question 9. Which of the following series of transitions in the spectrum of hydrogen atom tails in the visible region?

1. Brackett series
2. Lyman series
3. Balmer series
4. Paschen series

Answer: 3. Balmer series

Lyman series: UV region

Balmer series: Visible region

Paschen series: IR region

Brackett series: IR region

Question 10. Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck’s constant, h = 6.63 x 10^-34J s, speed of light, c = 3 x 10⁸ m s^-1)

Question 11. The value of Planck’s constant is 6.63 x 10^-34. The speed of light is 3 x 10¹⁷ nm s^-1. Which value is closest to the wavelength in the nanometer of a quantum of light with a frequency of 6 x 10¹⁵ s^-1?

1. 50
2. 75
3. 10
4. 25

Answer: 1. 50

c = vλ

λ = \(\frac{c}{v}=\frac{3 \times 10^{17}}{6 \times 10^{15}}=50 \mathrm{~nm}\)

Question 12. According to the law of photochemical equivalence, the energy absorbed (in ergs/mole) is given as (h = 6.62 x 10^-27 ergs, c = 3 x 10¹⁰ cm s^-1 N_A = 6.02 x 10²³ mol^-1)

1. \(\frac{1.196 \times 10^8}{\lambda}\)
2. \(\frac{2.859 \times 10^{16}}{\lambda}\)
3. \(\frac{2.859 \times 10^5}{\lambda}\)
4. \(\frac{1.196 \times 10^{16}}{\lambda}\)

Answer: 1. \(\frac{1.196 \times 10^8}{\lambda}\)

We know that, E = \(\frac{h c N_A}{\lambda}\)

= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda}\)

= \(\frac{1.1955 \times 10^8}{\lambda}=\frac{1.196 \times 10^8}{\lambda} \mathrm{ergs} \mathrm{mol} \mathrm{m}^{-1}\)

Question 13. The energies E₁ and E₂ of the two radiations are 25 eY and 50 eV respectively. The relation between their wavelengths i.e., λ₁ and λ₂ will be

1. \(\lambda_1=\lambda_2\)
2. \(\lambda_1=2 \lambda_2\)
3. \(\lambda_1=4 \lambda_2\)
4. \(\lambda_1=\frac{1}{2} \lambda_2\)

Answer: 2. \(\lambda_1=2 \lambda_2\)

∴ \(E_1=\frac{h c}{\lambda_1}\) and \(E_2=\frac{h c}{\lambda_2} ; \frac{E_1}{E_2}=\frac{h c}{\lambda_1} \times \frac{\lambda_2}{h c}=\frac{\lambda_2}{\lambda_1}\)

or \(\frac{25}{50}=\frac{\lambda_2}{\lambda_1}\) or \(\frac{1}{2}=\frac{\lambda_2}{\lambda_1} \Rightarrow \lambda_1=2 \lambda_2\)

Question 14. The value of Planck’s constant is 6.63 x 10_^-34 s. The velocity of light is 3.0 x 10⁸ m s^-1. Which value is closest to the wavelength in nanometers of a quantum of light with a frequency of 8 x 10¹⁵ s^-1?

1. \(2 \times 10^{-25}\)
2. \(4 \times 10^1\)
3. \(5 \times 10^{-18}\)
4. \(3 \times 10^7\)

Answer: 3. \(5 \times 10^{-18}\)

Applying v = c/λ

λ = \(\frac{c}{v}=\frac{3 \times 10^8}{8 \times 10^{15}}=37.5 \times 10^{-9} \mathrm{~m}=37.5 \mathrm{~nm}=4 \times 10^1 \mathrm{~nm}\)

Question 15. For given energy, E = 3.03 x 10^-19 joules corresponding wavelength is (h = 6.626 x 10^-34 J sec, c = 3 x 10⁸ m/sec)

1. 65.6 nm
2. 6.56 nm
3. 3.4 nm
4. 656 nm

Answer: 4. 656 nm

E = \(\frac{h c}{\lambda} \Rightarrow \lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.03 \times 10^{-19}}=656 \mathrm{~nm}\)

Question 16. What will be the longest wavelength line in the Balmer series of spectra?

1. 546 nm
2. 656 nm
3. 566 nm
4. 556 nm

Answer: 2. 656 nm

The longest wavelength means the lowest energy.

We know that relation for wavelength \(\frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

(\(R_{\mathrm{H}}\), Rydberg constant = \(109677 \mathrm{~cm}^{-1}\))

For \(n_1=2, n_2=3\)

∴ \(\frac{1}{\lambda}=109677\left(\frac{1}{(2)^2}-\frac{1}{(3)^2}\right)=15233\)

or, \(\lambda=\frac{1}{15233}=6.56 \times 10^{-5} \mathrm{~cm}\)

= \(6.56 \times 10^{-7} \mathrm{~m}\)

= \(656 \mathrm{~nm}\)

Question 17. If the radius of the second Bohr orbit of the He⁺ ion is 105.8 pm, what is the radius of the third Bohr orbit of the Li²⁺ ion?

1. 158.7 pm
2. 1587 pm
3. 1.587 pm
4. 158.7 Å

Answer: 1. 158.7 pm

Radius \(=r_0 \times \frac{n^2}{Z}\)

For \(\mathrm{He}^{+}, n=2 ; Z=2\)

∴ \(\mathrm{r}_{\mathrm{He}^{+}}=r_0 \times \frac{2 \times 2}{2} ; 105.8=r_0 \times 2\)

∴ \(r_0=\frac{105.8}{2} \mathrm{pm}\)

For \(\mathrm{Li}^{2+} ; n=3 ; Z=3\)

∴ \(r_{\mathrm{Li}^{2+}}=r_0 \times \frac{(3)^2}{3}=\frac{105.8}{2} \times 3=158.7 \mathrm{pm}\) or 1.587Å

Question 18. Based on equation E = \(-2.178 \times 10^{-18} \mathrm{~J}\left(\frac{Z^2}{n^2}\right),\) certain conclusions are written. Which of them is not correct?

1. The equation can be used to calculate the change in energy when the electron changes orbit.
2. For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
3. The negative sign in the equation simply means that the energy of the electron bound to the nucleus is lower than it would be if the electrons were at an infinite distance from the nucleus.
4. The larger the value of n, the larger is the orbit radius.

Answer: 2. For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.

The electron is more tightly bound in the smallest allowed orbit.

Question 19. According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

1. n = 6 to n = 1
2. n = 5 to n = 4
3. n = 6 to n = 5
4. n = 5 to n = 3

Answer: 3. n = 6 to n = 5

We know that \(\Delta E \propto\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\) where \(n_2>n_1\)

∴ n = 6 to n = 5 will give the least energetic photon.

Question 20. The energy of the second Bohr orbit of the hydrogen atom is -328 kJ mol^-1; hence the energy of the fourth Bohr orbit would be

1. \(41 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(-164 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer: 2. \(-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(E_n=-K\left(\frac{Z}{n}\right)^2\)

Z=1 for hydrogen; n=2

∴ \(E_2=\frac{-K \times 1}{4} \Rightarrow E_2=-328 \mathrm{~kJ} \mathrm{~mol}^{-1} ; K=4 \times 328\)

∴ \(E_4=\frac{-K \times 1}{16} \Rightarrow E_4=-4 \times 328 \times \frac{1}{16}=-82 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 21. The frequency of radiation emitted when the electron falls from n = 4ton=lina hydrogen atom will be (Given ionization energy of H = 2.18 x 10^-18 J atom^-1 and h = 6.626 x 10^-34 J s)

1. \(1.54 \times 10^{15} \mathrm{~s}^{-1}\)
2. \(1.03 \times 10^{15} \mathrm{~s}^{-1}\)
3. \(3.08 \times 10^{15} \mathrm{~s}^{-1}\)
4. \(2.00 \times 10^{15} \mathrm{~s}^{-1}\)

Answer: 3. \(3.08 \times 10^{15} \mathrm{~s}^{-1}\)

E = hv or v = E/h

For \(\mathrm{H}\) atom, \(E=\frac{-21.8 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{atom}^{-1}\)

ΔE= \(-21.8 \times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)=20.44 \times 10^{-19} \mathrm{Jatom}^{-1}\)

v = \(\frac{20.44 \times 10^{-19}}{6.626 \times 10^{-34}}=3.08 \times 10^{15} \mathrm{~s}^{-1}\)

Question 22. In a hydrogen atom, the energy of the first excited state is -3.4 eV. Then find out K.E. of the same orbit of the hydrogen atom.

1. +3.4 eV
2. +6.8 eV
3. -13.6 eV
4. +13.6 eV

Answer: 1. +3.4 eV

Kinetic energy = \(\frac{1}{2} m v^2\) = \(\left(\frac{\pi e^2}{n h}\right)^2 \times 2 m\)

(because \(v=\frac{2 \pi e^2}{\pi h}\))

Total energy, \(E_n=-\frac{2 \pi^2 m e^4}{n^2 h^2}=-\left(\frac{\pi e^2}{n h}\right)^2 \times 2 m=-K . E\).

∴ Kinetic energy = -E_n

Energy’s first excited state is -3.4 eV.

∴ The kinetic energy of the same orbit (n = 2) will be +3.4 eV

Question 23. Who modified Bohr’s theory by introducing elliptical orbits for electron path?

1. Rutherford
2. Thomson
3. Hund
4. Sommerfeld

Answer: 4. Sommerfeld

Sommerfeld modified Bohr’s theory considering that in addition to circular orbits electrons also move in elliptical orbits.

Question 24. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)

1. 4.77
2. 1.06
3. 0.13
4. 2.12

Answer: 4. 2.12

For n^th orbit of ‘H’ atom, r_n = n² x r₁

⇒ the radius of 2nd Bohr’s orbit.

r₂ = 4 x r₁ = 4 x 0.53 = 2.12Å

Question 25. In Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed?

1. \(2.389 \times 10^{-12} \text { ergs }\)
2. \(0.239 \times 10^{-10} \text { ergs }\)
3. \(2.15 \times 10^{-11} \mathrm{ergs}\)
4. \(0.1936 \times 10^{-10} \mathrm{ergs}\)

Answer: 4. \(0.1936 \times 10^{-10} \mathrm{ergs}\)

The energy of an atom when n = 1

∴ \(E_1=-\frac{1312}{(1)^2}=-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Similarly energy when n = 3, \((E_3)=-\frac{1312}{(3)^2}=-145.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The energy absorbed when an electron jumps from n = 1 to n = 3, \(E_3-E_1=-145.7-(-1312)=1166.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

= \(\frac{1166.3}{6.023 \times 10^{23}}=193.6 \times 10^{-23} \mathrm{~kJ}\)

= \(193.6 \times 10^{-20} \mathrm{~J} \quad\left[1 \text { joule }=10^7 \text { ergs }\right]\)

∴ \(193.6 \times 10^{-13} \text { ergs }=0.1936 \times 10^{-10} \text { ergs }\)

Question 26. The radius of a hydrogen atom in the ground state is 0.53 Å. The radius of Li²⁺ ion (atomic number = 3) in a similar state is

1. 0.53 Å
2. 1.06 Å
3. 0.17 Å
4. 0.265 Å

Answer: 3. 0.17 Å

Due to the ground state, the state of hydrogen atom (n) = 1;

The radius of hydrogen atom (r) = 0.53 Å

The atomic number of Li (Z) = 3

Now, radius of \(\mathrm{Li}^{2+} \text { ion }=r \times \frac{n^2}{Z}=0.53 \times \frac{(1)^2}{3}\) =0.17Å

Question 27. The energy of an electron in the n^th Bohr orbit of a hydrogen atom is

1. \(\frac{13.6}{n^4} \mathrm{eV}\)
2. \(\frac{13.6}{n^3} \mathrm{eV}\)
3. \(\frac{13.6}{n^2} \mathrm{eV}\)
4. \(\frac{13.6}{n} \mathrm{eV}\)

Answer: 3. \(\frac{13.6}{n^2} \mathrm{eV}\)

Energy of an electron in n^th Bohr orbit of hydrogen atom = \(\frac{-13.6}{n^2} \mathrm{eV}\)

Question 28. The spectrum of He is expected to be similar to that

1. H
2. Li⁺
3. Na
4. He⁺

Answer: 2. Li⁺

Both He⁺ and Li⁺ contain 2 electrons each.

Question 29. If r is the radius of the first orbit, the radius of the n^th orbit of H-atom is given by

1. rn²
2. rn
3. r/n
4. r²n²

Answer: 1. rn²

The radius of n^th orbit of H-atom = r₀n²

where r₀ = radius of the first orbit.

Question 30. In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, a₀ = 52.9 pm]

1. 211.6 pm
2. 211.6 π pm
3. 52.9 π pm
4. 105.8 pm

Answer: 2. 211.6 π pm

Bohr radius, ao = 52.9 pm

n = 2, r_n = n²a₀= (2)2a₀= 4x 52.9 pm = 211.6 pm

The angular momentum of an electron in a given stationary state can be expressed as in the equation,

mvr = \(n \cdot \frac{h}{2 \pi}=2 \times \frac{h}{2 \pi}=\frac{h}{\pi} \Rightarrow m v r \pi=h\)…..(1)

de-Broglie equation, \(\lambda=\frac{h}{m v} ; \lambda m v=h\)…..(2)

From equations (1) and (2), we get λ = πr

Putting the value of r, λ = 211.6 π pm

Question 31. A 0.66 kg ball is moving at a speed of 100 m/s. The associated wavelength will be (h = 6.6 x 10^-34 J s)

1. \(6.6 \times 10^{-32} \mathrm{~m}\)
2. \(6.6 \times 10^{-34} \mathrm{~m}\)
3. \(1.0 \times 10^{-35} \mathrm{~m}\)
4. \(1.0 \times 10^{-32} \mathrm{~m}\)

Answer: 3. \(1.0 \times 10^{-35} \mathrm{~m}\)

According to de-Broglie equation, \(\lambda=\frac{h}{m v}\)

Given, \(h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; m=0.66 \mathrm{~kg} ; v=100 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\lambda=\frac{6.6 \times 10^{-34}}{0.66 \times 100}=1 \times 10^{-35} \mathrm{~m}\)

Question 32. If uncertainty in position and momentum are equal, then uncertainty in velocity is

1. \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)
2. \(\sqrt{\frac{h}{\pi}}\)
3. \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
4. \(\sqrt{\frac{h}{2 \pi}}\)

Answer: 3. \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

From the Heisenberg uncertainty principle, \(\Delta p \cdot \Delta x \geq \frac{h}{4 \pi}\) or \(m \Delta v \times \Delta x \geq \frac{h}{4 \pi}\)

or \((m \Delta v)^2 \geq \frac{h}{4 \pi}\) (because \(\Delta x=\Delta p\)) or \(\Delta v \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 33. The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 x 10^-18 g cm s^-1. The uncertainty in electron velocity is (mass of an electron is 9 x 10^-28 g)

1. \(1 \times 10^5 \mathrm{~cm} \mathrm{~s}^{-1}\)
2. \(1 \times 10^{11} \mathrm{~cm} \mathrm{~s}^{-1}\)
3. \(1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)
4. \(1 \times 10^6 \mathrm{~cm} \mathrm{~s}^{-1}\)

Answer: 3. \(1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)

Uncertainty in momentum (mΔv) = 1 x 10^-18 g cm s^-1

Uncertainty in velocity (Δv) = \(\frac{1 \times 10^{-18}}{9 \times 10^{-28}}=1.1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}\)

Question 34. Given: The mass of an electron is 9.11 x 10^-31 kg, the Planck constant is 6.626 x 10^-34 J s, and the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is

1. \(5.79 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}\)
2. \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)
3. \(5.79 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\)
4. \(5.79 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 2. \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\Delta x \cdot m \Delta v=h / 4 \pi\)

0.1 x \(10^{-10} \times 9.11 \times 10^{-31} \times \Delta v=\frac{6.626 \times 10^{-34}}{4 \times 3.143}\)

∴ \(\Delta v=\frac{6.626 \times 10^{-34}}{0.1 \times 10^{-10} \times 9.11 \times 10^{-31} \times 4 \times 3.143}\)

= \(5.79 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 35. The uncertainty in the momentum of an electron is 1 x 10^-5 kg m/s. The uncertainty in its position will be (h = 6.62 x 10^-34 kg m²/s)

1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
2. \(80 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
3. \(50 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

∴ \(\Delta x \times \Delta p=\frac{h}{4 \pi}\) (Heisenberg uncertainty principle)

⇒ \(\Delta x=\frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 10^{-5}}=5.27 \times 10^{-30} \mathrm{~m}\)

Question 36. The de Broglie wavelength of a particle with a mass 1 g and velocity of 100 m/s is

1. \(6.63 \times 10^{-35} \mathrm{~m}\)
2. \(6.63 \times 10^{-34} \mathrm{~m}\)
3. \(6.63 \times 10^{-33} \mathrm{~m}\)
4. \(6.65 \times 10^{-35} \mathrm{~m}\)

Answer: 3. \(6.63 \times 10^{-33} \mathrm{~m}\)

λ = \(\frac{h}{m \nu}=\frac{6.63 \times 10^{-27} \mathrm{erg} \mathrm{sec}}{1 \mathrm{~g} \times 10^4 \mathrm{~cm} / \mathrm{s}}\)

= \(6.63 \times 10^{-31} \mathrm{~cm}=6.63 \times 10^{-33} \mathrm{~m}\)

Question 37. The position of both, an electron and a helium atom is known within 1.0 nm. Further, the momentum of the electron is known within 5.0 x 10^-26 kg m s^-1. The minimum uncertainty in the measurement of the momentum of the helium atom is

1. \(8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
2. \(80 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
3. \(50 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

Answer: 4. \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)

According to the uncertainty principle, the product of uncertainty in position and uncertainty in momentum is constant for a particle.

i.e., \(\Delta x \times \Delta p=\frac{h}{4 \pi}\)

As, Δx =1.0 nm for both electron and helium atom, so Δp is also the same for both the particles.

Thus, the uncertainty in the momentum of the helium atom is also 5.0 x 10^-26 kg m s^-1.

Question 38. Uncertainty in position of an electron (Mass = 9.1 x 10^-28 g) moving with a velocity of 3 x 10⁴ cm/s accurate upto 0.001% will be (Use h/(4π) in uncertainty expression where h = 6.626 x 10^-27 erg second)

1. 5.76 cm
2. 7.68 cm
3. 1.93 cm
4. 3.84 cm

Answer: 3. 1.93 cm

Mass of an electron (m) = 9.1 x 10^-28 g

Velocity of electron (v) = 3 x 10⁴ cm/s

Accuracy = 0.001% = \(\frac{0.001}{100}\) and

Planck constant (h)= 6.626 x 10^-27 erg-second.

We know that actual velocity of the electron \((\Delta v)=3 \times 10^4 \times \frac{0.001}{100}=0.3 \mathrm{~cm} / \mathrm{s}\)

Therefore, uncertainty in the position of the electron, \((\Delta x)=\frac{h}{4 \pi m \Delta v}=\frac{6.626 \times 10^{-27}}{4 \pi \times\left(9.1 \times 10^{-28}\right) \times 0.3}=1.93 \mathrm{~cm}\)

Question 39. Which of the following statements do not form a part of Bohr’s model of the hydrogen atom?

1. The energy of the electrons in the orbits is quantized.
2. The electron in the orbit nearest the nucleus has the lowest energy.
3. Electrons revolve in different orbits around the nucleus.
4. The position and velocity of the electrons in the orbit cannot be determined simultaneously.

Answer: 4. The position and velocity of the electrons in the orbit cannot be determined simultaneously.

It is Heisenberg’s uncertainty principle and not Bohr’s postulate.

Question 40. The relation between n_m (n_m = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number

1. \(n_m=2 l^2+1\)
2. \(n_m=l+2\)
3. \(l=\frac{n_m-1}{2}\)
4. \(l=2 n_m+1\)

Answer: 3. \(l=\frac{n_m-1}{2}\)

∴ \(n_m=2 l+1 \quad \text { or } \quad l=\frac{n_m-1}{2}\)

Question 41. Identify the incorrect statement from the following.

1. All the five 5d orbitals are different in size when compared to the respective 4d orbitals.
2. The five 4d orbitals have shapes similar to the respective 3d orbitals.
3. In an atom, all five 3d orbitals are equal in energy in a free state.
4. The shapes of d_xy, d_yz, and d_zx orbitals are similar to each other; and \(d_{x^2-y^2} \text { and } d_{z^2}\) are similar to each other

Answer: 4. The shapes of d_xy, d_yz, and d_zx orbitals are similar to each other; and \(d_{x^2-y^2} \text { and } d_{z^2}\) are similar to each other

The shapes of d_xy, d_xz,, d_yz, and \(d_{x^2-y^2}\) are similar to each other, whereas that of d; is different from others. AIl five 3d orbitals are equivalent in energy. The d-orbitals for which n is greater than 3 (4d, 5d, …) also have shapes similar to 3d-orbital but differ in energy and size.

Question 42. 4d, 5p, 5/and 6p orbitals are arranged in the order of decreasing energy. The correct option is

1. 5f> 6p > 4d > 5p
2. 5f> 6p> 5p> 4d
3. 6p > 5f> 5p > 4d
4. 6p > 5f> 4d > 5p

Answer: 2. 5f> 6p> 5p> 4d

The higher the value of (n + 1) for an orbital, the higher its energy. However, if two different types of orbitals have the same value of (n + 1), the orbital with the lower value of n has lower energy. Therefore, the decreasing order of energy of the given orbitals is 5f > 6p > 5p > 4d.

Question 43. Orbital having 3 angular nodes and 3 total nodes is

1. 5p
2. 3d
3. 4f
4. 6d

Answer: 3. 4f

Number of spherical/radial nodes in any orbital =n-l-1

Number of planar/angular nodes in orbital = l = 3

.’. Totalnumberofnodesinanyorbital= n -1=3

∴ n = 4

Thus, the orbital is 4f

Question 44. Which one is a wrong statement?

1. The total orbital angular momentum of an electron in the s-orbital is equal to zero.
2. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
3. The electronic configuration of the N atom is
4. The value of m for d_z² is zero.

Answer: 3. The electronic configuration of the N atom is

According to Hund’s rule of maximum multiplicity, the correct configuration of ‘N’ is

Question 45. Which one is the wrong statement?

1. The uncertainty principle is \(\Delta E \times \Delta t \geq \frac{h}{4 \pi}\)
2. Half-filled and fully-filled orbitals have greater stability due to greater exchange energy, greater symmetry, and a more balanced arrangement.
3. The energy of 2s-orbital is less than the energy of 2p-orbital in the case of hydrogen-like atoms.
4. de-Broglie’s wavelength is given by \(\lambda=\frac{h}{m v}\) where m = mass of the particle, v = group velocity of the particle.

Answer: 3. The energy of 2s-orbital is less than the energy of 2p-orbital in the case of hydrogen-like atoms

In the case of hydrogen-like atoms, energy depends on the principal quantum number only. Hence, 2s-orbital will have energy equal to 2p-orbital.

Question 46. Consider the following sets of quantum numbers:

Which of the following sets of quantum numbers is not possible?

1. (1), (2), (3) and (4)
2. (2), (4), and (5)
3. (1) and (3)
4. (2), (3), and (4)

Answer: 2. (2), (4) and (5)

1. Represents an electron in 3s orbital.
2. This is not possible as the value varies from 0, 1, … (n-1).
3. Represents an electron in 4/orbital.
4. Is not possible as the value of varies from -l… +l.
5. This is not possible as the value of z varies from -l …+l, It can never be greater than l.

Question 46. The following quantum numbers are possible for how many orbitals? n = 3, l = 2, m = +2

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

n = 3, l= 2, m = +2

It symbolizes one of the five d-orbitals (3d).

Question 47. For which of the following sets of four quantum numbers, an electron will have the highest energy?

Answer: 2.

The energy of an electron depends on the value of (n + 1). The subshell are 3d, 4d, 4p and 5s, out of which 4d has highest energy

Question 48. The electronic configuration of calcium atoms can be written as

1. [Ne]4p²
2. [Ar]4s²
3. [Ne]4s²
4. [Kr]4p²

Answer: 2. [Ar]4s²

The atomic number of Ca = 20

∴ Electronic configuration of Ca = [Ar]4s²

Question 49. For azimuthal quantum number l = 3, the maximum number of electrons will be

1. 2
2. 6
3. 0
4. 14

Answer: 4. 14

l = 3 means f subshell

Maximum no. of electrons in f subshell = 14