Chemistry Classification of Elements and Periodicity in Properties Question and Answers
Question 1. Write the name of the element which is diagonally related to the elementberyllium. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively. State the blocks in the periodic table to which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.
Answer: Aluminium (Al)
11A: 1, 14B: ls22s22p63s23p2,
17C: ls22s22p63s23p5
From the electronic configuration is evident that A is an s -a block element whereas C is a p -a block element.
The compound formed between B and C has the formula BC4 and the nature of the bond is covalent. A compound formed by a combination of A and C is AC (electrovalent).
Question 2. Classify as basic, amphoteric or acidic: BeO, A1203, CaO, Si02 State the modem periodic law.
Answer: BeO, A1203: amphoteric; CaO: basic; Si02: acidic
Question 3. Write the name of the element which is diagonally related to the elementberyllium.
Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.
State the blocks in the periodic table to which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.
Symbol of one transition element — Fe.
Symbol of one inner-transition element — Ce
Ionic radius is highest for O2- for the given case.
Electronegativity is lowest for Na for the given case.
Ionization energy is highest for Ar for the given case.
Question 4. State the group number in the modem periodic table where solid, liquid, and gaseous elements are present at room temperature. Identify solid, and liquid elements. Indicate the given elements as alkali metal, alkaline earth metal, coinage metal, or chalcogen: Li, Ca, S, Cu.
Answer: Group-17 of modem periodic table contains solid (iodine, I2), liquid (bromine, Br2), and gaseous (chlorine, Cl2) elements at the same time
Group 17 of modern periodic table contains solid
(iodine, I2), liquid (bromine, Br2) and gaseous
(chlorine, Cl2) elements at the same time.
Question 5. Among which of the following pairs of elements, the first has lesser ionization energy than the second—
- Na, K
- N, O
- B, Be
- Br, I
Answer: 3. B, Be
Question 6. Name a pair of elements that exhibit a diagonal relationship.
Answer: Li and Mg.
Question 7. Name a chalcogen and an alkaline earth metal.
Answer: Oxygen (O) is an example of chalcogen and magnesium (Mg) is an alkaline earth metal
Question 10. Mention the names of the most electropositive and most
electronegative stable elements in the periodic table. 2
Answer:
Question 11. What is a metalloid? Give one example.
Answer: The most electropositive stable element of the periodic table is cesium (Cs) and the most electronegative stable element of the periodic table is fluorine (F).
Question 11. Which has greater ionization energy and why—S or P?
Answer: Phas greater ionization energy than S. According to Hund’s rule the half-filled electronic configuration is highly stable and the outermost 3p -orbital of the P atom is half-filled.
To produce P+ ions by removing an electron from the 3p orbital ofPrequireshuge energyresulting a very high value of ionization energy.
On the other hand, the S atom has a partially filled 3p -orbital which attains a stable half-filled electronic configuration in the S+ ion.
Therefore, to produce S to S+ comparatively lower energy is required resulting in a lower value of ionization energy.
Question 13. Arrange as directed: 1. Si02, N02, A1203, C102 (Increasing acidic property) 2. Na+, Al3+, F-, Cl- (Increasing ionic radius) Arrange the following ions in the increasing order of their ionic radius. F-, Mg2+, Al3+, O2-
Answer: A1203 < Si02 < NOz < C102
Al3+ < Na+ < F- < Cl-
Question 14. Arrange the following ions in the increasing order of their ionic radius. F-, Mg2+, Al3+, O2-
Answer: Al3+ < Mg2+ < F- < O2-
Question 15. Which ofthe given elements has the lowest electron affinity
- C
- P
- O
- S
Answer: 2. P
Question 16. Arrange the following elements according to their decreasing oxidising ability: I, Br, F, Cl.
Answer: F > Cl > Br >I
Question 17. Arrange as per instruction: Na20, B203, A1203, MgO (increasing order of basicity) Be, 0, Cl, I (increasing order of electro-negativity)
Answer: B203 < A1203 < MgO < Na20
Be <I < Cl < O
Question 19. Two atoms X and Y have electronic configurations [He] 2sz2p3 and [Ne] 3s23p2 respectively. Which period and group of the periodic table do they belong to? The second electron affinity of an element is always endothermic. Explain.
Answer: X —1 Period 2, Group-15. Y —1 Period 3, Group-14.
Question 20. The Electron affinity of Cl is greater than that of. Explain. Indicate the basic/acidic behavior of the hydrides of the following elements: F, C, O, and N Metallic properties of elements in the periodic table.
Answer: HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increase).
Question 21. Metallic property of elements down the group in the periodic table
Answer: HF: acidic, CH4: neutral, H20: neutral, NH3: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increases).
Question 22. Arrange the following in increasing order of ionic radius: Na+, F-, O2-, Al3+, N3-.
Answer: Al3+ < Na+ < F- < O2- < N3
Question 23. Why does nitrogen have a higher ionization enthalpy than that of Oxygen? Arrange the following in increasing order of acidity: N02, A1203, Si02, C102
Answer: Na < A1 < Mg < Si
Question What will be the order of, Mg, A1, and Sn terms of the first ionisation enthalpy?
Answer: Mg2+ < Na+ < F- < O2-
Question 25. Arrange the following ions in order of increasing ionic radii: Na+, F-, O2-, Mg2+.
Answer: Mg2+ < Na+ < F- < O2-
Question 26. Why is the electron-gain enthalpy of oxygen less than
that of sulfur? Arrange the following metal oxides in terms of ascending order of basicity: ZnO, MgO, CaO, CuO.
Answer: Electron-gain enthalpy of O is less than that of S as the 2p -orbital of the outermost shell of O-atom is much smaller in size than the 3p -orbital of the outermost shell of the S-atom.
So, the additional electron-electron repulsive force produced due to the addition of one electron to the 2p -orbital of the oxygen atom from outside is more than the additional electron-electron repulsive force developed by the addition of one electron to the 3p -orbital of the S-atom.
So the electron-gain enthalpy of the O-atom becomes less than that S-atom.
\(\mathrm{CuO}<\mathrm{ZnO}<\mathrm{MgO}<\mathrm{CaO}\)Question 27. Why is the first ionization enthalpy of helium maximum among all the elements? Arrange the given compounds in terms of ascending order of oxidizing property: HC1, HBr, HI, HF
Answer: Configuration of He is Is2, i.e., all electrons of He are presenting Is -orbital. These electrons are attracted very strongly by the nucleus.
Further, there are no inner electronic orbitals to shield these electrons from the pull ofthe nucleus.
So removal of an electron from the Is -orbital requires a large amount of energy. Therefore, the first ionization enthalpy ofHe is maximum.
Hydrogen halides do not show oxidizing properties. However, the reducing power of the hydrogen halides follows the sequence: HF <HCl < HBr < H
Question 28. Determine the position of an element in the long form of the periodic table if its electronic configuration is [18Ar]3d104s2.
Answer: It is a d -block element. Hence period no. = principle quantum no. ofthe outermost shell = 4 and its group no. = total electronin 3d and 4s -orbital = 10 + 2 = 12.
Question 29. Mention the name and the position of two elements, one of which is the most electronegative and the most electropositive periodic table.
Answer: Most electronegative element =F (group 17, period-2)
Most electropositive element = Cs (group 1, period-6)
Question 30. Which of the following two elements has a diagonal relationship? Li, Be, A1, and Si Between 2gCu and igK which one has higher ionization enthalpy and why do Be and Al
Answer:
- Electronic configuration of 19K: ls22s22p63s23p64s1
- Electronic configuration of 2gCu: ls22s22p63s23p63d104s1
The nucleus of the Cu-atom contains 10 more protons than that of the K-atom.
Additional nuclear pull on the outermost electron (4s1) in Cu-atom is not counter-balanced by the shielding effect of ten 3d-electrons because orbitals have poor screening effects.
Thus the effective nuclear charge acting on the 4s-electron of copper is greater than that acting on the 4s-electron of
Question 31. Arrange the following ions in ascending order of radius: Na+, F-, O2-, Mg2+
Answer: Mg2+ < Na+ <F- < 02-
Question 32. Is the electronegativity of Sn2+ and Sn4+ equal or different?
Answer: Elements in a higher oxidation state have higher electronegativity compared to the elements in a lower oxidation state. Thus Sn4+ is more electronegative than Sn2
Question 33. What is the oxidation state of TI in the compound t1i3? Which is the stronger oxidizing agent between CO2 and PbO2 and why?
Answer: +1
PbO2. Due to the inert pair effect, Pb is very unstable in the +4 oxidation state. Thus Pb4+ can be easily reduced to Pb2.
Question 34. Which iw is more stable between BC13 and T1C13 and why? What is the oxidation state of Zn in Zn-Hg Arrange the following elements in the increasing order of their first ionisation enthalpy.IX Be. Na, Mg 56 Arrange the following elements in the decreasing order of their electro-negativity. Si N. F. Cl
Answer: Because of the poor shielding effect by the inner d – and/- electrons, the inert pair effect is maximum for Tl.
Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.
Question 37. The outermost electronic configuration of the atom of an element is 3s-3p3. Mention the position of the element in the long periodic table. Why is electron gain enthalpy of oxygen less than that of sulfur?
Answer: Because of the poor shielding effect by the inner d – and f- electrons, the inert pair effect is maximum for Tl. Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.