Gp 15 - Nitrogen family

Hydrides: - NH3 PH3 AsH3 SbH3 BiH3

Boiling points of Hydrides: PH3 < AsH3 < NH3 < SbH3 < BiH3 

Melting points of Hydrides: PH3 < AsH3 < SbH3 < NH3

Thermal stability: BiH3 < SbH3 < AsH3 < PH3 < NH3 (no effective overlapping in Bi - H) 

Reducing Character: NH3 < PH3 < AsH3 < SbH3 < BiH3 

Acidic Character: NH3 < PH3 < AsH3 < SbH3 < BiH3 

Basic Character: BiH3 < SbH3 < AsH3 < PH3 < NH3 

Anomalous Properties of Nitrogen: -

(i) exceptionally small size

(ii) high electronegativity

(iii) high ionisation enthalpy

(iv) absence of d-orbitals in its valence shell

- due to exceptionally small size and high electronegativity N forms pπ - pπ  multiple bonds

- P forms pπ  - dπ  multiple bonds

Gp 16 - Oxygen family

Hydrides: - H2O H2S H2Se H2Te

Volatility: H2O < H2S < H2Se < H2Te

Thermal stability: H2O < H2S < H2Se < H2Te (no effective overlapping of electron cloud in Te-H)

Acidic strength: H2O < H2S < H2Se < H2Te (no effective overlapping of electron cloud in Te-H) 

Reducing character: H2O < H2S < H2Se < H2Te  (however H2O is excluded from reducing character)

- Hybridisation of S in both SO2 and SO3 is sp2

Gp 17 - Fluorine family

Hydrides: - HF HCl HBr HI

- Melting point: HCl < HBr < HI < HF

- Boiling point: HCl < HBr < HI < HF

- Dipole moment: HI < HBr < HCl < HF

- percentage of ionic character: HI < HBr < HCl < HF

- bond length: HF < HCl < HBr < HI 

- bond strength: HI < HBr < HCl < HF (same for bond dissociation energies)

- thermal stability: HI < HBr < HCl < HF

- acidic strength: HF < HCl < HBr < HI 

- reducing power: HF < HCl < HBr < HI 

MCQs

1. Nitrogen does not form pentahalides due to 

a. small size

b. high ionisation energy

c. no d - orbital

d. high electronegativity

Ans. no d-orbital

2. maximum covalency of sulphur is

a. 2

b. 4

c. 6

d. 8

Ans. c.6

explanation 3s2 3p4 = 2 +4 = 6 e

3. which shows only -ve oxidation states

a. Cl

b. Br

c. I

d. F

Ans. d. F

4. liberate bromine when heated with a solution of KBr

a. Iodine

b. HI

c. F

d. SO2

Ans.c.F 

5. reddish brown colour gas formed when nitric oxide is oxidised by air is

a. N2O5

b. N2O4

c. NO2

d. N2O3

Ans.

6. among trihalides of nitrogen, which one is least basic (strong acid)

a. NF3

b. NCl3

c. NBr3

d. NI3

Ans. a. NF3 

explanation: acid = removal of  H+ = due to paucity of e- = higher oxidation state = less e- density

7. cyclic oxo acid

a. H4P2O7

b. H4P2O6

c. H3P3O9

d. H5P5O15

Ans. c. H3P3O9

8. most acidic

a. PCl3

b. SbCl3

c. BiCl3

d. CCl4

Ans. a. PCl3

explanation: acid = removal of  H+ = due to paucity of e- = higher oxidation state = less e- density

9. arrange in increasing acidity: w - H3PO4   x - H2CO3   y - HCl    z - HI

a. w < y < x < z

b. w < x < y < z

c. x < w < y < z

d. None

Ans. H2CO3 < H3PO4 < HCl < HI 

10. most acidic

a. N2O5

b. P2O5

c. As2O5

d. Sb2O5

Ans.a. a. N2O5

explanation: acid = removal of  H+ = due to paucity of e- = higher oxidation state = less e- density

11. oxidation no. of sulphur in S8 S2F2 and H2S

a. 0 +1 and -2 

b. +2 + 1 and -2

c. 0 +1 and +2

d. -2 +1 and -2

Ans. a. 0 +1 and -2

12. the no. of P-O-P bonds in cyclic metaphosphoric acid is 

a. zero

b. two 

c. three

d. four

Ans. c. three

13. Molecular shapes of SF4 CF4 and XeF4 are

a. the same with 2, 0 and 1 lone pair of electrons 

b. the same with 1, 1 and 1 lone pair of electrons

c. different with 0, 1 and 2 lone pair of electrons

d. different with 1, 0 and 2 lone pair of electrons 

Ans. d. different with 1, 0 and 2 lone pair of electrons.