Questions - Answers

I. Very important questions          

2 Marks questions.

1. Draw the structure of metaborate ion.
A: 

2. What is inert pair effect?
A: The reluctance of pair of s electrons of valency shell to take part in bonding.
                     eg: Tl⁺¹
 

3. Give the formula of Tincal & Colemanite
A: Tincal: Na₂B₄O₇. 10 H₂O
          Colemanite: Ca₂B₆O₁₁. 5 H₂O
                            

4. Give the formula of borazine. What is its common name?
A: B₃N₃H₆
It is also known as inorganic benzene.

5. Give the composition of Duralumin and Magnalium.
A: Duralumin: 2 -15% Mg & 85 - 98% Al
           Magnalium: 10 -12% Al & 88 - 90% Cu
 

6. Give two uses of borax.
A: 1. In leather industry for cleaning skin & hides
     2. In making pyrex glass & optical glass.
 

7. Why do BF₃ behave as a lewis acid?
A: Electron pair acceptor is lewis acid. BF₃ is electron deficient as it has only  in the valency shell of Boron. It can accept an electron pair to acquire octet configuration.
 

8. What happens when orthoboric acid is strongly heated?
A: 

 H₃BO₃ on strong heating, finally gives B₂O₃.
 

9. Give the reactions of B₂H₆ with
(a) NH₃ at 120°C, 200°C    (b) HCl, AlCl₃

10. Explain why the electronegativities of Ga, In, Tl do not vary much?
A: Ga, In, Tl comes after d block elements. The d-electrons can not shield the valence electrons from the nuclear attractions. So E.N. of Ga, In, Tl do not vary much.
 

11. Why is boric acid polymeric?
A: It is polymeric due to presence of inter molecular Hydrogen bonding.

12. What is the hybridization of B in diborane and borazine?
A: Hybridization of B

→ in diborane: sp³

→ in borazine : sp²
 

4 Marks Questions:

1. What are electron deficient molecules? Explain the structure of diborane.
A: Central atom of molecules having less number of electrons than that of Octet during the formation of bonds. eg. BF₃, B₂H₆.
Structure of Diborane:
                  B = 1S² 2S¹ 2P_x1 2P_y1 2P_z1 (1^st excited state)
 In diborane there are 12 valence electrons (B2H6 = (2 × 3) + 6 × 1 = 12 e^-) available to form only six bonds. When it is compared with C₂H₆ [(2 × 4) + 6 × 1 = 14 e- = 7 bonds] there is a shortage of 2e-to form 7 bonds. So B₂H₆ is called "electron deficient compound". In B₂H₆ both B atoms undergo sp³ hybridization and giving 3 sp³ orbitals with unpaired electrons and one is empty. 4 sp³ hybrid orbitals of 2B atoms overlap with 1s orbitals of H atoms to form 2 BH₂ groups. One sp3 hybrid orbital with one unpaired electron of one B atom and one sp³ hybrid orbital without unpaired electron overlaps with 1s orbital of H atom giving 3 centred 2 electron B- H-B bridge (banana) bond. 2 such non planar banana bonds are formed. 4 terminal hydrogens and 2 boron atoms lie in one plane and 2 bridged hydrogens lie above and below the plane.

2. Explain borax bead test with a suitable example
A: Transition elements can be identified by Borax bead test.
On heating borax it gives white, opaque mass Na₂B₄O₇ which on fusion gives borax glass (B₂O₃ + NaBO₂).
                 B₂O₃ on reaction with basic radical gives coloured metaborate. 
                
 

3. Write any two methods of preparation of diborane. How does it react with
a) Carbon monoxide   b) NH₃
A: Preparation methods:
* It can be prepared by reducing BF₃ by LiH
    2 BF₃ + 6 LiH → B₂H₆ + 6 LiF
* It can be prepared by passing silent electric discharge through the mixture of BCl₃ and H₂.
   2 BCl₃ + 6 H₂ → B₂H₆ + 6 HCl 
Reactions:
a) B₂H₆ reacts with CO at 1000°C and 2 atm pressure to give borane carbonyl.
     B₂H₆ + 2 CO → 2[BH₃ − CO]
b) B₂H₆ on heating with NH₃ upto 200°C gives Borazine.