(a) Explain why :
(i) BCl3 is a Lewis acid.
(ii) Boric acid is a monobasic acid.
(b) Compound ‘A’ of boron reacts with excess NH3 to give a compound ‘B’. Compound ‘B’ on heating gives cyclic compound ‘C’. Compound C is called inorganic benzene.
(i) Identify compounds ‘A’, ‘B’ and ‘C’
(ii) Give the reactions involved in these processes.
(a)
(i) As BCl3 is an electron deficient molecule with B having only 6 electrons and therefore it acts as Lewis acid and can accept a pair of electron to complete its octet. This is done on the basis of Lewis concept which states that any electron deficient or positively charged species acts as Lewis acid.
E.g. : NH3 + BCl3→ H3N : BCl3
(ii) Monobasic acids are those which donate one proton to a base in an acid base reaction. Boric acid in water, accepts OH- ion acting as lewis acid and water releases H+ ions. So instead of donating one proton, it accepts one pair of electron and thus it is a monobasic acid and the reaction is as follows:
B(OH)3 + H2O → H+ + [ B(OH)4]-
(b)
(i) A = B2H6 B = B2H6.2NH3 C=B3N3H6
B2H6 reacts with excess NH3 to give B2H6.2NH3 which is formulated as [BH2(NH3)2]+[BH4]-. This on heating gives B3N3H6 (borazine) which is called inorganic benzene.
(ii) Reactions involved in these processes are-
3B 2H6 +6NH3 → 3[BH2 (NH3 )2 ]+ [BH4 ]- → 2B3N3H6 +12H