Chemistry, asked by BrainlyHelper, 1 year ago

Question 11.20 What happens when

(a) Borax is heated strongly,

(b) Boric acid is added to water,

(c) Aluminium is treated with dilute NaOH,

(d) BF3 is reacted with ammonia?

Class XI The p-Block Elements Page 324

Answers

Answered by abhi178
3
(a) when Borax is heated strongly, a transparent glassy bead which consist of sodium metaborate and boric anhydride is formed .
e.g., Na2B4O7.10H2O---->Na2B4O7 +10H2O
Na2B4O7 ----->{2NaBO2 + B2O3 }- transparent glassy bead.

(b) We know, Boric acid is sparingly soluble in cold water , but fairy soluble in hot water . it acts as a weak monobasic acid. it isn't a protic acid but it acts as Lewis acid by accepting a hydroxide ion of water and releasing a proton into the solution .
H --OH + B(OH)3 ------> [B(OH)4]– + H+

(c) When Aluminum is treated with dilute NaOH, dihydrogen is evolved.
e.g., 2Al + 2NaOH + 6H2O ----->2Na+[Al(OH)4]– + 3H2

(d) BF3 being a Lewis acid accepts a pair of electrons from NH3 to form the corresponding complex .
F3B ( Lewis acid ) + :NH3 (Lewis base) -----> F3B ← NH3 ( complex )
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