Question

What happens when,

(a) borax is heated strongly,
(b) boric acid is added to water,
(c) aluminium is treated with dilute NaOH,
(d) $B{F}_3$ is reacted with ammonia?

Answer 1

(a) When borax is heated strongly, it loses water and swells into the white mass, which on further heating melts to form a transparent glassy solid called borax glass and borax bead.

$N{a}_2{B}_4{O}_7.10H_{2}O\stackrel{heat}{\to }N{a}_2{B}_4{O}_7+10H_{2}O$

$N{a}_2{B}_4{O}_7\stackrel{heat}{\to }\underset{\text{sodium meta borate}}{2NaB{O}_2+B_2{O}_3}$

(b) When boric acid is added to water, it accepts electrons from $–OH$ ion. Boric acid is sparingly soluble in cold water however fairly soluble in hot water.

$B(OH{)}_3+2H_{2}O\to [B(OH{)}_4{]}^{-}+H_3{O}^{+}$

(c) Al reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

$2Al+2NaOH+6H_{2}O\to \underset{Sodiumtetrahydroxoaluminate\left(III\right)}{2N{a}^{+}\left[Al\right(OH{)}_4{]}^{-}}+3H_2$

(d) $B{F}_3$ (a Lewis acid) reacts with $N{H}_3$ (a Lewis base) to form an adduct. This results in a complete octet around B in $B{F}_3$.

$F_{3}B+:N{H}_3\to F_{3}B\leftarrow :N{H}_3$