What happens when

$(a)$ Borax is heated strongly,

$(b)$ Boric acid is added to water,

$(c)$ Aluminium is treated with dilute $NaOH$,

$(d)$ $BF_3$ is reacted with ammonia?

$(a)$ When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

$\mathop {N{a_2}{B_4}{O_7}.10{H_2}O}\limits_{Borax} \xrightarrow{\Delta }\mathop {N{a_2}{B_4}{O_7}}\limits_{Sodium\,metaborate} \xrightarrow{\Delta }2NaB{O_2} + \mathop {{B_2}{O_3}}\limits_{Boric\,anhydride} $

$(b)$ When boric acid is added to water, it accepts electrons from $-OH$ ion.

$B ( OH )_{3}+2 HOH \longrightarrow\left[ B ( OH )_{4}\right]^{-}+ H _{3} O ^{+}$

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

$2 Al _{(s)}+2 NaOH _{(aq)}+6 H _{2} O _{(l)} \longrightarrow 2 Na ^{+}\left[ Al ( OH )_{4}\right]_{(aq)}+3 H _{2( g )}$

$(d)$ $BF_3$ (a Lewis acid) reacts with $NH_ 3$ (a Lewis base) to form an adduct. This results in a complete octet around $B$ in $ BF_3$.

$F _{3} B +: NH _{3} \longrightarrow F _{3} B \leftarrow : NH _{3}$