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 ? 

$(a)$ 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.

$\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}+10 \mathrm{H}_{2} \mathrm{O}$ Borax

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

$\mathrm{H}-\mathrm{OH}+\mathrm{B}(\mathrm{OH})_{3} \rightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}^{+}$

$(c)$ $Al$ reacts with dilute $\mathrm{NaOH}$ to form sodium tetrahydroxo aluminate$(III)$. Hydrogen gas is liberated in the process.

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

$(d)$ $\mathrm{BF}_{3}$ (a Lewis acid) reacts with $\mathrm{NH}_{3}$ (a Lewis base) to form an adduct. This results in a complete octet around $\mathrm{B}$ in $\mathrm{BF}_{3}$.

$\underset{\text { Lewis acid }}{\mathrm{F}_{3} \mathrm{~B}+: \mathrm{NH}_{3}} \underset{\text { Lewis base Complex }}{\rightarrow} \mathrm{F}_{3} \mathrm{~B} \leftarrow \mathrm{NH}_{3}$