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?

(N/A) When heated,borax loses water molecules and swells. On further heating,it turns into a transparent liquid,which solidifies to form a glass-like material called borax bead.
$Na_2B_4O_7 \cdot 10H_2O$ $\xrightarrow{\Delta} Na_2B_4O_7$ $\xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
$(b)$ Boric acid acts as a weak monobasic Lewis acid in water. It accepts an $OH^-$ ion from water molecules.
$B(OH)_3 + 2H_2O \longrightarrow [B(OH)_4]^- + H_3O^+$
$(c)$ Aluminium reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate$(III)$ and releases hydrogen gas.
$2Al_{(s)} + 2NaOH_{(aq)} + 6H_2O_{(l)} \longrightarrow 2Na[Al(OH)_4]_{(aq)} + 3H_{2(g)}$
$(d)$ $BF_3$ (a Lewis acid) reacts with $NH_3$ (a Lewis base) to form an adduct,completing the octet of boron.
$F_3B + :NH_3 \longrightarrow F_3B \leftarrow :NH_3$