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 of crystallization and swells. On further heating,it melts to form a transparent liquid,which solidifies into a glass-like material known as borax bead.
$Na_{2}B_{4}O_{7} \cdot 10H_{2}O \xrightarrow{\Delta} Na_{2}B_{4}O_{7} + 10H_{2}O$
$(b)$ Boric acid acts as a weak monobasic Lewis acid in water. It accepts an $OH^{-}$ ion from water to form the tetrahydroxoborate ion.
$B(OH)_{3} + 2H_{2}O \rightleftharpoons [B(OH)_{4}]^{-} + H_{3}O^{+}$
$(c)$ Aluminium reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate$(III)$ and releases hydrogen gas.
$2Al(s) + 2NaOH(aq) + 6H_{2}O(l) \rightarrow 2Na[Al(OH)_{4}](aq) + 3H_{2}(g)$
$(d)$ $BF_{3}$ acts as a Lewis acid and $NH_{3}$ acts as a Lewis base. They react to form an adduct,completing the octet of boron.
$F_{3}B + :NH_{3} \rightarrow F_{3}B \leftarrow NH_{3}$