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 ?

Explanation

(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.
`underset("Borax")underset()(Na_(2)B_(4)O_(7).10H_(2))Ooverset(Delta)rarrunderset("metaborate")underset("Sodium")underset()(Na_(2)B_(4)O_(7))overset(Delta)rarr2NaBO_(2)+underset("anhydride")underset("Boric")underset()(B_(2)O_(3))`
(b) When boric acid is added to water, it accepts electrons from `.^(–)OH` ion.
`B(OH)_(3)+2HOHrarr[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_((s))+2NaOH_((aq))+6H_(2)O_((l))rarr2Na^(+)[Al(OH)_(4)]_((aq))^(-)+3H_(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)rarrF_(3)Blarr:NH_(3)`