What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 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.
(b) When boric acid is added to water, it accepts electrons from –OH ion.
$\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}_{3} \mathrm{O}^{+}$
(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.
$2 \mathrm{Al}_{(s)}+2 \mathrm{NaOH}_{(a q)}+6 \mathrm{H}_{2} \mathrm{O}_{(\ell)} \longrightarrow 2 \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]_{(a q)}^{-}+3 \mathrm{H}_{2(g)}$
(d) BF3 (a Lewis acid) reacts with NH3 (a Lewis base) to form an adduct. This results in a complete octet around B in BF3.
$\mathrm{F}_{3} \mathrm{~B}+: \mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B} \leftarrow: \mathrm{NH}_{3}$