Question

What happens when,
a.) borax is heated strongly,
b.) boric acid is added to water,
c.) aluminum is treated with dilute NaOH,
d.) $B{{F}_{3}}$ is reacted with ammonia?

Answer 1

Hint : Borax and boric acid are two different compound formulations of the same element boron. In all actuality, there wouldn’t be any borax without boric acid as it a sodium salt of boric acid.

Complete step by step solution :
a.) When borax ($N{{a}_{2}}{{B}_{4}}{{O}_{7}}.10{{H}_{2}}O\,$) is strongly heated, it goes through many transformations. As a first step, it loses its water molecules and grows in volume and forms Sodium metaborate ($N{{a}_{2}}{{B}_{4}}{{O}_{7}}$). If we continue heating, it becomes anhydride and starts solidifying into a transparent liquid having an appearance similar to glass called borax bead ($NaB{{O}_{2}}$).

$N{{a}_{2}}{{B}_{4}}{{O}_{7}}.10{{H}_{2}}O\,\xrightarrow{heat}\,N{{a}_{2}}{{B}_{4}}{{O}_{7}}\,+\,10{{H}_{2}}O$
$N{{a}_{2}}{{B}_{4}}{{O}_{7}}\,\xrightarrow{heat}\,2NaB{{O}_{2}}+\,{{B}_{2}}{{O}_{3}}$

b.) Boric acid ($B{{(OH)}_{3}}$) when added to water, behaves as a weak monobasic acid by accepting a pair electron from hydroxyl ion $O{{H}^{-}}$ of water due to the small size of boron and presence of six electrons in its valence shell to form a hydrated species and releases a proton ${{H}^{+}}$in the solution. Hence, acting as a Lewis acid.
$B{{(OH)}_{3}}\,+\,2{{H}_{2}}O\,\to \,{{[B{{(OH)}_{4}}]}^{-}}\,+\,{{H}_{3}}{{O}^{+}}\,$

c.) Aluminium acts as an acid when treated with dilute NaOH and forms sodium tetrahydroxoaluminate (III), ($Na[Al{{(OH)}_{4}}]$). Hydrogen gas is liberated in the process.
$2Al\,+\,2NaOH\,+\,6{{H}_{2}}O\,\to \,2N{{a}^{+}}{{[Al{{(OH)}_{4}}]}^{-}}\,+\,3{{H}_{2}}$

d.) The reaction forms an adduct and results in a complete octet around boron. An additional complex (adduct) is formed as ammonia $N{{H}_{3}}$ acts as a Lewis base and $B{{F}_{3}}$ as a Lewis acid. This reaction results in formation of complete octet around B in $B{{F}_{3}}$ by sharing with lone pairs of $N{{H}_{3}}$.
${{F}_{3}}B\,+\,:\,N{{H}_{3}}\,\to \,{{F}_{3}}B\,\leftarrow \,:N{{H}_{3}}$

Note : This sodium aluminate salt ($Na[Al{{(OH)}_{4}}]$) produced is often used in water treatment plants for softening systems and to improve flocculation. It is also used in construction for accelerating the solidification of concrete.