A solution contains Mg^2+ in the presence of a high concentration of NH4Cl. On adding excess aqueous ammonia, no precipitate of Mg(OH)2 appears. The best explanation is

NH4+ suppresses ionisation of NH3 (common ion effect), so [OH-] is too low for the ionic product to reach K_sp of Mg(OH)2

Mg(OH)2 is amphoteric and dissolves in excess ammonia as [Mg(OH)4]^2-

NH4+ suppresses ionisation of NH3 (common ion effect), so [OH-] is too low for the ionic product to reach K_sp of Mg(OH)2

Mg^2+ forms the very stable ammine [Mg(NH3)6]^2+ in water

Mg(OH)2 is a freely soluble strong base

JEE Main levelSingle correct optionDifficulty 2 of 3qc130

A solution contains $Mg X^{2 +}$ in the presence of a high concentration of $NH X_{4} Cl$ . On adding excess aqueous ammonia, no precipitate of $Mg (OH) X_{2}$ appears. The best explanation is

A $Mg (OH) X_{2}$ is amphoteric and dissolves in excess ammonia as $[Mg (OH) X_{4}] X^{2 -}$
B $NH X_{4} X^{+}$ suppresses ionisation of $NH X_{3}$ (common ion effect), so $[OH X^{-}]$ is too low for the ionic product to reach $K_{s p}$ of $Mg (OH) X_{2}$
C $Mg X^{2 +}$ forms the very stable ammine $[Mg (NH X_{3}) X_{6}] X^{2 +}$ in water
D $Mg (OH) X_{2}$ is a freely soluble strong base

NH X_{3} + H X_{2} O NH X_{4} X^{+} + OH X^{-}

is shifted left by

NH X_{4} X^{+}

, which lowers

[OH X^{-}]

. The product

[Mg X^{2 +}] [OH X^{-}]^{2}

then stays below

K_{s p}

Mg (OH) X_{2}

is not amphoteric and

Mg X^{2 +}

does not form stable ammines in water.