Question

Ammonia acts as a very good ligand but ammonium ion does not form complexes because:
a) $N{H_3}$ is a gas white $N{H_4}^ + $ in liquid form:
b) $N{H_3}$ undergoes $s{p_3}$ hybridization while $N{H_4}^ + $ undergoes $s{p_3}d$ hybridization.
c) $N{H_4}^ + $ ion does not have any lone pair of electrons.
d) $N{H_4}^ + $ ion has one unpaired electron while $N{H_3}$ has two unpaired electrons.

Answer 1

Hint:Ammonia has formula $N{H_3}$. Nitrogen has five valence electrons, out of which three are used to make bonds with three hydrogen atoms. So, it has a lone pair also, whereas, ammonium ion has formula $N{H_4}^ + $, here one additional hydrogen takes up the lone pair to make it a cation. So, it is electron deficient.

Complete step by step answer: Ligands are generally electron donors, meaning they can donate electrons to electron deficient species. They are attached to the centre metal atom as metals can accept electrons. So, for a species to be a ligand, it should be either negatively charged, neutral or have a lone pair of electrons. Ammonia has one lone pair of electrons, so it can donate it from coordination complexes. Ammonia is converted into ammonium ion which does not have any lone pair to donate and form complexes.
Thus, Ammonia can be a very good ligand but ammonium ion does not form complexes.
Hence, option (C) is the correct explanation of the given statement.


Note:Formation of ammonium ions, $N{H_4}^ + $: When ammonia is treated with hydrogen chloride, the hydrogen nucleus of ${\text{HCl}}$ moves towards $N{H_3}$. During this the ${\text{H - }}$ atom leaves its electron behind to form $C{l^ - }$ anion, and itself becomes ${H^ + }$ ion to accept the lone pair of $N{H_3}$ to form $N{H_4}^ + $ ions.