Total number of free ${\mathrm{CN}}^{-}$ ion that are present in ${\mathrm{I}}_3(\mathrm{CN}{)}_3$where $\mathrm{ICN}$ is in liquid state.

There are no free ${\mathrm{CN}}^{-}$ ion that are present in ${\mathrm{I}}_3(\mathrm{CN}{)}_3$.Because ${\mathrm{CN}}_{ }^{+}$is sp3 hybridised and $(\mathrm{CN}{)}_2^{-}$is $s{p}^{3}d$ hybridised

The following reaction is result in aqueous solution

$CN⟶{\mathrm{I}}_2\left({\mathrm{CN}}_{ }^{+}+\mathrm{I}(\mathrm{CN}{)}_2^{-}\right.$