Question

$K_2\left[Ni\right(CN{)}_4]{\underset{N{H}_3}{\stackrel{Kinliq.}{\to }}}^{\prime }{X}^{\prime }$
Regarding this reaction correct statement is/ are:

• A. ${}^{\prime }{X}^{\prime }$ is $K_4\left[Ni\right(CN{)}_4]$
• B. The oxidation state of $Ni$ changed $+2$ to zero
• C. The structure of ${}^{\prime }{X}^{\prime }$ is tetrahedral
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar complex

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The oxidation state of $Ni$ changed $+2$ to zero
B ${}^{\prime }{X}^{\prime }$ is $K_4\left[Ni\right(CN{)}_4]$
C The structure of ${}^{\prime }{X}^{\prime }$ is tetrahedral
D $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar complex

$K_2\left[Ni\right(CN{)}_4]\underset{N{H}_3}{\overset{\text{K in liq.}}{\to }}\underset{{}^{\prime }{X}^{\prime }}{K_4\left[Ni\right(CN{)}_4]}$
Potassium in liquid ammonia generates solvated electrons.
These electrons reduce $N{i}^{2+}$ to Ni atom. Thus the oxidation state of nickel changes from +2 to zero.
The resulting product 'X' is $K_4\left[Ni\right(CN{)}_4]$. It involves $s{p}^3$ hybridization and has tetrahedral geometry.
$K_2\left[Ni\right(CN{)}_4]$ involves $ds{p}^2$ hybridization and has square planar geometry.