Question

$K_2\left[Ni\right(CN{)}_4]\stackrel{Kinliq.N{H}_3}{\to }$ "X", regarding this reaction correct statement is/are

• A. “X” is $K_4\left[Ni\right(CN{)}_4]$
• B. The oxidation state of Ni changed from +2 to zero
• C. The structure of “X” 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

“X” is $K_4\left[Ni\right(CN{)}_4]$

B

The oxidation state of Ni changed from +2 to zero

C

The structure of “X” is tetrahedral

D

$\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar complex

$K_2\left[Ni\right(CN{)}_4\left]\stackrel{Kinliq.N{H}_3}{\to }{K}_4\right[Ni\left(CN{)}_4\right]$

Let the oxidation number of $Ni$ be `y`.
Now, in $K_4\left[Ni\right(CN{)}_4]$, we know the oxidation numbers of the other species.
$C{N}^{-}$ as a whole is taken as -1 because it is a pseudohalide.
Potassium always has an oxidation number of +1. Overall, the complex is neutral.
Hence, we get, $1\times 4+y\times 1+4\times (-1)=0$
$\Rightarrow 4+y-4=0$
$\Rightarrow y=0$

Using this oxidation state, combined with the nature of the ligand, we get a tetrahedral structure for $Ni\left(0\right)$.