Question

The complex showing a spin only magnetic moment of $2.82$ BM is:

• A. $Ni(CO{)}_4$
• B. $[NiC{l}_4{]}^{2-}$
• C. $Ni(PP{h}_3{)}_4$
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$

Answer 1

The correct option is B $[NiC{l}_4{]}^{2-}$

The spin only magnetic moment is given by $\sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons.
Since here spin only magnetic moment is given as $2.82$, thus the come[lex has two unpaored electrons.
Again, in $Ni(CO{)}_4$ and in $Ni(PP{h}_3{)}_4$, the oxidation state of $Ni$ is $0$. So spin only magnetic moment cannot be $2.82$ in these cases.
In both $\left[Ni\right(CN{)}_4{]}^{2-}$ and $[NiC{l}_4{]}^{2-}$, $Ni$ is in $+2$ oxidation state. But since $C{N}^{-}$ is strong fied ligand thus its forms a square planner complex and all the electrons are paired here. Thus it is diamagnetic.
$[NiC{l}_4{]}^{2-}$ forms a tetrahedral complex and has an electronic configuration $e_g^4{t}_{2g}^4$. Thus it has two unpaired electron.
Hence the correct option is $[NiC{l}_4{]}^{2-}$