Question

Among $\left[Ni\right(CO{)}_4]$, $\left[Ni\right(CN{)}_4{]}^{-2}$ and $\left[Ni\right(Cl{)}_4{]}^{-2}$ :

• A. $\left[Ni\right(CO{)}_4]$, $\left[Ni\right(Cl{)}_4{]}^{-2}$are diamagnetic $\left[Ni\right(CN{)}_4{]}^{-2}$
  is paramegnetic
• B. $\left[Ni\right(CO{)}_4]$ is parmegnetic $\left[Ni\right(CN{)}_4{]}^{-2}$$\left[Ni\right(Cl{)}_4{]}^{-2}$ are diamegnetic
• C. $\left[Ni\right(CO{)}_4]$, $\left[Ni\right(CN{)}_4{]}^{-2}$are diamegenetic$\left[Ni\right(Cl{)}_4{]}^{-2}$
  is paramegenetic
• D. $\left[Ni\right(CO{)}_4]$, is diamagnetic $\left[Ni\right(CN{)}_4{]}^{-2}$ $\left[Ni\right(Cl{)}_4{]}^{-2}$are paramegenetic

Answer 1

Answer: (C)

$\left[Ni\right(CO{)}_4]$, $\left[Ni\right(CN{)}_4{]}^{-2}$are diamegenetic$\left[Ni\right(Cl{)}_4{]}^{-2}$

is paramegenetic

As CO and $C{N}^{-}$ are strong ligand so they form low spin complex and pairing happens so they are diamagnetic but $C{l}^{-}$ is a weak ligand so it higher spin outer orbital complex and in this complex Ni has 2 unpaired electron so it is paramagnetic.

Hence,option C is correct.