JEE Chemistry 2026 Question with Solution

Given: The complexes are $[Ni(CO)_4]$, $[NiCl_4]^{2-}$, $[PtCl_2(NH_3)_2]$, $[Ni(CN)_4]^{2-}$ and $[Pt(CN)_4]^{2-}$.

Find: The total number of unpaired electrons present in the central metal atoms/ions.

Use the idea that $d^8$ complexes show 2 unpaired electrons in tetrahedral geometry and 0 unpaired electrons in square planar geometry.

For $[Ni(CO)_4]$:

CO is a strong field ligand, so all electrons are paired.

For $[NiCl_4]^{2-}$:

$Cl^-$ is a weak field ligand, so the complex is tetrahedral and high spin.

For $[PtCl_2(NH_3)_2]$:

The complex is square planar and low spin.

For $[Ni(CN)_4]^{2-}$:

$CN^-$ is a strong field ligand, so the complex is square planar and low spin.

For $[Pt(CN)_4]^{2-}$:

The complex is square planar and low spin.

Therefore,

So, the total number of unpaired electrons is $2$.

Given: Five coordination compounds containing central metals Ni or Pt.

Find: The sum of unpaired electrons over all the central metal atoms/ions.

1. Identify the metal oxidation state and corresponding $d$-electron count.
2. Use ligand strength to decide whether the $d^8$ complex is tetrahedral or square planar.
3. Assign unpaired electrons accordingly.

Only $[NiCl_4]^{2-}$ is tetrahedral among the listed $d^8$ complexes, so it contributes $2$ unpaired electrons. All the remaining complexes are either $d^{10}$ or square planar low-spin $d^8$, so they contribute $0$.

Hence the total is $2$.