Question

Total number of unpaired electrons present in the central metal atoms/ions of $[\text{Ni}(\text{CO})_4]$, $[\text{NiCl}_4]^{2-$, $[\text{PtCl}_2(\text{NH}_3)_2]$, $[\text{Ni}(\text{CN})_4]^{2-}$ and $[\text{Pt}(\text{CN})_4]^{2-}$ is ___.}

Hint:

$d^8$ complexes show 2 unpaired electrons in tetrahedral geometry, and 0 in square planar geometry.

Answer 1

Correct Answer: 2

1. $[\mathrm{Ni(CO)_4}]$
Oxidation state of Ni $=0$
Electronic configuration: $d^{10}$
CO is a strong field ligand $\Rightarrow$ all electrons paired
Unpaired electrons $=0$
2. $[\mathrm{NiCl_4}]^{2-$}
Oxidation state of Ni $=+2$
Electronic configuration: $d^{8}$
$\mathrm{Cl^-}$ is a weak field ligand
Geometry: tetrahedral (high spin)
Unpaired electrons $=2$
3. $[\mathrm{PtCl_2(NH_3)_2}]$
Oxidation state of Pt $=+2$
Electronic configuration: $d^{8}$
Pt is a $3d$-series metal $\Rightarrow$ strong crystal field
Geometry: square planar (low spin)
Unpaired electrons $=0$
4. $[\mathrm{Ni(CN)_4}]^{2-$}
Oxidation state of Ni $=+2$
Electronic configuration: $d^{8}$
$\mathrm{CN^-}$ is a strong field ligand
Geometry: square planar (low spin)
Unpaired electrons $=0$
5. $[\mathrm{Pt(CN)_4}]^{2-$}
Oxidation state of Pt $=+2$
Electronic configuration: $d^{8}$
$3d$-series metal with strong field ligand
Geometry: square planar (low spin)
Unpaired electrons $=0$
\[ \text{Total unpaired electrons} = 0 + 2 + 0 + 0 + 0 = \boxed{2} \]