Question

How many of the following complexes have unpaired electrons $[\text{Ni}(\text{CO})_4]$, $[\text{NiCl}_4]^{2-}$, $[\text{PtCl}_4]^{2-}$, $[\text{Pt}(\text{CN})_4]^{2-}$, $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For $4d$ and $5d$ transition metals, the crystal field splitting energy ($\Delta$) is very large, leading predominantly to low-spin complexes regardless of the ligand field strength, especially for $d^8$ configuration in square planar geometry.

Answer 1

The Correct Option is A

1. $[\text{Ni}(\text{CO})_4]$: $\text{Ni}(0)$, $d^{10}$. Tetrahedral $sp^3$. All electrons paired. $n=0$.
2. $[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+}$, $d^8$. $\text{Cl}^-$ is weak field. Tetrahedral $sp^3$. High spin $t_2^4 e^4$. Two unpaired electrons. $n=2$.
3. $[\text{PtCl}_4]^{2-}$: $\text{Pt}^{2+}$, $5d^8$. $\text{Pt}$ ($5d$ series) forms square planar $dsp^2$ complexes which are always low spin. All electrons paired. $n=0$.
4. $[\text{Pt}(\text{CN})_4]^{2-}$: $\text{Pt}^{2+}$, $5d^8$. $\text{CN}^-$ is strong field. Square planar $dsp^2$. All electrons paired. $n=0$.
5. $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$: $\text{Pt}^{2+}$, $5d^8$. Square planar $dsp^2$. All electrons paired. $n=0$.
Only $[\text{NiCl}_4]^{2-}$ has unpaired electrons.
The number of complexes with unpaired electrons is 1.