Total number of unpaired electrons present in the central metal atoms/ions of
\([Ni(CO)_4]\), \([NiCl_4]^{2-}\), \([PtCl_2(NH_3)_2]\), \([Ni(CN)_4]^{2-}\) and
\([Pt(CN)_4]^{2-}\) is
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$d^8$ complexes show 2 unpaired electrons in tetrahedral geometry, and 0 in square planar geometry.
To determine the number of unpaired electrons in the complexes, we need to evaluate the electron configuration and bonding nature of the central metal in each complex.
For $[\text{Ni}(\text{CO})_4]$: Nickel (Ni) has an atomic number of 28, so its electronic configuration is $[Ar]3d^{8}4s^{2}$. CO is a strong field ligand, causing pairing of electrons. Ni is in the 0 oxidation state: $3d^{10}$. All electrons are paired. Unpaired electrons = 0.
For $[\text{NiCl}_4]^{2-}$: Ni is in the +2 oxidation state, $[Ar]3d^{8}$. Cl⁻ is a weak field ligand, not causing pairing. Electronic configuration remains as $3d^{8}$, with two unpaired electrons. Unpaired electrons = 2.
For $[\text{PtCl}_2(\text{NH}_3)_2]$: Pt is in the +2 oxidation state, $[Xe]4f^{14}5d^{8}$. NH₃ is a stronger field ligand than Cl⁻ but doesn't affect pairing of electrons in $5d^{8}$. All electrons are paired. Unpaired electrons = 0.
For $[\text{Ni}(\text{CN})_4]^{2-}$: Ni is in the +2 oxidation state, $[Ar]3d^{8}$. CN⁻ is a strong field ligand, causing pairing. Configuration becomes $3d^{8}$, all paired. Unpaired electrons = 0.
For $[\text{Pt}(\text{CN})_4]^{2-}$: Pt is in the +2 oxidation state, $[Xe]4f^{14}5d^{8}$. CN⁻ is a strong field ligand, causing pairing. Electrons are paired. Unpaired electrons = 0.
The total number of unpaired electrons is: 0 + 2 + 0 + 0 + 0 = 2. This result falls within the expected range of 2,2.
