Predict the number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion.
(0) In the complex $[Pt(CN)_{4}]^{2-}$,the central metal ion is $Pt^{2+}$.
The atomic number of $Pt$ is $78$. The electronic configuration of $Pt$ is $[Xe] 4f^{14} 5d^{9} 6s^{1}$.
For $Pt^{2+}$,the configuration is $[Xe] 4f^{14} 5d^{8}$.
$CN^{-}$ is a strong field ligand,which forces the pairing of electrons in the $5d$ orbitals to facilitate $dsp^{2}$ hybridization,which is characteristic of square planar geometry.
After pairing,all $8$ electrons in the $5d$ orbitals are paired.
Therefore,the number of unpaired electrons is $0$.
The atomic number of $Pt$ is $78$. The electronic configuration of $Pt$ is $[Xe] 4f^{14} 5d^{9} 6s^{1}$.
For $Pt^{2+}$,the configuration is $[Xe] 4f^{14} 5d^{8}$.
$CN^{-}$ is a strong field ligand,which forces the pairing of electrons in the $5d$ orbitals to facilitate $dsp^{2}$ hybridization,which is characteristic of square planar geometry.
After pairing,all $8$ electrons in the $5d$ orbitals are paired.
Therefore,the number of unpaired electrons is $0$.
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| Coordination number and type of hybridisation | Distribution of hybrid orbitals in space |
|---|---|
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| $(b)$. $3, sp^2$ | $(ii)$. octahedral |
| $(c)$. $5, sp^3 d$ | $(iii)$. tetrahedral |
| $(d)$. $6, sp^3 d^2$ | $(iv)$. trigonal planar |
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$B$. $[MnCl_{6}]^{3-}$: Outer orbital complex; $sp^{3}d^{2}$ hybridized
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