Find the number of unpaired electrons present in the d orbital (whose lobes are present along the axis) for the complex $\ce{[Co(SCN)4]^{2-}}$.

Since $\ce{{SCN}^{-}}$ is a weak ligand I did not pair the electrons, but still I got the wrong answer. The correct answer is 0. But how?

  • $\begingroup$ "whose lobes are present along the axis" seems very ambiguous to me. Could you clarify what it means? $\endgroup$ – chipbuster Nov 3 '13 at 9:14
  • $\begingroup$ @ssavec i applied crystal field theory ...... i filled the d orbitals according to hunds rule..... since Cobalt is in +2 oxidation state hence $Co=[Ar]4S^2 3d^7$ hence $Co^{2+}=[Ar]3d^7$ Now i filled up the orbitals. $\endgroup$ – user2619 Nov 3 '13 at 10:32
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    $\begingroup$ @chipbuster i think this refers to the $e_g$ set of orbitals because $e_g$ set contains $d_{x^2-y^2}$ and $d_{z^2}$ whose lobes are directed along the axis as the lobes of $d_{xy}$ $d_{yz}$ and $d_{xz}$ are present between the axis. $\endgroup$ – user2619 Nov 3 '13 at 10:36
  • $\begingroup$ @Hardik which geometry you assumed, having $\ce{M L_4}$ complex? $\endgroup$ – ssavec Nov 3 '13 at 11:43
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    $\begingroup$ @ssavec i got my mistake...... i filled up the orbitals according to octahedral splitting...... instead i had to do it with tetrahedral. Thankyou. $\endgroup$ – user2619 Nov 3 '13 at 12:14

According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.


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