In a square planar complex of Co^{2+} ion,the | vedclass

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(C) In a square planar complex,the $d$-orbitals split into four energy levels: $d_{x^2-y^2}$ (highest energy),$d_{xy}$,$d_{z^2}$,and the degenerate pa

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$d_{xy}$

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(C) In a square planar complex,the $d$-orbitals split into four energy levels: $d_{x^2-y^2}$ (highest energy),$d_{xy}$,$d_{z^2}$,and the degenerate pair $d_{zx}$ and $d_{yz}$ (lowest energy).For a $Co^{2+}$ ion ($d^7$ configuration) in a square planar field,the electrons fill these orbitals starting from the lowest energy level.The filling order is: $(d_{zx}, d_{yz})^4, (d_{z^2})^2, (d_{xy})^1, (d_{x^2-y^2})^0$.Thus,the single unpaired electron is present in the $d_{xy}$ orbital.

In a square planar complex of $Co^{2+}$ ion,the unpaired electron is present in which orbital?

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