The correct order of energies of d- orbitals | vedclass

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Question

(D) In a square planar complex,the ligands approach along the $x$ and $y$ axes. This leads to a significant increase in the energy of the $d_{x^2-y^2}

Vedclass · Accepted Answer

$d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{yz} = d_{zx}$

Vedclass · Answer

(D) In a square planar complex,the ligands approach along the $x$ and $y$ axes. This leads to a significant increase in the energy of the $d_{x^2-y^2}$ orbital. The $d_{xy}$ orbital also experiences repulsion,but less than $d_{x^2-y^2}$. The $d_{z^2}$ orbital is lower in energy than $d_{xy}$ because the ligands are absent along the $z$-axis. The $d_{yz}$ and $d_{zx}$ orbitals are the lowest in energy and are degenerate. Thus,the correct order of energy is $d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{yz} = d_{zx}$.

The correct order of energies of $d-$ orbitals of a metal ion in a square planar complex is:

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Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex?
$A$. $d_{xy} = d_{xz} > d_{x^2-y^2}$
$B$. $d_{xy} = d_{yz} > d_{z^2}$
$C$. $d_{x^2-y^2} > d_{z^2} > d_{xz}$
$D$. $d_{x^2-y^2} = d_{z^2}$
Choose the correct answer from the given below:

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