Question

What should be possible $ d $-orbital energy levels of $ Ni $ in $ [Ni(CN)_4]^{2-} $ ?

• A. $ d_{z^{2}} < d_{xy} < d_{xy} = d_{yz} < d_{x^{2}-y^{2}} $
• B. $ d_{x^{2}-y^{2}} < d_{xy} < d_{z^{2}} < d_{xy} = d_{yz} $
• C. $ d_{xy} = d_{yz} < d_{z^{2}} < d_{xy} < d_{x^{2}-y^{2}} $
• D. $ d_{z^{2}} < d_{xy} = d_{yz} < d_{xy} < d_{x^{2}-y^{2}} $

Answer 1

The Correct Option is C

$[Ni(CN)_{4}]^{2-}$ has square planar geometry with $dsp^{2}$ hybridisation
The order of energies of $d$-orbitals in square planar geometry is:
$d_{xy}=d_{yz} < d_{z^{2}} < d_{xy} < d_{x^{2} -y^{2}}$