What should be possible d-orbital energy levels of $ \mathrm{Ni} $ in $ \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} ? $

Question

What should be possible d-orbital energy levels of $ \mathrm{Ni} $ in $ \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} ? $, $ \mathrm{d}_{\mathrm{z}^{2}} < \mathrm{d}_{\mathrm{zx}} < \mathrm{d}_{\mathrm{xy}}=\mathrm{d}_{\mathrm{yz}} < \mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}} $, $ \mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}} < \mathrm{d}_{\mathrm{xy}} < \mathrm{d}_{\mathrm{z}^{2}} < \mathrm{d}_{\mathrm{xy}}=\mathrm{d}_{\mathrm{yz}} $, $ \mathrm{d}_{\mathrm{zx}}=\mathrm{d}_{\mathrm{yz}} < \mathrm{d}_{\mathrm{z}^{2}} < \mathrm{d}_{\mathrm{xy}} < \mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}} $, $ \mathrm{d}_{\mathrm{z}^{2}} < \mathrm{d}_{\mathrm{xy}}=\mathrm{d}_{\mathrm{yz}} < \mathrm{d}_{\mathrm{xy}} < \mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}} $

MARKS App · Accepted Answer

Square planar complex is formed by removal of ligand from z-axis in octahedral complex. So, ligands are present in x-y plane and directed towards axis. Therefore, orbitals which are pointed towards x and y axis encounters very high repulsion therefore, energy level of d x 2 - y 2 is highest and d xy has lower energy because its lobes are at 45 ° from axis. d z 2 has one third portion in xy plane so, its energy is higher than that of d zx and d yz . Energy of d zx = d yz , because they encounter same repulsion. Thus, overall order of energy of d- orbitals are d zx = d yz d z 2 d xy d x 2 - y 2

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