# Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? ( relative orbital energies not on scale).Option 1)Option 2)Option 3)Option 4)

Valence bond theory (VBT) -

1. Ligands should have atleast one donor atom with a lone pair of elements.
2. Transition metal atom provide required number of vacant orbitals.
3. These vacant orbitals undergo hybridiation.
4. These hybridised orbitals overlap with filled orbitals of ligands to form coordinaate compound.

-

Hybridisation -

sp3d2 - square bipyramidal or octahedral

d2sp3 - octahedral

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - $[Ni(Cl)_{4}]^{2-}$

dsp2 - $[Pt(CN)_{4}]^{2-}$

Crystal field splitting in octahedral complex -

$e_{g}$ orbitals are more energetic as compared to $t_{2g}$

- wherein

The correct increasing order:

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

Now, we are removing the ligands that are pretend along 2-direction & all the orbitals containg two component will have a gradual decrease in energy.

Therefore, Orbital splitting will be done as follows

$\therefore$ option (2) is correct.

Option 1)

Option 2)

Option 3)

Option 4)

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