Question

Draw the d-orbital splitting diagrams for the octahedral complex ions of each of the following

1. ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$(low spin and high spin)
2. ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$(high spin)
3. ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$
   
4. ${\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}$
   
5. role="math" localid="1664016939463" ${\mathbf{Co}}^{\mathbf{2}\mathbf{+}}$ (low spin and high spin)

Short answer

Here all the six electrons remain paired in ${\mathrm{t}}_{2\mathrm{g}}$ orbitals for low spin case while 4 electrons are unpaired in${\mathrm{t}}_{2\mathrm{g}}$ and${\mathrm{e}}_{\mathrm{g}}$ orbitals for high spin case.

Step-by-step solution

The electronic configuration of Fe2+is  3d6.

Here all the six electrons remain paired in ${\mathrm{t}}_{2\mathrm{g}}$ orbitals for low spin case while 4 electrons are unpaired in${\mathrm{t}}_{2\mathrm{g}}$ and${\mathrm{e}}_{\mathrm{g}}$ orbitals for high spin case.

The electronic configuration of Fe3+  is 3d5

Here all the 5 electrons remain unpaired in ${\mathrm{t}}_{2\mathrm{g}}$and ${\mathrm{e}}_{\mathrm{g}}$ orbitals for high spin case.

The electronic configuration of Ni2+ is 3d8

Here 2 electrons remain unpaired in ${\mathrm{e}}_{\mathrm{g}}$ orbitals and 6 electrons remain paired in${\mathrm{t}}_{2\mathrm{g}}$ orbitals of the nickel complex.

The electronic configuration of  Zn2+is 3d10

Here all 10 electrons remain paired in${\mathrm{t}}_{2\mathrm{g}}$ and${\mathrm{e}}_{\mathrm{g}}$ orbitals of the zinc complex.

The electronic configuration of  Co2+is 3d7

Here all the six electrons remain paired in ${\mathrm{t}}_{2\mathrm{g}}$ orbitals with 1 unpaired electron in${\mathrm{e}}_{\mathrm{g}}$ orbital for low spin case while 3 electrons are unpaired in ${\mathrm{t}}_{2\mathrm{g}}$and ${\mathrm{e}}_{\mathrm{g}}$ orbitals for high spin case.