Question

In an octahedral complex, CFSE (in ${\Delta }_{0}$ units) will be zero in a strong ligand field for a:
A.${{{d}}^{{5}}}$ case
B.${{{d}}^{{6}}}$ case
C.${{{d}}^{{3}}}$ case
D.${{{d}}^{{{10}}}}$ case

Answer 1

Under normal circumstances all the five d-orbitals are equivalent in energy. But in an octahedral complex when ligands come close to the vicinity of the metal centre these degenerate sets of five d-orbitals split.
Two orbitals gets raised in energy from the equivalent point and known as ${{{e}}_{{g}}}$ energy levels, they gain energy by a magnitude of 0.6 ${{{\Delta }}_{{0}}}$.
While three orbitals lose some energy, called ${{{t}}_{{{2g}}}}$ energy levels. They fall in energy by a magnitude of 0.4 ${{{\Delta }}_{{0}}}$.
Now the value of ${{{\Delta }}_{{0}}}$, whether it will be comparatively less or more will depend on the nature of the ligand.
If the ligand is a strong field ligand the separation between the two sets of energy level will be more and hence it will be very difficult for electrons to cross this energy barrier and due to reason, in the presence of a strong field ligand all the ${{{t}}_{{{2g}}}}$ orbitals will get doubly occupied each and then the electrons will move to ${{{e}}_{{g}}}$ orbitals.
The other type of ligands are weak field ligands, in their presence the separation between the two sets of orbitals is comparatively less and hence electrons can overcome this energy barrier, and therefore in such a scenario, all the five orbitals will first get singly occupied.
Complete step by step answer:
Since we are provided with a strong field ligand, firstly all the ${{{t}}_{{{2g}}}}$ orbitals will get doubly occupied and then we can start filling ${{{e}}_{{g}}}$ orbitals. Let us look into each case specifically keeping the previous statement in mind:
${{{d}}^{{5}}}$case:
There are 5 electrons in the d-orbital. And there are three ${{{t}}_{{{2g}}}}$ orbitals each having occupancy of 2 electrons and thus all the electrons will be in ${{{t}}_{{{2g}}}}$ energy level. Thus the energy will be = ${{5 \times ( - 0}}{{.4}}{{{\Delta }}_{{0}}}{{) = - 2}}{{.0}}{{{\Delta }}_{{0}}}$
${{{d}}^{{6}}}$ case:
Similar as the above, all electrons will be in ${{{t}}_{{{2g}}}}$ energy level. Thus the energy will be: ${{6 \times ( - 0}}{{.4}}{{{\Delta }}_{{0}}}{{) = - 2}}{{.4}}{{{\Delta }}_{{0}}}$
${{{d}}^{{3}}}$ case:
Energy = ${{3 \times ( - 0}}{{.4}}{{{\Delta }}_{{0}}}{{) = - 1}}{{.2}}{{{\Delta }}_{{0}}}$

${{{d}}^{{{10}}}}$ case:
Here, we have 10 electrons but the capacity of ${{{t}}_{{{2g}}}}$ orbitals is 6, therefore the remaining 4 electrons will now go to ${{{e}}_{{g}}}$ level. Thus, the energy in this case will be:
${{6 \times ( - 0}}{{.4}}{{{\Delta }}_{{0}}}{{) + 4}} \times {{(0}}{{.6) = 0}}{{{\Delta }}_{{0}}}$
Thus in the case of ${{{d}}^{{{10}}}}$, we will have CFSE value equal to zero in terms of ${{{\Delta }}_{{0}}}$ if we have a strong field ligand.

Hence, the correct option is D.

Note: CFSE stands for Crystal Field Stabilisation Energy.
This can be defined as the energy of the electronic configuration in a ligand field minus the energy of the electronic structure in the isotropic field.
An isotropic field is one when there is no separation between the d-orbitals. In such a case the energy is taken to be 0 ${{{\Delta }}_{{0}}}$ by convention.
CFSE is usually measured in terms of ${{{\Delta }}_{{0}}}$.
It is the sum of energy of electrons in both the energy levels (${{{e}}_{{g}}}$ and ${{{t}}_{{{2g}}}}$)