Question

Calculate CFSE (in terms of ${{\Delta }_{0}}$) for ${{d}^{5}}$- high spin (octahedral).
(A) $3.2{{\Delta }_{0}}$
(B) $0{{\Delta }_{0}}$
(C) $1{{\Delta }_{0}}$
(D) $2{{\Delta }_{0}}$

Answer 1

Hint: If the field is strong, it will have few unpaired electrons and thus low spin. If the field is weak, it will have more unpaired electrons and thus high spin. The orbitals are filled with respect to the energy of the orbitals.


Complete step by step answer: The Crystal Field stabilization energy is the gain in the energy achieved by preferential filling up of orbitals by electrons. CFSE is a negative value.
When the crystal field splitting energy is greater than the pairing energy, first electrons will be filled in all the lower energy orbitals and only then pair with electrons in these orbitals before going to the higher energy orbitals. Electrons tend to fall in the lowest possible energy state, and when the pairing energy is lower than the crystal field splitting energy, it is more energetically favorable for the electrons to pair up and completely fill up the low energy orbitals when there is no room left at all, and then high energy orbitals are filled.
At ${{d}^{5}}$ configuration, it has 5 unpaired electrons. With 5 electrons in d-orbital, the configuration is ${{t}_{2g}}^{3}{{e}_{g}}^{2}$.
The CFSE will be : $\dfrac{3}{5}\times 2-\dfrac{2}{5}\times 3=0$
Therefore, the CAFE will be $0{{\Delta }_{0}}$.


The correct answer is the B option.


Note: The ligand field theory states that electron-electron repulsion causes the energy splitting between orbitals. The magnitude of CFSE depends on the number and nature of ligands and the geometry of the complex.