Question

Calculate the CFSE values for the following system.
${d^5}$ - low spin octahedral:
A.2.4 ${\Delta _0}$
B.-0.4 ${\Delta _0}$
C.-2.0 ${\Delta _0}$
D.0.6 ${\Delta _0}$

Answer 1

In order to solve the given question, we must follow these steps: draw a basic skeleton for a low spin octahedral complex, then fill the corresponding number of electrons in the structure. Calculate the number of ${t_{2g}}$ and ${e_g}$ electrons present and then proceed with the calculations.

Complete step by step answer:
In order to calculate the CFSE value for ${d^5}$ - low spin octahedral, we need to first write the splitting pattern and the electronic configuration for both isotropic and octahedral ligand fields. This can be represented as below:
${e_g}$
$- 0.4{\Delta _0}$
${t_{2g}}$
From this, the CFSE vale of the given low spin octahedral system can be calculated as:
Number of ${t_{2g}}$ electrons = 5
Number of ${e_g}$ electrons = 0
Hence, the CFSE value is:
$\Delta$ for given low – spin octahedral complex = [ (number of electrons in ${t_{2g}}$) (+0.4)] + [(Number of ${e_g}$ electrons) (+0.6)]
$= \left( 0 \right)\left( { + 0.6} \right) + \left( 5 \right)\left( { - 0.4} \right)$
= - 2.0 ${\Delta _0}$
Hence the value of CFSE for ${d^5}$ - low spin octahedral is - 2.0 ${\Delta _0}$

Hence, Option C is the correct option.

Note:
CFSE stands for Crystal field stabilization energy and is basically the total energy of the electronic configuration in the given ligand field without the energy of the electronic configuration in the isotropic field.