Question

Which of the following is/are correct relations for crystal field splitting energy of octahedral $\left({\Delta }_o\right)$, square planar $\left({\Delta }_{sp}\right)$ and tetrahedral $\left({\Delta }_t\right)$ complexes?

• A. ${\Delta }_t=\frac{4}{9}{\Delta }_0$
• B. ${\Delta }_{sp}=1.3{\Delta }_0$
• C. ${\Delta }_o=1.3{\Delta }_{sp}$
• D. (a) and (b) above.

Answer 1

The correct option is D (a) and (b) above.

Since splitting in tetrahedral complex is$\frac{2}{3}$rd of octahedral complex, so for one ligand splitting on $Oh=\frac{{\Delta }_0}{6}$, then for one lignd splitting in tetrahedral is $\frac{2}{3}\left(\frac{{\Delta }_0}{6}\right).$ So for 4 ligand, $\Delta t=4\times \frac{2}{3}\times \frac{{\Delta }_0}{6}$$=\frac{4}{9}{\Delta }_0$
$\Delta t=\frac{4}{9}{\Delta }_0$

The square planar arrangement of ligands is derived from octahedral ligand field by removing two trans ligands located along the z-axis. Thus, lower of energy of $d_{x}z$ and $d_{y}z$ due to less ligand interaction with these. As the result the spilliting of d-orbital increases.
${\Delta }_{sp}>1.3{\Delta }_o$

Hence, option (d) is correct.