# Homoleptic octahedral complexes of a metal ion '

Question:

Homoleptic octahedral complexes of a metal ion ' $\mathrm{M}^{3+}$ ' with three monodentate ligands and $\mathrm{L}_{1}, \mathrm{~L}_{2}, \mathrm{~L}_{3}$ absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is :

1. $\mathrm{L}_{2}<\mathrm{L}_{1}<\mathrm{L}_{3}$

2. $\mathrm{L}_{3}<\mathrm{L}_{2}<\mathrm{L}_{1}$

3. $\mathrm{L}_{3}<\mathrm{L}_{1}<\mathrm{L}_{2}$

4. $\mathrm{L}_{1}<\mathrm{L}_{2}<\mathrm{L}_{3}$

Correct Option: , 3

Solution:

Order of $\lambda_{a b s}-L_{3}>L_{1}>L_{2}$

So $\Delta_{\mathrm{O}}$ order will be $\mathrm{L}_{2}>\mathrm{L}_{1}>\mathrm{L}_{3}$ (as $\Delta_{\mathrm{O}} \propto \frac{1}{\lambda_{\mathrm{abs}}}$ )

So order of ligand strength will be $\mathrm{L}_{2}>\mathrm{L}_{1}>\mathrm{L}_{3}$