Question

For a high spin metal ion in the octahedral and tetrahedral fields, the values of the crystal field stabilisation energies are

• A. $-2.4{∆}_0\mathrm{and}-0.6{∆}_{\mathrm{t}}$
• B. $–1.6{∆}_0\mathrm{and}–0.4{\mathrm{\Delta }}_{\mathrm{t}}$
• C. $–0.4{∆}_0\mathrm{and}–0.27{∆}_{\mathrm{t}}$
• D. $–0.4{∆}_0\mathrm{and}–0.6{∆}_{\mathrm{t}}$

Answer 1

The correct option is D

$–0.4{∆}_0\mathrm{and}–0.6{∆}_{\mathrm{t}}$

Explanation for correct option:

D. $–0.4{∆}_0\mathrm{and}–0.6{∆}_{\mathrm{t}}$

For Octahedral Fields:

• $\begin{array}{rcl}{\mathrm{d}}^6\left(\mathrm{Octahedral}\right)& =& \mathrm{high}\mathrm{spin}\mathrm{complex}\\ & =& {{\mathrm{t}}_{2\mathrm{g}}}^4{\mathrm{eg}}^2\end{array}$
• $\begin{array}{c}\mathrm{CFSE}=\left[-\frac{2}{5}\times 4+\frac{3}{5}\times 2\right]{\mathrm{\Delta }}_0\\ =\left[\frac{-8+6}{5}\right]{∆}_0\\ =0.4{∆}_0\end{array}$

For Tetrahedral Fields:

• $\begin{array}{rcl}{\mathrm{d}}^6\left(\mathrm{Tetrahedral}\right)& =& \mathrm{high}\mathrm{spin}\mathrm{complex}\\ & =& {{\mathrm{t}}_{2\mathrm{g}}}^3{\mathrm{eg}}^3\end{array}$
• $\begin{array}{c}\mathrm{CFSE}=\left[-\frac{3}{5}\times 3+\frac{2}{5}\times 3\right]{\mathrm{\Delta }}_{\mathrm{t}}\\ =\begin{array}{l}-0.6{\mathrm{\Delta }}_{\mathrm{t}}\end{array}\end{array}$

Final Answer: Hence, the correct option is D$–0.4{∆}_0\mathrm{and}–0.6{∆}_{\mathrm{t}}$.