JEE Main 2023 — Coordination Compounds Question with Solution

In ${\left[\mathrm{Ti}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$, oxidation state of Titanium is +3.
Thus, electronic configuration:

${\mathrm{Ti}}^{+3}:3{\mathrm{d}}^1\Rightarrow {\mathrm{t}}_{2\mathrm{g}}^{1,0,0},{{\mathrm{e}}_{\mathrm{g}}}^{0,0}$

We know CFSE $=\left[-0.4{\mathrm{n}}_{{\mathrm{t}}_{2\mathrm{g}}}+0.6{\mathrm{n}}_{{\mathrm{e}}_{\mathrm{g}}}\right]\mathrm{\Delta o}$       (      $\mathrm{\Delta o} = \mathrm{Octahedral} \mathrm{splitting} \mathrm{energy}$)

$\mathrm{\Delta o}=\frac{96\times {10}^3}{0.4\times 6\times {10}^{23}}=4\times {10}^{-19} \mathrm{J}$

Using planck's equation and fact that octahedral splitting energy per complex is:
$\Rightarrow \mathrm{\Delta o}=\frac{\mathrm{hc}}{\mathrm{\lambda }}=4\times {10}^{-19} \mathrm{J}$

$\Rightarrow \mathrm{\lambda }=\frac{6.4\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-19}}=4.8\times {10}^{-7} \mathrm{m}$

$\Rightarrow \lambda =480\times {10}^{-9} \mathrm{m}$

$\therefore \mathrm{\lambda }=480 \mathrm{nm}$