JEE Main 2023 — Chemistry Inorganic Chemistry

In ${[\mathrm{Ti}{({H}_{2}O)}_{6}]}^{3+}$[Ti(H2​O)6​]3+, oxidation state of Titanium is +3.

Thus, electronic configuration:

$${\mathrm{Ti}}^{+3}:3{d}^{1}\Rightarrow {t}_{2g}^{1,0,0},{{e}_{g}}^{0,0}$$Ti+3:3d1⇒t2g1,0,0​,eg​0,0

We know CFSE $=[-0.4{n}_{{t}_{2g}}+0.6{n}_{{e}_{g}}]\Delta o$=[−0.4nt2g​​+0.6neg​​]Δo ( $\Delta o=\mathrm{Octahedral}\mathrm{splitting}\mathrm{energy}$Δo=Octahedralsplittingenergy)

$$\Delta o=\frac{96\times {10}^{3}}{0.4\times 6\times {10}^{23}}=4\times {10}^{-19}J$$Δo=0.4×6×102396×103​=4×10−19J

Using planck's equation and fact that octahedral splitting energy per complex is:

$$\Rightarrow \Delta o=\frac{\mathrm{hc}}{\lambda }=4\times {10}^{-19}J$$⇒Δo=λhc​=4×10−19J

$$\Rightarrow \lambda =\frac{6.4\times {10}^{-34}\times 3\times {10}^{8}}{4\times {10}^{-19}}=4.8\times {10}^{-7}m$$⇒λ=4×10−196.4×10−34×3×108​=4.8×10−7m

$$\Rightarrow \lambda =480\times {10}^{-9}m$$⇒λ=480×10−9m

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