JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 21)
If the CFSE of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is $-96.0 \mathrm{~kJ} / \mathrm{mol}$, this complex will absorb maximum at wavelength ___________ $\mathrm{nm}$. (nearest integer)
Assume Planck's constant (h) $=6.4 \times 10^{-34} \mathrm{Js}$, Speed of light $(\mathrm{c})=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and Avogadro's
Constant $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} / \mathrm{mol}$
Assume Planck's constant (h) $=6.4 \times 10^{-34} \mathrm{Js}$, Speed of light $(\mathrm{c})=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and Avogadro's
Constant $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} / \mathrm{mol}$
Answer
480
Explanation
$$
\begin{aligned}
& {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},} \\\\
& \mathrm{CFSE}=-0.4 \Delta_0 =-96.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\
& \therefore \Delta_0=\frac{-96.0}{-0.4} \\\\
& \therefore \Delta_0=240 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\
\end{aligned}
$$
$$ \begin{aligned} \frac{\mathrm{hc}}{\lambda} & =\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}} \\\\ \lambda & =\frac{0.4 \times 6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3 \times 10^8}{96 \times 10^3} \\\\ & =0.48 \times 10^{-6} \mathrm{~m} \\\\ & =480 \times 10^{-9} \mathrm{~m} \\\\ & =480 \mathrm{~nm} \end{aligned} $$
$$ \begin{aligned} \frac{\mathrm{hc}}{\lambda} & =\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}} \\\\ \lambda & =\frac{0.4 \times 6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3 \times 10^8}{96 \times 10^3} \\\\ & =0.48 \times 10^{-6} \mathrm{~m} \\\\ & =480 \times 10^{-9} \mathrm{~m} \\\\ & =480 \mathrm{~nm} \end{aligned} $$
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