JEE MAIN - Chemistry (2022 - 25th June Morning Shift - No. 22)
If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be ____________ $$\times$$ 10$$-$$21 J. [Nearest Integer]
(Given : h = 6.63 $$\times$$ 10$$-$$34 Js and c = 3.08 $$\times$$ 108 ms$$-$$1)
Answer
766
Explanation
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}$ is tetrahedral
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is octahedral
$$ \because \Delta_{\mathrm{t}}=\frac{4}{9} \times \Delta_{0} $$
$$ \begin{aligned} &\Delta_{\mathrm{t}}=\frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \\\\ &\Delta_{0}=\frac{9}{4} \times \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \approx 765.7 \times 10^{-21} \mathrm{~J} \end{aligned} $$
$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is octahedral
$$ \because \Delta_{\mathrm{t}}=\frac{4}{9} \times \Delta_{0} $$
$$ \begin{aligned} &\Delta_{\mathrm{t}}=\frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \\\\ &\Delta_{0}=\frac{9}{4} \times \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}} \approx 765.7 \times 10^{-21} \mathrm{~J} \end{aligned} $$
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