If CuH2O42+ absorbs a light of wavelength 600 n m for d-d transition, then the value of octahedral crystal field splitting energy for CuH2O62+ will be(Given : h=6.63×10-34JS and c=3.08×108 ms-1 )

Solution:

$\begin{aligned} Δ_{t} = \frac{hc}{λ} = \frac{6 .63 \times 10^{- 34} \times 3 .08 \times 10^{8}}{600 \times 10^{- 9}} \\ = \frac{6 .63 \times 3 .08 \times 10^{- 17}}{600} \\ = 0 .034034 \times 10^{- 17} \\ = 340 .34 \times 10^{- 21} J \\ Δ_{0} = \frac{9}{4} Δ_{t} \\ = \frac{9}{4} \times 340 .34 \times 10^{- 21} \\ = 765 .765 \times 10^{- 21} J \\ \approx 766 \times 10^{- 21} J \end{aligned} Answer = 766$

If ${[Cu {(H_{2} O)}_{4}]}^{2 +}$ absorbs a light of wavelength 600 n m for d-d transition, then the value of octahedral crystal field splitting energy for ${[Cu {(H_{2} O)}_{6}]}^{2 +}$ will be
(Given : $h = 6.63 \times 10^{- 34} JS$ and $c = 3.08 \times 10^{8} {ms}^{- 1}$ )

Solution:

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