Question

If $[Cu(H_2O)_4]^{2+}$ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [$Cu(H_2O)_6]^{2+}$ will be _______ $×10^{–21} J$. [Nearest integer]
(Given : h = $6.63 × 10^{–34} Js$ and $c = 3.08×10^8 ms^{–1}$)

Answer 1

$[Cu(H_2O)_4]^{2+}$ is tetrahedral
$[Cu(H_2O)_6]^{2+}$ is octahedral
Because,
$△_t = \frac 49 \times△_0$

$△_t = \frac {6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}$

$△_0 = \frac 94 \times \frac {6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}$
$≃ 765 \times 10^{-21}\ J$

So, the answer is $765$.