If the CFSE of ${[Ti {(H_{2} O)}_{6}]}^{3 +} is - 96.0 kJ / mol$ , this complex will absorb maximum at wavelength _____nm. (nearest integer)
Assume Planck’s constant $(h) = 6.4 \times 10^{- 34} Js$ , speed of light $(c) = 3.0 \times 10^{8} m / s$ and
Avogadro’s Constant $(N_{A}) = 6 \times 10^{23} / mol$ .

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answer is 480.

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Detailed Solution

$\begin{array}{l} CFSE = - 0.4 Δ \\ - 96.0 kJ / mol = - 0.4 Δ \\ Δ = \frac{- 96 \times 10^{3}}{0.4} J / mol \\ Δ = \frac{- 96 \times 10^{3}}{0.4 \times 6 \times 10^{23}} J / molecule \\ λ = \frac{hc}{Δ} \\ = \frac{6.4 \times 10^{- 34} \times 3 \times 8 \times 0.4 \times 6 \times 10^{23}}{96 \times 10^{3}} \\ = 0.48 \times 10^{- 6} m \\ = 480 nm \end{array}$