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 x 10^-34 Js, Speed of light ( c)=3.0 x 10^8 m / s and Avogadro's Constant ( N_ A)=6 x 10^23 / mol

[ Ti( H_2 O)_6]^3+, \ CFSE=-0.4 Delta_0 =-96.0 kJ mol^-1 \ Therefore Delta_0=(-96.0 / -0.4) \ Therefore Delta_0=240 kJ mol^-1 \ frac hclambda =frac96 x 10^30.4 x 6 x 10^23 \ lambda =frac0.4 x 6 x 10^23 x 6.4 x 10^-34 x 3 x 10^896 x 10^3 \ =0.48 x 10^-6 m \ =480 x 10^-9 m \ =480 nm

If the CFSE of TiH2 O63 is 960 kJ mol this complex will... | null (Chemistry)

\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}

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