Question:
$\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ absorbs light of wavelength 498 $\mathrm{nm}$ during a $\mathrm{d}-\mathrm{d}$ transition. The octahedral splitting energy for the above complex is__________
$\times 10^{-19} \mathrm{~J}$. (Round off to the Nearest Integer). $\mathrm{h}$ $=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$.
Solution:
$\lambda_{\text {absorbed }}=498 \mathrm{~nm}$ (given)
The octahedral spilitting energy
$\Delta_{0}$ or $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{498 \times 10^{-9}}$
$=0.0399 \times 10^{-17} \mathrm{~J}$
$=3.99 \times 10^{-19} \mathrm{~J}$
$=4.00 \times 10^{-19} \mathrm{~J}$ (round off)