$$[Ti(H_2O)_6]^{3+}$$ absorbs light of wavelength 498 nm during a d-d transition. The octahedral splitting energy for the above complex is ________ $$\times 10^{-19}$$ J. (Round off to the Nearest Integer).
$$h = 6.626 \times 10^{-34}$$ Js; $$c = 3 \times 10^8$$ ms$$^{-1}$$.

The octahedral splitting energy ($$\Delta_o$$) corresponds to the energy of the photon absorbed during the d-d transition. The wavelength of light absorbed is $$\lambda = 498$$ nm $$= 498 \times 10^{-9}$$ m.

Using the relation $$E = \frac{hc}{\lambda}$$, where $$h = 6.626 \times 10^{-34}$$ J s and $$c = 3 \times 10^8$$ m s$$^{-1}$$:

$$\Delta_o = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{498 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{498 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{4.98 \times 10^{-7}}$$.

$$\Delta_o = 3.992 \times 10^{-19}$$ J $$\approx 4 \times 10^{-19}$$ J.

The octahedral splitting energy is $$4 \times 10^{-19}$$ J.