If $$[CuH_2O_4]^{2+}$$ absorbs a light of wavelength $$600$$ nm for d-d transition, then the value of octahedral crystal field splitting energy for $$[CuH_2O_6]^{2+}$$ will be ______ $$\times 10^{-21}$$ J [Nearest integer]
(Given : $$h = 6.63 \times 10^{-34}$$ Js and $$c = 3.08 \times 10^{8}$$ ms$$^{-1}$$)

We need to find the octahedral crystal field splitting energy for $$[Cu(H_2O)_6]^{2+}$$, given that the tetrahedral complex $$[Cu(H_2O)_4]^{2+}$$ absorbs light at 600 nm.

First, note that $$[Cu(H_2O)_4]^{2+}$$ has coordination number 4 and adopts a tetrahedral geometry, whereas $$[Cu(H_2O)_6]^{2+}$$ has coordination number 6 and adopts an octahedral geometry.

Since the d-d transition absorbs a photon whose energy equals the crystal field splitting energy, we can write
$$\Delta_{tet} = \frac{hc}{\lambda}$$
Substituting the given values ($$h = 6.63 \times 10^{-34}$$ Js, $$c = 3.08 \times 10^{8}$$ m s$$^{-1}$$, $$\lambda = 600$$ nm $$= 600 \times 10^{-9}$$ m) leads to
$$\Delta_{tet} = \frac{6.63 \times 10^{-34} \times 3.08 \times 10^{8}}{600 \times 10^{-9}}$$

The numerator is calculated as $$6.63 \times 3.08 = 20.4204$$, giving $$20.4204 \times 10^{-26}$$ J·m, and the denominator is $$600 \times 10^{-9} = 6.0 \times 10^{-7}$$ m. Therefore,
$$\Delta_{tet} = \frac{20.4204 \times 10^{-26}}{6.0 \times 10^{-7}} = 3.4034 \times 10^{-19} \text{ J}$$

According to crystal field theory, the tetrahedral and octahedral splitting energies for the same metal ion and ligand are related by
$$\Delta_{tet} = \frac{4}{9}\Delta_{oct}$$
because the tetrahedral crystal field is inherently weaker with fewer ligands. Consequently,
$$\Delta_{oct} = \frac{9}{4}\Delta_{tet}$$
Using this relationship yields
$$\Delta_{oct} = \frac{9}{4} \times 3.4034 \times 10^{-19} = 2.25 \times 3.4034 \times 10^{-19}$$
$$= 7.6577 \times 10^{-19} \text{ J} = 765.77 \times 10^{-21} \text{ J}$$
Finally, rounding to the nearest integer gives $$\Delta_{oct} \approx 766 \times 10^{-21} \text{ J}$$.

The correct answer is 766.