If \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) absorbs a light of wavelength 600 nm for \(\mathrm{d}-\mathrm{d}\) transition, then the value of octahedral crystal field splitting energy for \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will be \(\_\_\_\_\) \(\times 10^{-21} \mathrm{~J}\). (Nearest Integer) (Given : \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) and \(\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}\) )

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Coordination Compounds

[Cu (H_{2} O)_{4}]^{2 +}

absorbs a light of wavelength 600 nm for

d - d

transition, then the value of octahedral crystal field splitting energy for

[Cu (H_{2} O)_{6}]^{2 +}

will be

____

\times 1 0^{- 21} J

. (Nearest Integer)
(Given :

h = 6.63 \times 1 0^{- 34} Js

and

c = 3.0 \times 1 0^{8} ms^{- 1}

)

Answer: 746

The tetrahedral aqua complex absorbs at

λ = 600 nm

, so its crystal field splitting is

Δ_{t} = \frac{h c}{λ} = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{600 \times 1 0 ^{- 9}} \approx 3.315 \times 1 0^{- 19} J = 331.5 \times 1 0^{- 21} J

. For the same metal and ligands the octahedral and tetrahedral splittings are related by

Δ_{t} = \frac{4}{9} Δ_{o}

, that is

Δ_{o} = \frac{9}{4} Δ_{t}

, so

Δ_{o} = \frac{9}{4} \times 331.5 \times 1 0^{- 21} \approx 745.9 \times 1 0^{- 21} J

, which rounds to

746 \times 1 0^{- 21} J

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer