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Coordination Compounds question

2022 · 25 Jun · Shift 1 · Q22
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Coordination Compounds question

2022 · 25 Jun · Shift 1 · Q22

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
If [Cu(H2O)4]2+[Cu(H_2O)_4]^{2+}[Cu(H2​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(H2O)6]2+[Cu(H_2O)_6]^{2+}[Cu(H2​O)6​]2+ will be ‾×\underline{\hspace{2cm}}\times​× 10 −-− 21 J. [Nearest Integer] (Given : h = 6.63 ×\times× 10 −-− 34 Js and c = 3.08 ×\times× 108 ms −-− 1)
Numerical answer
View written solutionFree

Correct answer: 766

  1. Relation between absorbed light and crystal field splitting

For a ddd-ddd transition,

Δ=hcλ\Delta = \frac{hc}{\lambda}Δ=λhc​

where:

  • h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s
  • c=3.08×108 m s−1c = 3.08 \times 10^8\,\text{m s}^{-1}c=3.08×108m s−1
  • λ=600 nm=600×10−9 m\lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m}λ=600nm=600×10−9m

So for [Cu(H2O)4]2+[Cu(H_2O)_4]^{2+}[Cu(H2​O)4​]2+,

Δt=(6.63×10−34)(3.08×108)600×10−9\Delta_t = \frac{(6.63 \times 10^{-34})(3.08 \times 10^8)}{600 \times 10^{-9}}Δt​=600×10−9(6.63×10−34)(3.08×108)​
  1. Calculate the transition energy

First compute the numerator:

6.63×3.08=20.42046.63 \times 3.08 = 20.42046.63×3.08=20.4204

so,

hc=20.4204×10−26=2.04204×10−25hc = 20.4204 \times 10^{-26} = 2.04204 \times 10^{-25}hc=20.4204×10−26=2.04204×10−25

Now,

Δt=2.04204×10−256.00×10−7=3.4034×10−19 J\Delta_t = \frac{2.04204 \times 10^{-25}}{6.00 \times 10^{-7}} = 3.4034 \times 10^{-19}\,\text{J}Δt​=6.00×10−72.04204×10−25​=3.4034×10−19J

This is the crystal field splitting for the tetrahedral complex [Cu(H2O)4]2+[Cu(H_2O)_4]^{2+}[Cu(H2​O)4​]2+.

  1. Relation between tetrahedral and octahedral splitting

For the same metal ion and ligand,

Δt=49Δo\Delta_t = \frac{4}{9}\Delta_oΔt​=94​Δo​

Therefore,

Δo=94Δt\Delta_o = \frac{9}{4}\Delta_tΔo​=49​Δt​

So,

Δo=94(3.4034×10−19)\Delta_o = \frac{9}{4}(3.4034 \times 10^{-19})Δo​=49​(3.4034×10−19) Δo=7.65765×10−19 J\Delta_o = 7.65765 \times 10^{-19}\,\text{J}Δo​=7.65765×10−19J
  1. Express in the required form

We need

Δo=‾×10−21 J\Delta_o = \underline{\hspace{1cm}} \times 10^{-21}\,\text{J}Δo​=​×10−21J

Now,

7.65765×10−19=765.765×10−217.65765 \times 10^{-19} = 765.765 \times 10^{-21}7.65765×10−19=765.765×10−21

Nearest integer:

766766766
  1. Final answer
766\boxed{766}766​
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