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Coordination Compounds question

2021 · 16 Mar · Shift 2 · Q15
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Coordination Compounds question

2021 · 16 Mar · Shift 2 · Q15

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
[Ti(H2O)6]3+[Ti(H_2O)_6]^{3+}[Ti(H2​O)6​]3+ absorbs light of wavelength 498 nm during a d −-− d transition. The octahedral splitting energy for the above complex is ‾\underline{\hspace{2cm}}​×\times× 10 −-− 19 J. (Round off to the Nearest Integer). h = 6.626 ×\times× 10 −-− 34 Js; c = 3 ×\times× 108 ms −-− 1
Numerical answer
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Correct answer: 4

  1. For an octahedral complex, the energy absorbed in the d−dd-dd−d transition is equal to the octahedral splitting energy:

Δo=E=hcλ\Delta_o = E = \frac{hc}{\lambda}Δo​=E=λhc​

  1. Given:

h=6.626×10−34 J sh = 6.626 \times 10^{-34}\, \text{J s}h=6.626×10−34J s c=3×108 m s−1c = 3 \times 10^8\, \text{m s}^{-1}c=3×108m s−1 λ=498 nm=498×10−9 m\lambda = 498\, \text{nm} = 498 \times 10^{-9}\, \text{m}λ=498nm=498×10−9m

  1. Substitute into the formula:

E=(6.626×10−34)(3×108)498×10−9E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{498 \times 10^{-9}}E=498×10−9(6.626×10−34)(3×108)​

  1. First calculate the numerator:

(6.626×10−34)(3×108)=19.878×10−26=1.9878×10−25(6.626 \times 10^{-34})(3 \times 10^8) = 19.878 \times 10^{-26} = 1.9878 \times 10^{-25}(6.626×10−34)(3×108)=19.878×10−26=1.9878×10−25

  1. Now divide by 498×10−9498 \times 10^{-9}498×10−9:

E=1.9878×10−25498×10−9E = \frac{1.9878 \times 10^{-25}}{498 \times 10^{-9}}E=498×10−91.9878×10−25​

E=(1.9878498)×10−16E = \left(\frac{1.9878}{498}\right) \times 10^{-16}E=(4981.9878​)×10−16

E≈0.00399×10−16=3.99×10−19 JE \approx 0.00399 \times 10^{-16} = 3.99 \times 10^{-19}\, \text{J}E≈0.00399×10−16=3.99×10−19J

  1. Rounding to the nearest integer in the form ‾×10−19\underline{\hspace{0.5cm}} \times 10^{-19}​×10−19 J:

3.99×10−19 J≈4×10−19 J3.99 \times 10^{-19}\, \text{J} \approx 4 \times 10^{-19}\, \text{J}3.99×10−19J≈4×10−19J

So, the required integer is:

4\boxed{4}4​

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