Coordination Compounds Result Question 6
absorb light in the visible region. The correct order of the wavelength of light absorbed by them is
(2019 Main, 10 April I)
(a) II $>$ I $>$ III
(b) I $>$ II $>$ III
(c) III $>$ I $>$ II
(d) III $>$ II $>$ I
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Answer:
Correct Answer: 6. ( b )
Solution:
- Key Idea The wavelength ( ) of light absorbed by the complexes is inversely proportional to its ${ } _0$ CFSE (magnitude). ${ } _0$ (CFSE) $1 /$
The complexes can be written as:
I. $\left.\left[\mathrm{CoCl}\left(\mathrm{NH} _3\right) _5\right]^2 \quad\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5(\mathrm{Cl})\right]^2\right]$
II. $\left[\mathrm{Co}\left[\mathrm{NH} _3\right] _5 \mathrm{H} _2 \mathrm{O}\right]^3 \quad\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5\left(\mathrm{H} _2 \mathrm{O}\right)\right]^3$
III. $\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5\right]^3 \quad\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5\left(\mathrm{NH} _3\right)\right]^3$
So, the differentiating ligands in the octahedral complexes of $\mathrm{Co}$ (III) in I, II and III are $\mathrm{Cl}^{\ominus}, \mathrm{H} _2 \mathrm{O}$ and $\mathrm{NH} _3$ respectively. In the spectrochemical series, the order of this power for crystal field splitting is $\mathrm{Cl} \quad \mathrm{H} _2 \mathrm{O} \quad \mathrm{NH} _3$.
So, the crystal field splitting energy (magnitude) order will be ${ } _0^{\text {CFSE }}$ (I) $\quad{ } _0^{\text {CFSE }}$ (II) $\quad{ } _0^{\text {CFSE }}$ (III) and the order of wavelength ( ) of light absorbed by the complexes will be
(I) (II) (III)
BETA
