Coordination Compounds Result Question 6
absorb light in the visible region. The correct order of the wavelength of light absorbed by them is from shortest to longest.
(2019 Main, 10 April)
(a) II $>$ I $>$ III
(b) I $>$ II $>$ III
(c) III $>$ I $>$ II
(d) III $>$ II $>$ I
Show Answer
Answer:
Correct Answer: 6. ( b )
Solution:
- Key Idea The wavelength (λ) of light absorbed by the complexes is inversely proportional to its Δ₀ CFSE (magnitude). Δ₀ (CFSE) $1 /$
The complexes can be written as:
I. $\left.\left[\mathrm{CoCl}\left(\mathrm{NH} _3\right) _5\right]^+\quad\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5\mathrm{Cl}\right]^+\right]$
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. $\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _5\right]^{3+} \quad\left[\mathrm{Co}\left(\mathrm{NH} _3\right) _6\right]^{3+}$
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 $\mathrm{Cl}^{\ominus} \quad \mathrm{H} _2 \mathrm{O} \quad \mathrm{NH} _3$.
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 (from shortest to longest) of light absorbed by the complexes will be
(I) (II) (III)
BETA
