Coordination Compounds 2 Question 6-1
6. Three complexes,
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+} $ (I),
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} $ (II)
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ (III)
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 $(\lambda)$ of light absorbed by the complexes is inversely proportional to its $\Delta_0$ CFSE (magnitude). $\Delta_0(\mathrm{CFSE}) \propto 1 / \lambda$
The complexes can be written as:
I. $[\mathrm{CoCl}(\mathrm{NH}_3)_5]^{2+} \equiv[\mathrm{Co}(\mathrm{NH}_3)_5(\mathrm{Cl})]^{2+}]$
II. $\left[\mathrm{Co}\left[\mathrm{NH}_3\right]_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} \equiv\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+}$
III. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right]^{3+} \equiv\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NH}_3\right)\right]^{3+}$
So, the differentiating ligands in the octahedral complexes of 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 $\mathrm{Cl}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3$.
So, the crystal field splitting energy (magnitude) order will be
$ \Delta_0^{\mathrm{CFSE}} \text { (I) }<\Delta_0^{\mathrm{CFSE}} \text { (II) }<\Delta_0^{\mathrm{CFSE}} \text { (III) } $
and the order of wavelength $(\lambda)$ of light absorbed by the complexes will be
$ \lambda(\mathrm{I})>\lambda \text { (II) }>\lambda(\mathrm{III}) \quad\left[\because \text { Energy }\left(\Delta_0^{\mathrm{CFSE}}\right) \propto \frac{1}{\lambda}\right] $
BETA
