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If a didentate ligand ethane-1, 2-diamine is progressively added in the molar ratio en : $\mathrm{Ni}$ $:: 1: 1,2: 1,3: 1$ to $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ aq solution, following co-ordination entities are formed.
I. $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{en}\right]^{2+}(a q)-$ pale blue
II. $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{en})_2\right]^{2+}(a q)-$ blue/purple
III. $\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+}(\mathrm{aq})$ - violet
The wavelength in $\mathrm{nm}$ of light absorbed in case of I and III are respectively
Options:
I. $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{en}\right]^{2+}(a q)-$ pale blue
II. $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{en})_2\right]^{2+}(a q)-$ blue/purple
III. $\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+}(\mathrm{aq})$ - violet
The wavelength in $\mathrm{nm}$ of light absorbed in case of I and III are respectively
Solution:
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Verified Answer
The correct answer is:
$600 \mathrm{~nm}$ and $535 \mathrm{~nm}$
Given that, compound I shows pale blue
colour. It means that, absorbed colour is orange whereas compound III shows violet colour then the absorbed colour is yellow. The wavelength of light absorbed by I and III are nearly $600 \mathrm{~nm}$ and $535 \mathrm{~nm}$ respectively.
colour. It means that, absorbed colour is orange whereas compound III shows violet colour then the absorbed colour is yellow. The wavelength of light absorbed by I and III are nearly $600 \mathrm{~nm}$ and $535 \mathrm{~nm}$ respectively.
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