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Among the following $\mathrm{Cr}(\mathrm{III})$ complexes, which one will have the highest octahedral crystal field splitting?
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Verified Answer
The correct answer is:
$\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$
A strong field ligand will enable higher crystal field splitting. It can be predicted from the relative position of a ligand in spectrochemical series.
In the given octahedral complexes of $\mathrm{Cr}(\mathrm{II})$ ion, the ligands are $\mathrm{F}^{-}, \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CN}^{-}$and their increasing order of field strength in spectrochemical series is
$\mathrm{F}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{CN}^{-} \text {. }$
So, highest octahedral crystal field splitting is present in $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$.
In the given octahedral complexes of $\mathrm{Cr}(\mathrm{II})$ ion, the ligands are $\mathrm{F}^{-}, \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CN}^{-}$and their increasing order of field strength in spectrochemical series is
$\mathrm{F}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{CN}^{-} \text {. }$
So, highest octahedral crystal field splitting is present in $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$.
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