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In which of the following coordination entities the magnitude of $\Delta_o$ (CFSE in octahedral field) will be maximum?
(Atomic number $\mathrm{Co}=27$ )
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(Atomic number $\mathrm{Co}=27$ )
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Verified Answer
The correct answer is:
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$
Key Idea: The magnitude of $\Delta_{\propto \mathrm{c} t}$ (the orbital splitting energy) is decided by the nature of ligand Strong field ligand has highest $\Delta_{\text {oct }}$. The increasing field strength is as
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-} < \mathrm{F}^{-} < \mathrm{OH}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{C}_2 \mathrm{O}_4^{2-} < \mathrm{NH}_3 < \mathrm{en} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$
The $\mathrm{CN}^{-}$is the strongest ligand among these, hence the magnitude of $\Delta_{\text {oct }}$ will be maximum in $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$.
Note : The magnitude of $\Delta_{\text {oct }}$ is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of $\Delta_{\text {oct }}$. But here the oxidatinon state of Co metal ion is same in all complexes.
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-} < \mathrm{F}^{-} < \mathrm{OH}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{C}_2 \mathrm{O}_4^{2-} < \mathrm{NH}_3 < \mathrm{en} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$
The $\mathrm{CN}^{-}$is the strongest ligand among these, hence the magnitude of $\Delta_{\text {oct }}$ will be maximum in $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$.
Note : The magnitude of $\Delta_{\text {oct }}$ is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of $\Delta_{\text {oct }}$. But here the oxidatinon state of Co metal ion is same in all complexes.
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