Which of the following will have maximum stab | vedclass

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(C) The crystal field stabilization energy $(CFSE)$ depends on the nature of the ligand and the oxidation state of the metal ion.
$CN^-$ is a strong

Vedclass · Accepted Answer

$[Co(CN)_6]^{3-}$

Vedclass · Answer

(C) The crystal field stabilization energy $(CFSE)$ depends on the nature of the ligand and the oxidation state of the metal ion.
$CN^-$ is a strong field ligand,which causes a large splitting of $d$-orbitals $(\Delta_o)$.
In $[Co(CN)_6]^{3-}$,$Co$ is in the $+3$ oxidation state ($d^6$ configuration).
Since $CN^-$ is a strong field ligand,it leads to a low-spin $t_{2g}^6 e_g^0$ configuration,resulting in a very high $CFSE$ value compared to the other complexes listed,which involve weak field ligands like $H_2O$ or have lower oxidation states.

$I$. $H_2O$	$II$. $CO$
$III$. $NH_3$	$IV$. $I^{-}$
$V$. $F^{-}$

Which of the following will have maximum stabilization due to crystal field?

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The crystal field splitting energy of $[Co(oxalate)_3]^{3-}$ complex is '$n$' times that of the $[Cr(oxalate)_3]^{3-}$ complex. Here '$n$' is [Assume $\Delta_0 \gg P$].

What is the electronic arrangement of a metal atom/ion in an octahedral complex with $d^4$ configuration,if $\Delta_0 < P$ (pairing energy)?

The correct order for wavelengths of absorption in the visible region for the following complexes will be

Arrange the following ligands in the order of increasing field strength.
$I$. $H_2O$ $II$. $CO$
$III$. $NH_3$ $IV$. $I^{-}$
$V$. $F^{-}$

The Crystal Field Stabilisation Energy $(CFSE)$ for $[CoCl_{6}]^{4-}$ is $18000 \; cm^{-1}$. The $CFSE$ for $[CoCl_{4}]^{2-}$ will be $...... \ cm^{-1}$.

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