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(D) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.For a high spin $d^4$ configuration,th

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$- 0.6 \Delta_o$

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(D) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.For a high spin $d^4$ configuration,the electrons occupy the orbitals according to Hund's rule because the crystal field splitting energy $\Delta_o$ is less than the pairing energy $P$.The electronic configuration is $t_{2g}^3 e_g^1$.The Crystal Field Stabilization Energy $(CFSE)$ is calculated as:$CFSE = (n_{t_{2g}} 	imes -0.4 \Delta_o) + (n_{e_g} 	imes 0.6 \Delta_o)$$CFSE = (3 	imes -0.4 \Delta_o) + (1 	imes 0.6 \Delta_o)$$CFSE = -1.2 \Delta_o + 0.6 \Delta_o = -0.6 \Delta_o$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Cu(NH_3)_6]^{2+}$	$I$. $-0.6$
$B$. $[Ti(H_2O)_6]^{3+}$	$II$. $-2.0$
$C$. $[Fe(CN)_6]^{3-}$	$III$. $-1.2$
$D$. $[NiF_6]^{4-}$	$IV$. $-0.4$

Crystal field stabilization energy for high spin $d^4$ octahedral complex is

Similar Questions

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If the crystal field splitting energy of a tetrahedral complex $\Delta_t$ of the type $[ML_4]^{n+}$ is $x \ eV$,what is the crystal field splitting energy with respect to an octahedral complex,$[ML_6]^{n+}$?

Match List-$I$ with List-$II$:

List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$)

$A$. $[Cu(NH_3)_6]^{2+}$ $I$. $-0.6$

$B$. $[Ti(H_2O)_6]^{3+}$ $II$. $-2.0$

$C$. $[Fe(CN)_6]^{3-}$ $III$. $-1.2$

$D$. $[NiF_6]^{4-}$ $IV$. $-0.4$

Choose the correct answer from the options given below:

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