The crystal field stabilisation energy (CFSE) | vedclass

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(B) The oxidation state of $Fe$ in $K_3[Fe(CN)_6]$ is $+3$. Thus,the electronic configuration for $Fe^{3+}$ is $[Ar] 3d^5$.Since $CN^-$ is a strong fi

Vedclass · Accepted Answer

$-2.0 \Delta_o$ and $\sqrt{3} BM$

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(B) The oxidation state of $Fe$ in $K_3[Fe(CN)_6]$ is $+3$. Thus,the electronic configuration for $Fe^{3+}$ is $[Ar] 3d^5$.Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $d$-orbitals.The configuration in the crystal field is $t_{2g}^5 e_g^0$.$CFSE = (-0.4 	imes n(t_{2g}) + 0.6 	imes n(e_g)) \Delta_o + mP$$CFSE = (-0.4 	imes 5 + 0.6 	imes 0) \Delta_o + 2P = -2.0 \Delta_o + 2P$.However,considering only the crystal field splitting energy contribution:$CFSE = -0.4 	imes 5 = -2.0 \Delta_o$.Number of unpaired electrons $(n) = 1$.Magnetic moment,$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} BM$.Therefore,the correct option is $B$.

The crystal field stabilisation energy $(CFSE)$ and the spin-only magnetic moment in Bohr Magneton $(BM)$ for the complex $K_3[Fe(CN)_6]$ are,respectively

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