Given below are two statements:Statement I: C | vedclass

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(A) For $[Mn(H_2O)_6]^{2+}$, the configuration is $d^5$ with a weak field ligand $(H_2O)$, resulting in $t_{2g}^3 e_g^2$. The $CFSE$ is $(3 	imes -0.

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(A) For $[Mn(H_2O)_6]^{2+}$, the configuration is $d^5$ with a weak field ligand $(H_2O)$, resulting in $t_{2g}^3 e_g^2$. The $CFSE$ is $(3 	imes -0.4 \Delta_o) + (2 	imes 0.6 \Delta_o) = 0 \Delta_o$.For $[Cr(H_2O)_6]^{2+}$, the configuration is $d^4$ with a weak field ligand $(H_2O)$, resulting in $t_{2g}^3 e_g^1$. The $CFSE$ is $(3 	imes -0.4 \Delta_o) + (1 	imes 0.6 \Delta_o) = -0.6 \Delta_o$. Since $|-0.6| > 0$, Statement $I$ is true.For potassium ferricyanide, $K_3[Fe(CN)_6]$, the iron is in $+3$ oxidation state $(d^5)$. $CN^-$ is a strong field ligand, leading to $t_{2g}^5 e_g^0$ with $1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$For sodium ferrocyanide, $Na_4[Fe(CN)_6]$, the iron is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand, leading to $t_{2g}^6 e_g^0$ with $0$ unpaired electrons. $\mu = 0 \ B.M.$Since $\sqrt{3} > 0$, Statement $II$ is true.

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