Match the electronic configurations in List-I | vedclass

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Question

(A) The electronic configurations are determined based on the crystal field splitting of the metal ions:$1$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$.

Vedclass · Accepted Answer

$P$ $\rightarrow 3; Q$ $\rightarrow 2; R$ $\rightarrow 4; S$ $\rightarrow 1$

Vedclass · Answer

(A) The electronic configurations are determined based on the crystal field splitting of the metal ions:$1$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand $(WFL)$,so the configuration is $t_{2g}^4 e_g^2$ $(S \rightarrow 1)$.$2$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. $H_2O$ is a $WFL$,so the configuration is $t_{2g}^3 e_g^2$ $(Q \rightarrow 2)$.$3$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand $(SFL)$,so the configuration is $t_{2g}^6 e_g^0$ $(P \rightarrow 3)$.$4$. $[FeCl_4]^{-}$: $Fe^{3+}$ is $3d^5$. It is a tetrahedral complex with a $WFL$,so the configuration is $e^2 t_2^3$ $(R \rightarrow 4)$.$5$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. It is a tetrahedral complex with a $WFL$,so the configuration is $e^4 t_2^3$.Thus,the correct matching is $P$ $\rightarrow 3, Q$ $\rightarrow 2, R$ $\rightarrow 4, S$ $\rightarrow 1$.

List-$I$	List-$II$
$P$. $t_{2g}^6 e_g^0$	$1$. $[Fe(H_2O)_6]^{2+}$
$Q$. $t_{2g}^3 e_g^2$	$2$. $[Mn(H_2O)_6]^{2+}$
$R$. $e^2 t_2^3$	$3$. $[Co(NH_3)_6]^{3+}$
$S$. $t_{2g}^4 e_g^2$	$4$. $[FeCl_4]^{-}$
	$5$. $[CoCl_4]^{2-}$

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