The correct match is: Complexes No. of | vedclass

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Question

(A) $1$. For $A. [CrF_6]^{4-}$: $Cr^{2+}$ is $3d^4$. $F^-$ is a weak field ligand,so electrons remain unpaired. Unpaired electrons = $4$ $(S)$.$2$. Fo

Vedclass · Accepted Answer

$A-S, B-P, C-Q, D-R$

Vedclass · Answer

(A) $1$. For $A. [CrF_6]^{4-}$: $Cr^{2+}$ is $3d^4$. $F^-$ is a weak field ligand,so electrons remain unpaired. Unpaired electrons = $4$ $(S)$.$2$. For $B. [MnF_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $F^-$ is a weak field ligand,so electrons remain unpaired. Unpaired electrons = $5$ $(P)$.$3$. For $C. [Cr(CN)_6]^{4-}$: $Cr^{2+}$ is $3d^4$. $CN^-$ is a strong field ligand,causing pairing. $t_{2g}^4 e_g^0$ configuration leads to $2$ unpaired electrons $(Q)$.$4$. For $D. [Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. $t_{2g}^5 e_g^0$ configuration leads to $1$ unpaired electron $(R)$.Thus,the correct match is $A-S, B-P, C-Q, D-R$.

The correct match is:

Complexes No. of unpaired electrons

$A. [CrF_6]^{4-}$ $P. 5$

$B. [MnF_6]^{4-}$ $Q. 2$

$C. [Cr(CN)_6]^{4-}$ $R. 1$

$D. [Mn(CN)_6]^{4-}$ $S. 4$

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Complexes	No. of unpaired electrons
$A. [CrF_6]^{4-}$	$P. 5$
$B. [MnF_6]^{4-}$	$Q. 2$
$C. [Cr(CN)_6]^{4-}$	$R. 1$
$D. [Mn(CN)_6]^{4-}$	$S. 4$

The correct match is: Complexes No. of unpaired electrons $A. [CrF_6]^{4-}$ $P. 5$ $B. [MnF_6]^{4-}$ $Q. 2$ $C. [Cr(CN)_6]^{4-}$ $R. 1$ $D. [Mn(CN)_6]^{4-}$ $S. 4$