If an iron (III) complex with the formula [Fe | vedclass

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(B) The complex is $[Fe(NH_3)_{x}(CN)_{y}]^{-}$.The oxidation state of $Fe$ is $+3$, so the electronic configuration of $Fe^{3+}$ is $3d^5$.For the $e

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$6$

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(B) The complex is $[Fe(NH_3)_{x}(CN)_{y}]^{-}$.The oxidation state of $Fe$ is $+3$, so the electronic configuration of $Fe^{3+}$ is $3d^5$.For the $e_g$ orbital to have no electrons, all $5$ electrons must occupy the $t_{2g}$ orbitals.This requires a strong field ligand environment to cause large crystal field splitting $(\Delta_o > P)$.$CN^-$ is a strong field ligand, while $NH_3$ is a moderate field ligand.For the complex to be low-spin with $t_{2g}^5 e_g^0$, the coordination number must be $6$ to satisfy the geometry of an octahedral complex.Thus, $x+y = 6$.For example, in $[Fe(CN)_6]^{3-}$, $x=0$ and $y=6$, which satisfies the condition.

List-$I$ (Complex/ion)	List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$	$I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$	$II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$	$III$. Square planar
$D$. $[Fe(CO)_5]$	$IV$. Tetrahedral

If an iron $(III)$ complex with the formula $[Fe(NH_3)_{x}(CN)_{y}]^{-}$ has no electron in its $e_g$ orbital, then the value of $x+y$ is

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Match List-$I$ with List-$II$:
List-$I$ (Complex/ion) List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$ $I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$ $II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$ $III$. Square planar
$D$. $[Fe(CO)_5]$ $IV$. Tetrahedral

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