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.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

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

Similar Questions

Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

The $Cl-Co-Cl$ bond angle values in a $fac-[Co(NH_3)_3Cl_3]$ complex is/are:

The hybridisation involved in $[CoF_6]^{3-}$ is :

The hybridization and magnetic nature of $[Mn(CN)_6]^{4-}$ and $[Fe(CN)_6]^{3-}$,respectively are

The hybridization and magnetic behaviour of the cobalt ion in the $[Co(NH_3)_6]^{3+}$ complex,respectively,are:

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