The hybridization of Fe in K_3[Fe(CN)_6] is | vedclass

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(D) . In $K_3[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. The electronic config

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$d^2sp^3$

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(D) . In $K_3[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The $5$ electrons in $3d$ orbitals rearrange to occupy $3$ orbitals,leaving $2$ empty $3d$ orbitals. These $2$ empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals hybridize to form $6$ equivalent $d^2sp^3$ hybrid orbitals. Thus,the hybridization of $Fe$ is $d^2sp^3$.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$(a)$. $4, sp^3$	$(i)$. trigonal bipyramidal
$(b)$. $3, sp^2$	$(ii)$. octahedral
$(c)$. $5, sp^3 d$	$(iii)$. tetrahedral
$(d)$. $6, sp^3 d^2$	$(iv)$. trigonal planar

The hybridization of $Fe$ in $K_3[Fe(CN)_6]$ is

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