In the formation of K_4[Fe(CN)_6],the hybridi | vedclass

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Question

(B) In $K_4[Fe(CN)_6]$,the central metal ion is $Fe^{2+}$.The atomic number of $Fe$ is $26$,so the electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6

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$d^2sp^3$

Vedclass · Answer

(B) In $K_4[Fe(CN)_6]$,the central metal ion is $Fe^{2+}$.The atomic number of $Fe$ is $26$,so the electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.After pairing,two $3d$ orbitals become vacant.These two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals undergo $d^2sp^3$ hybridisation to form six equivalent octahedral hybrid orbitals.Thus,the hybridisation involved is $d^2sp^3$.

In the formation of $K_4[Fe(CN)_6]$,the hybridisation involved is

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