The complex ion [Fe(CN)_6]^{4-} contains | vedclass

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(D) In $[Fe(CN)_6]^{4-}$,the oxidation state of $Fe$ is $+2$. The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$. Since $CN^-$ is a strong field

Vedclass · Accepted Answer

six sigma bonds

Vedclass · Answer

(D) In $[Fe(CN)_6]^{4-}$,the oxidation state of $Fe$ is $+2$. The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. This results in $d^2sp^3$ hybridization,which corresponds to an octahedral geometry. There are $6$ $Fe-C$ sigma bonds formed by the donation of electron pairs from $6$ $CN^-$ ligands. The total number of electrons on $Fe^{2+}$ is $26 - 2 = 24$. Thus,the statement regarding $6$ sigma bonds is correct.

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$

The complex ion $[Fe(CN)_6]^{4-}$ contains

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