Which of the following is an incorrect pair?

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(C) $1$. For $[Cu(NH_3)_4]^{2+}$,the geometry is square planar ($dsp^2$ hybridization).$2$. In $[Ni(CO)_4]$,$CO$ is a neutral ligand.$3$. For $[Fe(CN)

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$[Fe(CN)_6]^{3-}$: $sp^3d^2$ hybridization

Vedclass · Answer

(C) $1$. For $[Cu(NH_3)_4]^{2+}$,the geometry is square planar ($dsp^2$ hybridization).$2$. In $[Ni(CO)_4]$,$CO$ is a neutral ligand.$3$. For $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is $3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing,resulting in $d^2sp^3$ hybridization (inner orbital complex),not $sp^3d^2$.$4$. For $[Co(en)_3]^{3+}$,$Co^{3+}$ has $Z=27$,so $Co^{3+} = 24$ electrons. $en$ donates $2 	imes 3 = 6$ electrons. Total $EAN = 24 + 12 = 36$,which follows the $EAN$ rule (Krypton configuration).

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