Which of the following orders is not correct?

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(B) $1$. For $\Delta_0$ order: $[Co(H_2O)_6]^{+2}$ has a lower oxidation state than $[Fe(H_2O)_6]^{+3}$,and $CN^-$ is a stronger field ligand than $H_

Vedclass · Accepted Answer

$[Fe(CN)_6]^{-4} > [Fe(CN)_6]^{-3}$ $Fe-C$ bond length

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(B) $1$. For $\Delta_0$ order: $[Co(H_2O)_6]^{+2}$ has a lower oxidation state than $[Fe(H_2O)_6]^{+3}$,and $CN^-$ is a stronger field ligand than $H_2O$,so the order $[Co(H_2O)_6]^{+2} $2$. For $Fe-C$ bond length: In $[Fe(CN)_6]^{-4}$,$Fe$ is in $+2$ oxidation state,leading to more back-bonding compared to $[Fe(CN)_6]^{-3}$ ($Fe$ in $+3$). Stronger back-bonding increases the double bond character of $Fe-C$,shortening the bond. Thus,$[Fe(CN)_6]^{-4} $3$. For $\pi$-accepting tendency: $CO$ is a better $\pi$-acceptor than $CN^-$,so $CN^- $4$. For $C-C$ bond length: In Zeise's salt $K[PtCl_3(C_2H_4)]$,back-bonding from $Pt$ to the $\pi^*$ orbital of $C_2H_4$ reduces the $C-C$ bond order,making it longer than in free $C_2H_4$. Thus,$C_2H_4 Note: Both $B$ and $D$ are technically incorrect. However,in standard chemistry problems of this type,$B$ is the primary focus regarding back-bonding effects in coordination complexes.

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