The shortest ${\rm{C}} - {\rm{O}}$ bond order exists in:

${\left[ {{\rm{V}}{{\left( {{\rm{CO}}} \right)}_6}} \right]^ - }$

${\left[ {{\rm{Mn}}{{\left( {{\rm{CO}}} \right)}_6}} \right]^ + }$

$\left[ {{\rm{Fe}}{{\left( {{\rm{CO}}} \right)}_5}} \right]$

$\left[ {{\rm{Cr}}{{\left( {{\rm{CO}}} \right)}_6}} \right]$

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Analyzing +4 | -1

The shortest

\mathrm{C} - \mathrm{O}

bond order exists in:

View answer and explanation

Correct option: D

{\left[ {\mathrm{V}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ - }

Option D (

{\left[ {\mathrm{V}{{\left( {\mathrm{CO}} \right)}_6}} \right]^ - }

) is correct because the negative charge on the complex increases back-bonding, which weakens the C-O bond and thus shortens the C-O bond length.

Practice All Chapter MCQs →

\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

\left[\mathrm{Cr}(\mathrm{Ox})_{2} \mathrm{ClBr}\right]^{3-}

\text{ox =}

x

{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}

\mathrm{M}{\mathrm{n}^{2 + }}

\mathrm{CuS}{\mathrm{O}_4}

{\mathrm{K}_2}\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]