In which of the following octahedral complexes of ${{\rm{Co}}}$ (at. No. 27), will be magnitude of ${\Delta _0}$ be the highest?

${\left[ {{\rm{Co}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}$

${\left[ {{\rm{Co}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]^{3 - }}$

${\left[ {{\rm{Co}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{3 + }}$

${\left[ {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]^{3 + }}$

JEE CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Understanding +4 | -1

In which of the following octahedral complexes of

{\mathrm{Co}}

(at. No. 27), will be magnitude of

{\Delta _0}

be the highest?

View answer and explanation

Correct option: A

{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}

Option A (

{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}

) is correct because cyanide (CN

^-

) is a strong field ligand, resulting in a larger crystal field splitting energy (

{\Delta _0}

Practice All Chapter MCQs →

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

\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{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\left[\mathrm{PtCl}_4\right]

\left[{Co}\left({NH}_3\right)_5\left({NO}_2\right)\right] {Cl}_2

\left[{Co}\left({NH}_3\right)_5({ONO})\right] {Cl}_2 \ldots \ldots \ldots

\left[{Cu}\left({NH}_3\right)_4\right]\left[{PtCl}_4\right] \,\, \& \,\, \left[{Pt}\left({NH}_3\right)_4\right]\left[{CuCl}_4\right] \ldots \ldots \ldots

\left[{Pt}\left({NH}_3\right)_4 {Cl}_2\right] {Br}_2 \, \, \& \, \, \left[{Pt}\left({NH}_3\right)_4 {Br}_2\right]{Cl}_2 \ldots \ldots \ldots