The effective atomic number of $Cr{\rm{ }}\left( {{\rm{At}}{\rm{. No}}{\rm{.}} = 24} \right){\rm{in }}\left[ {{\rm{Cr}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]{\rm{C}}{{\rm{l}}_3}{\rm{\;}}$ is

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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The effective atomic number of

Cr\mathrm{ }\left( {\mathrm{At}\mathrm{. No}\mathrm{.} = 24} \right)\mathrm{in }\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}\mathrm{\;}

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Correct option: C

The correct answer is 33 because the EAN of Cr in

\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3}

is calculated as 24 (atomic number) - 3 (oxidation state) + 2(6) (electrons from ligands) = 33.

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{\;}

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\text{ox =}

x

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

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

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