EAN of ${\rm{Cr}}$ in $\left[ {{\rm{Cr}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]{\rm{C}}{{\rm{l}}_3}$ is:

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Application +4 | -1

EAN of

\mathrm{Cr}

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

is:

View answer and explanation

Correct option: B

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{\;}

\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]