The coordination number and oxidation state of $Cr$ in ${{\rm{K}}_3}\left[ {{\rm{Cr}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]$ are respectively

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

1 Easy

1 of 1

Application +4 | -1

The coordination number and oxidation state of

Cr

{\mathrm{K}_3}\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]

are respectively

View answer and explanation

Correct option: A

+6 and +3

Option A (+6 and +3) is correct because in

{\mathrm{K}_3}\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]

, the coordination number of Cr is 6 (since each oxalate ligand is bidentate), and the oxidation state of Cr is +3.

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]