The oxidation number of ${\rm{Fe}}$ in ${{\rm{K}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]\,{\rm{is}}$

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

1 Easy

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Application +4 | -1

The oxidation number of

\mathrm{Fe}

{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]\,\mathrm{is}

View answer and explanation

Correct option: C

The correct answer is 2 because the oxidation state of Fe in

{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]

is +2.

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]