Oxidation number of ${\rm{Fe}}$ in ${{\rm{K}}_3}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]$ is:

NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Oxidation number of

\mathrm{Fe}

{\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]

is:

View answer and explanation

Correct option: A

Option A (3) is correct because the oxidation state of Fe is calculated based on the charge of the complex and the ligands.

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\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 =}

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