NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

1 Hard

1 of 1

Understanding +4 | -1

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

reacts with

{FeCl}_3

to form:

View answer and explanation

Correct option: B

{KFe}\left[{Fe}({CN})_6\right]

Option B (

{KFe}\left[{Fe}({CN})_6\right]

) is correct because

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

reacts with

{FeCl}_3

to form Prussian blue, which has the formula

{KFe}\left[{Fe}({CN})_6\right]

More Questions from CO-ORDINATION COMPOUNDS

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]