NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Application +4 | -1

Consider the following complex compounds.

(i)

\left[{Pt}\left({NH}_3\right)_2({SCN})_2\right]\quad\quad

(ii)

\left[{Co}\left({NH}_3\right)_3\left({NO}_2\right)_3\right]

(iii)

\left[{Pt}({en}) {Cl}_2\right]\quad\quad\quad

(iv)

\left[{Cr}({en})_2 {Br}_2\right]^{+}

(v)

\left[{Rh}({en})_3\right]^{3+}\quad\quad\quad

(vi)

\left[{CoCl}_2 {Br}_2\right]^{2-}

Then calculate sum of total number of geometrical isomers in all above complex compounds.

View answer and explanation

Correct option: A

Option A (6) is correct because the sum of geometrical isomers for the given complexes is 6.

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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{\;}

\left[\mathrm{Cr}(\mathrm{Ox})_{2} \mathrm{ClBr}\right]^{3-}

\text{ox =}

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{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}

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