JEE CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

1 Hard

1 of 1

Analyzing +4 | -1

For an octahedral complex, which of the following

d

- electron configuration will give maximum CFSE?

View answer and explanation

Correct option: C

Low spin

d^5

Option D (

\mathrm{MgO}

) is correct because MgO has the highest lattice energy due to the higher charges of Mg2+ and O2- compared to the other options.

More Questions from CO-ORDINATION COMPOUNDS

Practice All Chapter MCQs →

\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

\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{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\left[\mathrm{PtCl}_4\right]

\left[{Co}\left({NH}_3\right)_5\left({NO}_2\right)\right] {Cl}_2

\left[{Co}\left({NH}_3\right)_5({ONO})\right] {Cl}_2 \ldots \ldots \ldots

\left[{Cu}\left({NH}_3\right)_4\right]\left[{PtCl}_4\right] \,\, \& \,\, \left[{Pt}\left({NH}_3\right)_4\right]\left[{CuCl}_4\right] \ldots \ldots \ldots

\left[{Pt}\left({NH}_3\right)_4 {Cl}_2\right] {Br}_2 \, \, \& \, \, \left[{Pt}\left({NH}_3\right)_4 {Br}_2\right]{Cl}_2 \ldots \ldots \ldots