MCQ
The color of $KMnO_4$ is due to :
- ✓$L\to M$ charge transfer transition
- B$\sigma - \sigma ^*$ transition
- C$M\to L$ charge transfer transition
- D$d - d$ transition
$\therefore \ln M n O_{4}^{-},$ Mn has $+7$ oxidation state having no electron in d-orbitals. It ts considered that higher the oxidation state of metal, greater is the tendency to occur $L \rightarrow M$ charge transfer,
because ligand is able to donate the electron into the vacant ( - orbital of metal.
since, charge transfer is laporate as well as spin allowed, therefore, it shows colour
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$(a)$ They might be configurational isomers $(b)$ They might be diastereomers
$(c)$ They might be constitutional isomers $(d)$ They might be tautomers
$(e)$ They might be conformational isomers $(f)$ They might be enantiomers
$(g)$ They might be positional isomers
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
$\mathrm{H}_3 \mathrm{PO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_4 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{CO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 $
$ \mathrm{H}_3 \mathrm{BO}_3 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_2 \ \ \ \ \mathrm{H}_2 \mathrm{CrO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_3 $