Oxidation state of ' $\mathrm{V}^{\prime}$ in $\mathrm{Rb}_4 \mathrm{Na}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right]$ is

JEE CHEMISTRY MCQ: REDOX REACTIONS

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Application +4 | -1

Oxidation state of '

\mathrm{V}^{\prime}

\mathrm{Rb}_4 \mathrm{Na}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right]

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Correct option: A

+5

The correct answer is +5. The oxidation state of V is calculated by setting the sum of oxidation states of all elements in the compound to zero.

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\mathrm{H}

2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2

\mathrm{HNO}_3

\mathrm{HNO}_3

\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_2(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})

\mathrm{Br}

-1

+5.