Q29 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

The correct order of oxidizing power of the given ions is

  • Correct answer
    A
    $\mathrm{VO}_2^{+} \lt \mathrm{Cr}_2 \mathrm{O}_7^{2-} \lt \mathrm{MnO}_4^{-}$
  • B
    $\mathrm{VO}_2^{+} \lt \mathrm{MnO}_4^{-} \lt \mathrm{Cr}_2 \mathrm{O}_7^{2-}$
  • C
    $\mathrm{MnO}_4^{-} \lt \mathrm{Cr}_2 \mathrm{O}_7^{2-} \lt \mathrm{VO}_2^{+}$
  • D
    $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \lt \mathrm{VO}_2^{+} \lt \mathrm{MnO}_4^{-}$

Solution

In $\mathrm{VO}_2^{+}, \mathrm{V}$ is in +3 oxidation state. $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \mathrm{Cr}$ is in +6 oxidation state. $\mathrm{MnO}_4^{-} \mathrm{Mn}$ is in +7 oxidation state. Oxidising power means substance have more tendency to accept electron. Mn is in highest oxidation state so it have more oxidising power. $\mathrm{MnO}_4^{-}\gt\mathrm{Cr}_2 \mathrm{O}_7^{2-}\gt\mathrm{VO}_2^{+}$

Hence, the correct answer is option: 'A'.

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