Question

The correct descending order of oxidizing power of the following is

• A. Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2--} $>$ MnO\textsubscript{4}\textsuperscript{--} $>$ VO\textsubscript{2}\textsuperscript{+}
• B. MnO\textsubscript{4}\textsuperscript{--} $>$ Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2--} $>$ VO\textsubscript{2}\textsuperscript{+}
• C. VO\textsubscript{2}\textsuperscript{+} $>$ MnO\textsubscript{4}\textsuperscript{--} $>$ Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2--}
• D. MnO\textsubscript{4}\textsuperscript{--} $>$ VO\textsubscript{2}\textsuperscript{+} $>$ Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2--}
• E. Cr\textsubscript{2}O\textsubscript{7}\textsuperscript{2--} $>$ VO\textsubscript{2}\textsuperscript{+} $>$ MnO\textsubscript{4}\textsuperscript{--}

Hint:

In the first transition series oxoanions, oxidizing power generally increases across the period from Left to Right (V to Cr to Mn). More protons in the nucleus make the high oxidation state more "desperate" for electrons.

Answer 1

The Correct Option is B