Among the given first-row transition metal oxides (VO2+, Cr2O72−, MnO4−), the metal oxide(s) with the lowest oxidizing power will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → VO2+
An oxoion is a strong oxidising agent when the metal is in a high oxidation state and is easily reduced. The higher the standard reduction potential, the stronger the oxidising power.
Oxidation state of the metal in each ion (O = −2):
VO2+: x + 2(−2) = +1, so x = +5.
Cr2O72−: 2x + 7(−2) = −2, so x = +6.
MnO4−: x + 4(−2) = −1, so x = +7.
Their standard reduction potentials in acid are:
\( \displaystyle \ce{VO2^+ / VO^2+}\ (1.00\ \mathrm{V}) \lt \ce{Cr2O7^2- / Cr^3+}\ (1.33\ \mathrm{V}) \lt \ce{MnO4^- / Mn^2+}\ (1.51\ \mathrm{V}) \)
So the oxidising power increases in the order \( \ce{VO2^{+}} \) < \( \ce{Cr2O7^{2-}} \) < \( \ce{MnO4^{-}} \). This follows the rise in oxidation state (+5, +6, +7) and the decreasing stability of the highest oxidation state across the series. The lowest oxidising power belongs to VO2+.
Why the other options are incorrect
Option (2) Cr2O72− — Incorrect. Cr(+6) is a stronger oxidant than V(+5) (1.33 V against 1.00 V).
Option (3) MnO4− — Incorrect. Mn(+7) is the strongest oxidant of the three.
Option (4) Cr2O72− and MnO4− have the same oxidizing power, but lower than VO2+ — Incorrect. Their potentials differ (1.33 V and 1.51 V), and both are stronger oxidants than \( \ce{VO2^{+}} \), not weaker.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – …
