Arrange the following oxides in decreasing order of oxidation state of the metal ion.
\( (A)\text{Cu}_2\text{O} \)
\( (B)\text{CrO}_3 \)
\( (C)\text{Mn}_2\text{O}_7 \)
\( (D)\text{V}_2\text{O}_5 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (B), (D), (A)
The oxidation state of a metal in its oxide follows from the charge balance: oxygen is −2 and the compound is neutral. For the metals from scandium to manganese, the highest oxidation state in the oxides equals the group number: +5 for vanadium (group 5), +6 for chromium (group 6) and +7 for manganese (group 7), because oxygen is very good at stabilising high oxidation states.
(A) \( \ce{Cu2O} \): \( 2x + (-2) = 0 \Rightarrow x = +1 \)
(B) \( \ce{CrO3} \): \( x + 3(-2) = 0 \Rightarrow x = +6 \)
(C) \( \ce{Mn2O7} \): \( 2x + 7(-2) = 0 \Rightarrow x = +7 \)
(D) \( \ce{V2O5} \): \( 2x + 5(-2) = 0 \Rightarrow x = +5 \)
In decreasing order of the oxidation state of the metal:
\( \displaystyle \ce{Mn2O7}\,(+7) \gt \ce{CrO3}\,(+6) \gt \ce{V2O5}\,(+5) \gt \ce{Cu2O}\,(+1) \)
Shortcut: \( \ce{Mn2O7} \) has the highest and \( \ce{Cu2O} \) the lowest oxidation state, so the order starts with (C) and ends with (A); only Option (4) does both.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – “The ability of oxygen to stabilise the highest oxidation state is demonstrated in the oxides. The highest oxidation number in the oxides … coincides with the group number and is attained in Sc2O3 to Mn2O7.”