Increasing order of oxidation states of transition metal oxides will be:
(A) \(TiO_2\)
(B) \(MnO_4 ^-\)
(C) \(VO_2^+\)
(D) \(CrO_4^{2-}\)
(E) \(Ni(CO)_4\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (E) < (C) < (A) < (D) < (B)
Note: The given answer is as per the official NTA Answer Key. However, the correct answer should be Option (3) (E) < (A) < (C) < (D) < (B), as explained below.
The oxidation state of the metal in each species follows from the charge balance; placing them from the lowest up:
(E) \( \ce{Ni(CO)4} \) — first. CO is a neutral ligand, so Ni = 0.
(A) \( \ce{TiO2} \) — second. x + 2(−2) = 0, so Ti = +4.
(C) \( \ce{VO2+} \) — third. x + 2(−2) = +1, so V = +5.
(D) \( \ce{CrO4^2-} \) — fourth. x + 4(−2) = −2, so Cr = +6.
(B) \( \ce{MnO4-} \) — last. x + 4(−2) = −1, so Mn = +7.
Option (2) (E) < (C) < (A) < (D) < (B) — The answer given by NTA, but incorrect. It places \( \ce{VO2+} \) (+5) before \( \ce{TiO2} \) (+4).
Option (3) (E) < (A) < (C) < (D) < (B) — Correct as per NCERT. 0 < +4 < +5 < +6 < +7.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – …
