The transition elements exhibit variable oxidation states because (n−1)d electrons can also participate in bonding, as the energy difference between (n−1)d and ns orbitals is very small. The maximum oxidation state that can be exhibited by a transition element is equal to the total number of electrons present in both ns and (n−1)d orbitals. The elements in the middle exhibit more oxidation states, e.g., manganese exhibits all oxidation states from +2 to +7. In the second series, a stable maximum oxidation state is exhibited by technetium, and an unstable maximum oxidation state of +8 is exhibited by ruthenium. In the third series, a stable maximum oxidation state of +8 is exhibited by osmium. Transition metal ions having a completely filled or an exactly half-filled d subshell are stable. The Fe3+ ion is more stable than the Fe2+ ion because of the stable half-filled 3d5 electronic configuration of Fe3+.
Consider the following statements about the oxidation states shown by the 3d transition metals in their compounds.
(A) Titanium(IV) is more stable than titanium(III) or titanium(II).
(B) Among the halides, vanadium(V) is represented only by VF5; the other halides undergo hydrolysis to give oxohalides, VOX3.
(C) All the copper(II) halides are known except the fluoride.
(D) Beyond manganese, no metal of the 3d series has a trihalide except FeX3 and CoF3.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
The stability of an oxidation state changes along the 3d series. Early metals have too few electrons to lose for many states, and after manganese the higher oxidation states become much less stable; the halides a metal forms show this.
(A) Titanium(IV) is more stable than titanium(III) or titanium(II). — True. Titanium (3d24s2) reaches its most stable state by using all four of its outer electrons; its highest halides are the tetrahalides, TiX4.
(B) Among the halides, vanadium(V) is represented only by VF5; the other halides undergo hydrolysis to give oxohalides, VOX3. — True. Fluorine can hold vanadium in the +5 state; the other halogens cannot, and their V(V) compounds appear only as oxohalides.
(C) All the copper(II) halides are known except the fluoride. — False. The missing copper(II) halide is the iodide, not the fluoride. Cu2+ oxidises iodide ions to iodine, so CuI2 cannot be obtained:
\( \displaystyle \ce{2Cu^2+ + 4I- -> Cu2I2(s) + I2} \)
(D) Beyond manganese, no metal of the 3d series has a trihalide except FeX3 and CoF3. — True. After manganese the +3 state is hard to hold in halides; only iron (FeX3) and cobalt (CoF3, the fluoride only) form trihalides.
NCERT (2026-27 edition):
4.3.4 Oxidation States – “…Thus, early in the series scandium(II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II).”
4.3.7 Trends in Stability of Higher Oxidation States – “…The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 and CrF6. The +7 state for Mn is not represented in simple halides but MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 and CoF3.” … “Although V+5 is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3.”
“…On the other hand, all CuII halides are known except the iodide. In this case, Cu2+ oxidises I– to I2: 2Cu2+ + 4I− → Cu2I2(s) + I2”