Key Fact
Variable oxidation states
Multiple stable oxidation states from sequential removal of (n-1)d and ns electrons. Mn: +2 to +7 (d⁵ ground); Cu: +1 (d¹⁰), +2 (d⁹). Highest OS often in oxide/fluoride.
-- NCERT Class 12 Chemistry, Ch. 4, p. 100Oxidation States Transition
8 MCQs3 revision cards9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026
Oxidation States Transition, explained for NEET
The trap that costs marks here: assuming every transition metal shows every oxidation state from +2 to its group number. They don't. The pattern is specific, and NEET tests whether you know which element breaks the expected trend.
Why transition metals show variable oxidation states: The energy gap between (n–1)d and ns electrons is small. Both sets participate in bonding, allowing multiple stable states. This is the defining feature that separates d-block from s-block or p-block elements (NCERT Class 12 Chemistry Chapter 4, page 90).
The oxidation-state catalogue you must know:
| Element | Common oxidation states | Maximum | Key note |
|---|---|---|---|
| Ti | +2, +3, +4 | +4 | +4 most stable |
| V | +2, +3, +4, +5 | +5 | All states accessible |
| Cr | +2, +3, +6 | +6 | +3 most stable (d³ half-filled t₂g) |
| Mn | +2, +3, +4, +6, +7 | +7 | Widest range; +2 most stable (d⁵) |
| Fe | +2, +3, (+6 rare) | +6 | +3 slightly more stable than +2 |
| Co | +2, +3 | +3 | +2 more stable in simple salts |
| Ni | +2 | +2 (rare +3/+4) | Almost exclusively +2 |
| Cu | +1, +2 | +2 | +2 more stable in aqueous |
| Zn | +2 | +2 | Only +2 (d¹⁰ — full shell) |
Key patterns for NEET:
- Maximum oxidation state increases from Ti (+4) to Mn (+7), then decreases. Mn shows the highest maximum because it can lose all seven electrons (2 from 4s + 5 from 3d).
- After Mn, electrons pair in d-orbitals and are harder to remove — maximum oxidation state drops.
- The +2 state is universal (loss of 4s²). The +3 state is common for early members but rare for Cu and absent for Zn.
- Zn shows only +2 because d¹⁰ is fully filled — no d-electron participation in bonding.
Watch-out: A common distractor exploits the d⁰/d¹⁰ stability concept. When asked "which element shows the maximum number of oxidation states," students pick Fe or Cr (common elements) instead of Mn.
Can you answer these Oxidation States Transition MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which first-row transition metal shows the maximum number of oxidation states?
Show answer and why every option is right or wrong
Answer: C. Mn shows oxidation states from +2 to +7 — the widest range among first-row transition metals. This is because Mn has the configuration [Ar] 3d⁵ 4s², and all seven electrons (5d + 2s) can participate in bonding (NCERT Class 12 Chemistry Chapter 4, page 97).
Why A is wrong: A is wrong because Cr shows states +2 to +6 (maximum +6), which is fewer than Mn's +2 to +7 range. Cr has 6 outer electrons (3d⁵4s¹) but its maximum is +6, not +7.
Why B is wrong: B is wrong because Fe commonly shows only +2 and +3 (with rare +6 in ferrate). Its range is far narrower than Mn's seven accessible states.
Why D is wrong: D is wrong because V shows +2 to +5 (maximum +5, with 5 outer electrons from 3d³4s²). This is fewer oxidation states than Mn.
MCQ 2Easy RecallPractice
Zn shows only the +2 oxidation state among first-row d-block elements because:
Show answer and why every option is right or wrong
Answer: B. Zn has the configuration [Ar] 3d¹⁰ 4s². The fully filled d¹⁰ shell is exceptionally stable and these electrons are not available for bonding. Only the two 4s electrons are lost, giving exclusively +2 (NCERT Class 12 Chemistry Chapter 4, page 97).
Why A is wrong: A is wrong because while Zn does have relatively high Zeff, the reason is specifically the d¹⁰ stability, not just 'tight binding.' Earlier elements like Cu also have high nuclear charge but still show +1 and +2.
Why C is wrong: C is wrong because absence of unpaired electrons does not prevent ion formation — Zn²⁺ forms readily. The issue is specifically that d-electrons in a filled d¹⁰ shell won't participate in bonding, not a general inability to bond.
Why D is wrong: D is wrong because while Zn is debatably a transition metal (d¹⁰ in ground state), it absolutely can lose d-electrons under extreme conditions — the point is that under normal chemical conditions the d¹⁰ stability prevents variable oxidation states.
MCQ 3Easy RecallPractice
The most stable oxidation state of Mn in aqueous solution is:
Show answer and why every option is right or wrong
Answer: C. Mn²⁺ has the configuration [Ar] 3d⁵ — a half-filled d-shell giving extra stability through maximum exchange energy. This makes +2 the most stable state in aqueous solution (NCERT Class 12 Chemistry Chapter 4, page 98).
Why A is wrong: A is wrong because Mn⁴⁺ (as in MnO₂) is stable as a solid but not the most stable state in aqueous solution. The d³ configuration lacks the exchange-energy stabilisation of d⁵.
Why B is wrong: B is wrong because +7 (as in MnO₄⁻) is a strong oxidising agent — it readily accepts electrons to reach lower states. It is highly reactive, not stable in the thermodynamic sense.
Why D is wrong: D is wrong because Mn³⁺ (d⁴) is unstable in aqueous solution and readily disproportionates to Mn²⁺ and MnO₂. It lacks the half-filled d⁵ stabilisation.
MCQ 4Direct ApplicationPractice
The maximum oxidation state equals the total number of (n–1)d + ns electrons for early transition metals. For which element does the maximum observed oxidation state first fall below this expected value?
Show answer and why every option is right or wrong
Answer: A. Fe has configuration [Ar] 3d⁶ 4s² — total outer electrons = 8, so the "expected" maximum would be +8. But the observed maximum is +6 (in FeO₄²⁻, ferrate). Fe is the first element where the maximum OS falls short of the total (n–1)d + ns electron count. This happens because after the d-shell is more than half-filled, the increasing nuclear charge and electron pairing make it energetically unfavourable to remove all d-electrons.
Why B is wrong: B is wrong because Co (3d⁷4s², total 9) also falls short (max +3 commonly, +4 rarely), but the question asks for the first element where this happens. Fe (total 8, max +6) comes before Co in the periodic table.
Why C is wrong: C is wrong because Mn achieves its full expected maximum: 3d⁵4s² = 7 outer electrons, and Mn shows +7 (in MnO₄⁻). The rule holds perfectly for Mn.
Why D is wrong: D is wrong because Cr (3d⁵4s¹ = 6 outer electrons) achieves +6 (in CrO₄²⁻/Cr₂O₇²⁻), which equals its total outer electron count. The rule still holds for Cr.
MCQ 5Direct ApplicationPractice
Among Ti²⁺, V²⁺, Cr²⁺, Mn²⁺, Fe²⁺, which ion has zero contribution to paramagnetism from d-electrons?
Show answer and why every option is right or wrong
Answer: C. In the +2 state, the configuration is [Ar] 3dⁿ where n = (atomic number − 20). Ti²⁺ is d², V²⁺ is d³, Cr²⁺ is d⁴, Mn²⁺ is d⁵, Fe²⁺ is d⁶. Every one of these has at least one unpaired electron (Ti²⁺ has 2, V²⁺ has 3, Cr²⁺ has 4, Mn²⁺ has 5, Fe²⁺ has 4). None has zero unpaired electrons; all contribute to paramagnetism.
Why A is wrong: A is wrong because Mn²⁺ is d⁵ with ALL five electrons unpaired (one in each d-orbital) — it is actually the most paramagnetic ion in this list, not the least.
Why B is wrong: B is wrong because Ti²⁺ is d² with 2 unpaired electrons. It is paramagnetic.
Why D is wrong: D is wrong because Fe²⁺ is d⁶ with 4 unpaired electrons (one orbital doubly occupied, four singly occupied in weak-field/free ion). It is paramagnetic.
MCQ 6Direct ApplicationPractice
Cu commonly shows +1 and +2 oxidation states. The reason Cu⁺ is unstable in aqueous solution (disproportionates to Cu and Cu²⁺) is:
Show answer and why every option is right or wrong
Answer: C. Cu⁺ (d¹⁰) is stable in the gas phase, but in aqueous solution the much higher hydration enthalpy of the smaller, doubly-charged Cu²⁺ ion provides enough energy to offset the second IE. The disproportionation 2Cu⁺ → Cu + Cu²⁺ is thermodynamically favoured in water because the hydration energy gain of Cu²⁺ drives the equilibrium (NCERT Class 12 Chemistry Chapter 4, page 100).
Why A is wrong: A is wrong because while Jahn-Teller distortion does occur in Cu²⁺ (d⁹), it is NOT the thermodynamic driving force for Cu⁺ disproportionation. The key factor is hydration enthalpy difference, not crystal-field distortion effects.
Why B is wrong: B is wrong because this is the opposite of what happens — the second IE of Cu is indeed high, but the hydration enthalpy of Cu²⁺ OVERCOMES it. If B were true, Cu²⁺ would not form at all.
Why D is wrong: D is wrong because Cu⁺ DOES disproportionate in aqueous solution — the premise of the option contradicts the observed fact. While d¹⁰ is intrinsically stable, aqueous thermodynamics (hydration) override gas-phase electronic stability.
MCQ 7Concept TrapPractice
Consider the series Sc to Zn. The +2 oxidation state is exhibited by all elements in this series EXCEPT:
Show answer and why every option is right or wrong
Answer: A. Sc has configuration [Ar] 3d¹ 4s². When it loses 2 electrons (the 4s²), Sc²⁺ would be d¹ — but Sc preferentially loses the single d-electron as well to form the noble-gas-core Sc³⁺ (d⁰). The +2 state of Sc is virtually non-existent in normal chemistry. All other elements from Ti to Zn readily form M²⁺ compounds by losing the 4s² electrons.
Why B is wrong: B is wrong because Zn shows +2 as its ONLY oxidation state (losing both 4s electrons from [Ar]3d¹⁰4s²). Zn²⁺ is extremely stable and common.
Why C is wrong: C is wrong because Cu readily forms Cu²⁺ compounds (d⁹ configuration). CuSO₄, CuCl₂ are standard examples. Cu shows both +1 and +2.
Why D is wrong: D is wrong because Ti readily forms Ti²⁺ (d² configuration), though Ti³⁺ and Ti⁴⁺ are more common. The +2 state exists for Ti in TiCl₂ and similar compounds.
MCQ 8CalculationPractice
An element X from the first transition series forms an oxide where X is in its maximum oxidation state. The oxide has the formula X₂O₇. Identify X and its ground-state electronic configuration.
Show answer and why every option is right or wrong
Answer: B. In X₂O₇, oxygen is –2, so: 2(OS of X) + 7(–2) = 0 → OS of X = +7. Among first-row transition metals, only Mn achieves +7 as its maximum oxidation state (using all 5d + 2s = 7 electrons). Mn₂O₇ is the anhydride of permanganic acid. Mn's ground-state configuration is [Ar] 3d⁵ 4s².
Why A is wrong: A is wrong because Cr's maximum oxidation state is +6 (not +7). Cr has [Ar] 3d⁵ 4s¹ giving 6 outer electrons. Cr forms CrO₃ (Cr in +6) but cannot form Cr₂O₇ with Cr in +7.
Why C is wrong: C is wrong because Fe's maximum common oxidation state is +6 (in ferrate, FeO₄²⁻) — not +7. Fe has 8 outer electrons but cannot utilise all of them. Fe₂O₇ does not exist.
Why D is wrong: D is wrong because V's maximum oxidation state is +5 (3d³ + 4s² = 5 electrons). V forms V₂O₅ as its highest oxide, not V₂O₇.
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Oxidation States Transition: quick recall before you leave
How do you solve a Oxidation States Transition question? A worked example
Pattern: Predict highest stable oxidation state for a transition metal from its electronic configuration (P.CHE.U11.OXIDATION_STATE_TRANSITION).
- 1
Given
Element: Vanadium (V), atomic number 23. Ground-state configuration: [Ar] 3d³ 4s².
- 2
Required
Determine the maximum oxidation state of V and the formula of its highest oxide.
- 3
Concept
For early transition metals (up to Mn), the maximum oxidation state equals the number of (n–1)d electrons plus ns electrons. All these electrons can participate in bonding because the d-orbitals are not yet more than half-filled.
- 4
Formula
Maximum OS = number of 3d electrons + number of 4s electrons = 3 + 2 = +5.
- 5
Substitution
V: 3d³ gives 3 d-electrons; 4s² gives 2 s-electrons. Total = 5.
- 6
Calculation
Maximum oxidation state = +5. For the highest oxide: 2(+5) + x(–2) = 0 → x = 5 → formula V₂O₅.
Note: The numbers 3 and 2 (electron counts) and 5 (their sum) are exact counting integers; they do not limit significant figures. - 7
Final answer
Maximum oxidation state of V = +5. Highest oxide = V₂O₅ (vanadium pentoxide).
- 8
Common trap
A distractor might claim V shows +6 or +7 by analogy with Cr or Mn. But V has only 5 outer electrons (3d³ 4s²) — it cannot exceed +5. Another distractor offers V₂O₃ (which is V in +3, not the maximum).
- 9
Similar NEET-style question
"An element from the first transition series forms an oxide XO₃. Identify the element and its d-electron count in the ground state." (Answer: Cr, since CrO₃ has Cr in +6; ground state Cr is [Ar] 3d⁵ 4s¹ with 5 d-electrons.)
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What to remember before solving Oxidation States Transition questions
Which Oxidation States Transition formulas do you need for NEET?
Common oxidation states (first-row TM)
Catalogues common stable oxidation states across first-row transition metals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| OS | oxidation state | - |
Valid when
- First-row d-block
- Common (not exotic) compounds
More in The d and f Block Elements: 3 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Oxidation States Transition questions from past NEET papers
3 questions from NEET 2023, 2024. Answers verified against NTA official keys.
NEET 2024Revised key
1d5 to d4 configuration
2d5 to d2 configuration
3d4 to d5 configuration
4d3 to d5 configuration
NTA Answer: Option 3(revised_final)
NEET 2023
The stability of Cu2+ is more than Cu+ salts in aqueous solution due to
1Enthalpy of atomization
2Hydration energy
3Second ionisation enthalpy
4First ionisation enthalpy
NTA Answer: Option 2(final)
NEET 2023
1B and D only
2C and D only
3B and C only
4A and E only
NTA Answer: Option 2(final)
All 15 past-paper questions from The d and f Block Elements →
How does NEET ask about Oxidation States Transition?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict highest stable oxidation state for transition metal. Mn: +7 in MnO4-. Fe: +3 stable. Cu: +2.
InterpretationMedium
Common distractors
misses d0 d10 stability
Doesn't account for closed-shell stability
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.90
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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