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. 12Oxidation States Transition
8 MCQs3 revision cards9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 12).
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.
Practice 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?
MCQ 2Easy RecallPractice
Zn shows only the +2 oxidation state among first-row d-block elements because:
MCQ 3Easy RecallPractice
The most stable oxidation state of Mn in aqueous solution is:
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?
MCQ 5Direct ApplicationPractice
Among Ti²⁺, V²⁺, Cr²⁺, Mn²⁺, Fe²⁺, which ion has zero contribution to paramagnetism from d-electrons?
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:
MCQ 7Concept TrapPractice
Consider the series Sc to Zn. The +2 oxidation state is exhibited by all elements in this series EXCEPT:
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.
Quick recall before you leave
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.) ---
Before solving, remember these
Formulas
Spin-only magnetic moment
Magnetic moment from n unpaired electrons. 1 unpaired: 1.73 BM; 5: 5.92 BM.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | Bohr magneton |
Valid when
- Spin-only contribution (no orbital contribution)
- Octahedral or tetrahedral complex
Spin-only magnetic moment for transition metal
Predicts paramagnetic moment of d-block ion. n unpaired electrons in d-orbitals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | BM |
Valid when
- Spin-only contribution
- Octahedral or tetrahedral complex
- First-row d-block
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
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
When it triggers
Question gives KMnO4 oxidation in unspecified or specific medium.
How to avoid
Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.
Root cause: concept gap
Correction
Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
Root cause: concept gap
Correction
Lanthanoid contraction (imperfect 4f shielding) makes 5d elements similar in size to 4d. Zr/Hf, Nb/Ta, Mo/W have nearly identical chemistry.
Past Year Questions
10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
The correct order of decreasing basic strength of the given amines is:
1benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
2N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
3N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
4N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
NTA Answer: Option 4(final)
NEET 2024Revised key
1B and D only
2A and E only
3B and C only
4A and D only
NTA Answer: Option 1(revised_final)
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 2024Revised key
The pair of lanthanoid ions which are diamagnetic is
1Ce4+ and Yb2+
2Ce3+ and Eu2+
3Gd3+ and Eu3+
4Pm3+ and Sm3+
NTA Answer: Option 1(revised_final)
NEET 2023
Which one of the following statements is correct?
1All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor
2The bone in human body is an inert and unchanging substance
3Mg plays roles in neuromuscular function and interneuronal transmission
4The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g
NTA Answer: Option 4(final)
NEET 2023
1B and D only
2C and D only
3B and C only
4A and E only
NTA Answer: Option 2(final)
NEET 2022
12.57
25.57
33.57
44.57
NTA Answer: Option 2(final)
NEET 2022
Gadolinium has a low value of third ionisation enthalpy because of
1high basic character
2small size
3high exchange enthalpy
4high electronegativity
NTA Answer: Option 3(final)
NEET 2022
1+6 to +5
2+7 to +4
3+6 to +4
4+7 to +3
NTA Answer: Option 2(final)
NEET 2021
Tritium, a radioactive isotope of hydrogen, emits which of the following particles? (β–)
1Neutron (n)
2Beta (α) (γ)
3Alpha
4Gamma
NTA Answer: Option 2(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Use KMnO4 or K2Cr2O7 in acidic/neutral/basic medium for oxidation. Different products by medium.
InterpretationMedium
Common distractors
uses acidic formula in basic medium
Assumes Mn²⁺ regardless of medium
Effect of lanthanoid contraction on properties of 4d/5d transition metals. Closer atomic/ionic radii, similar properties (Zr/Hf, Nb/Ta).
RecallEasy
Common distractors
predicts large difference due to period shift
Expects 5d to be much larger than 4d
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.12
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