Key Fact
Catalytic behavior
Variable oxidation states + ability to form complexes → catalysis. V₂O₅ (contact process), Fe (Haber), Ni (hydrogenation), TiCl₃ (Ziegler-Natta).
-- NCERT Class 12 Chemistry, Ch. 4, p. 104Catalytic Behaviour
8 MCQs9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 27 Sep 2026
Catalytic Behaviour, explained for NEET
Transition metals act as catalysts in many industrial and biological reactions — and NEET asks about the why, not just the what. The core fact is straightforward: d-block elements exhibit catalytic behaviour because of (a) variable oxidation states allowing intermediate formation with reactants, and (b) large surface area in finely divided form (NCERT Class 12 Chemistry Chapter 4, page 104).
The mechanism link: A catalyst provides an alternative reaction pathway of lower activation energy. Transition metals achieve this by forming short-lived intermediates — cycling between oxidation states during the reaction and returning to the original state at the end. Example: Fe³⁺/Fe²⁺ in the Contact process; V₂O₅ cycling between V⁵⁺ and V⁴⁺ in sulfuric acid manufacture.
What NEET actually tests:
- Why transition metals (and not s-block metals) make effective catalysts — the answer hinges on variable oxidation states and partially filled d-orbitals enabling bonding with reactant molecules.
- Specific examples: Fe in Haber process (N₂ + 3H₂ → 2NH₃), V₂O₅ in Contact process (2SO₂ + O₂ → 2SO₃), Ni in hydrogenation, MnO₂ in KClO₃ decomposition.
- Finely divided form increases surface area → more active sites → better catalytic activity.
Watch-out: Do not confuse "variable oxidation states" with "high oxidation states." Zinc (+2 only) and scandium (+3 only) are poor catalysts precisely because they lack multiple accessible oxidation states — yet both are d-block elements. NEET uses this distinction as a distractor.
Can you answer these Catalytic Behaviour 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 property of transition metals is primarily responsible for their catalytic behaviour?
Show answer and why every option is right or wrong
Answer: D. Transition metals catalyse reactions by forming intermediates through variable oxidation states, providing an alternative lower-energy pathway (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: High melting point is a physical property related to strong metallic bonding, not to the mechanism of catalysis. Students confuse general d-block properties with the specific feature enabling catalysis.
Why B is wrong: High density is a bulk physical property arising from atomic mass and packing. It does not explain how a catalyst lowers activation energy.
Why C is wrong: Metallic lustre relates to free electrons reflecting light — it has no connection to providing alternative reaction pathways or intermediate formation.
MCQ 2Easy RecallPractice
V₂O₅ is used as a catalyst in the Contact process. During catalysis, vanadium cycles between which oxidation states?
Show answer and why every option is right or wrong
Answer: B. In the Contact process (2SO₂ + O₂ → 2SO₃), V₂O₅ is reduced to V₂O₄ (V⁴⁺) by SO₂, then reoxidised to V⁵⁺ by O₂, completing the catalytic cycle (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: V²⁺/V³⁺ are lower oxidation states not involved in this industrial process. The Contact process operates at high oxidation state cycling.
Why C is wrong: V³⁺ is not a documented intermediate in the Contact process mechanism. The reduction goes from +5 directly to +4.
Why D is wrong: A jump from +2 to +5 would require a 3-electron change in one step, which is energetically unfavourable under Contact process conditions.
MCQ 3Direct ApplicationPractice
Which of the following d-block elements is a poor catalyst due to lack of variable oxidation states?
Show answer and why every option is right or wrong
Answer: B. Zinc has electronic configuration [Ar] 3d¹⁰ 4s² and shows only +2 oxidation state (d¹⁰ is stable). Without variable oxidation states, it cannot form the intermediate complexes needed for catalytic cycling (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: Fe shows +2, +3, and +6 oxidation states, making it an effective catalyst (e.g., Haber process). Students may pick Fe thinking the question asks for a good catalyst.
Why C is wrong: Mn exhibits the widest range of oxidation states (+2 to +7) among first-row d-block elements, making it an excellent catalyst.
Why D is wrong: Cr shows +2, +3, and +6 oxidation states and acts as catalyst in several reactions. Its partially filled d-orbitals enable intermediate formation.
MCQ 4Direct ApplicationPractice
Finely divided iron is used in the Haber process rather than a large iron block because:
Show answer and why every option is right or wrong
Answer: A. Catalytic activity depends on available surface area for reactant adsorption. Finely divided form maximises the number of active sites per unit mass (NCERT Class 12 Chemistry Chapter 4, page 104).
Why B is wrong: Melting point is an intrinsic property — it does not change with particle size at the scale relevant to industrial catalysis. This confuses a physical constant with a surface phenomenon.
Why C is wrong: The oxidation states available to iron are the same regardless of physical form. Subdivision does not create new electronic configurations.
Why D is wrong: Finely divided iron is chemically identical to bulk iron (same element, same electronic structure). The difference is purely physical — increased surface area.
MCQ 5Direct ApplicationPractice
Which of the following correctly explains why scandium (Sc) is not commonly used as a catalyst?
Show answer and why every option is right or wrong
Answer: A. Sc has configuration [Ar] 3d¹ 4s². In compounds, it almost exclusively shows +3 (d⁰). Without accessible multiple oxidation states, it cannot provide the intermediate-forming mechanism required for catalysis (NCERT Class 12 Chemistry Chapter 4, page 104).
Why B is wrong: Sc has configuration [Ar] 3d¹ 4s² — its d-orbitals are NOT completely filled. Zinc (3d¹⁰) has completely filled d-orbitals. Students confuse Sc with Zn.
Why C is wrong: Sc IS a transition metal (forms Sc³⁺ with no d-electrons, but the neutral atom has partially filled d-orbitals in the ground state). This is a common definitional confusion.
Why D is wrong: Sc shows essentially only +3. Saying it has 'too many' oxidation states is the opposite of reality.
MCQ 6Easy RecallPractice
In a catalysed reaction, the transition metal catalyst:
Show answer and why every option is right or wrong
Answer: D. A catalyst lowers activation energy by providing an alternative mechanism (via intermediate formation). It does not alter thermodynamic quantities — ΔH and Keq remain unchanged (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: A catalyst is regenerated at the end of the reaction cycle — it is NOT consumed. If it were consumed, it would be a reactant, not a catalyst.
Why B is wrong: A catalyst affects kinetics (rate), not thermodynamics. The equilibrium constant depends only on ΔG°, which is unchanged by a catalyst.
Why C is wrong: Enthalpy of reaction (ΔH) is a state function depending on initial and final states only. A catalyst changes the pathway, not the endpoints — ΔH is unaffected.
MCQ 7CalculationPractice
Fe³⁺ catalyses the reaction 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻ through the two steps below:
Step 1: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Step 2: 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻
What role does the Fe³⁺/Fe²⁺ pair play across these two steps?
Show answer and why every option is right or wrong
Answer: B. Across the two steps, Fe³⁺ is reduced to Fe²⁺ while oxidising I⁻ to I₂ (Step 1), then the Fe²⁺ is reoxidised back to Fe³⁺ by S₂O₈²⁻ (Step 2) — restoring the original oxidation state so Fe³⁺/Fe²⁺ cycles catalytically instead of being consumed (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: Fe³⁺ IS regenerated — Step 2 reoxidises the Fe²⁺ produced in Step 1 back to Fe³⁺, so it is not permanently consumed overall; this option ignores the second step of the cycle.
Why C is wrong: Fe³⁺ (as its reduced form Fe²⁺) is also essential to Step 2 — the Fe²⁺ generated in Step 1 reduces S₂O₈²⁻ back to Fe³⁺, completing the catalytic cycle.
Why D is wrong: Fe²⁺ does not accumulate as a final product; it is reoxidised to Fe³⁺ by S₂O₈²⁻ in Step 2, restoring the starting oxidation state for another cycle.
MCQ 8Concept TrapPractice
A student claims: "All d-block elements are good catalysts." Which pair of elements best disproves this claim?
Show answer and why every option is right or wrong
Answer: C. Zn (d¹⁰, only +2) and Sc (effectively only +3, d⁰ in ion) both lack variable accessible oxidation states. They are d-block elements but poor catalysts — disproving the blanket claim (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: Fe (+2/+3/+6) and Co (+2/+3) both have variable oxidation states and ARE effective catalysts. They support the claim rather than disproving it.
Why B is wrong: Cr (+2/+3/+6) and Mn (+2 to +7) have extensive variable oxidation states and are well-known catalysts. They support the claim.
Why D is wrong: Ni and Pd both catalyse hydrogenation reactions effectively due to variable oxidation states. They support rather than refute the claim.
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How do you solve a Catalytic Behaviour question? A worked example
- 1
Given
• V₂O₅: vanadium in +5 oxidation state• ZnO: zinc in +2 oxidation state• Reaction catalysed: 2SO₂ + O₂ → 2SO₃ (Contact process)
- 2
Required
Identify the property enabling V₂O₅ to act as a catalyst and explain why ZnO fails.
- 3
Concept
Catalytic behaviour in transition metals requires variable oxidation states to form intermediates with reactants, creating an alternative low-energy pathway.
- 4
Formula
No numerical formula required — this is a conceptual application of the variable-oxidation-state criterion.
- 5
Substitution / Application
• Vanadium: accessible oxidation states include +5, +4, +3, +2. In the Contact process, V⁵⁺ is reduced to V⁴⁺ by SO₂ (forming the intermediate), then reoxidised by O₂ back to V⁵⁺.• Zinc: only stable oxidation state is +2 (d¹⁰ configuration). Cannot be reduced or oxidised under Contact process conditions to form an intermediate.
- 6
Analysis
V₂O₅ cycles: V⁵⁺ + SO₂ → V⁴⁺ + SO₃, then V⁴⁺ + ½O₂ → V⁵⁺. The catalyst is regenerated. ZnO cannot undergo this cycling because Zn²⁺ has no accessible lower or higher oxidation state under these conditions.
- 7
Final answer
V₂O₅ is effective because vanadium has multiple accessible oxidation states (+4/+5 cycling), enabling intermediate formation and catalyst regeneration. ZnO fails because Zn²⁺ (d¹⁰) has only one stable oxidation state (+2) — the fundamental requirement for catalytic activity (variable oxidation states) is absent.
- 8
Common trap
Students may argue "ZnO is a d-block oxide, so it should catalyse reactions." The trap is equating "d-block membership" with "catalytic ability." The criterion is not membership in the d-block but the availability of multiple oxidation states for intermediate cycling.
- 9
Similar NEET-style question
"Among TiO₂, V₂O₅, ZnO, and CuO, which oxide(s) can act as effective catalysts for redox reactions and why?"
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What to remember before solving Catalytic Behaviour questions
More in The d and f Block Elements: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
Catalytic Behaviour questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
1A-I, B-IV, C-III, D-II
2A-I, B-II, C-IV, D-III
3A-II, B-III, C-I, D-IV
4A-I, B-II, C-III, D-IV
NTA Answer: Option 4(final)
All 15 past-paper questions from The d and f Block Elements →
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.104
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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