NEETAceToday's NEET Daily

Transition-element trends, colour, catalysis, magnetism and complexes: NEET MCQs with answers

54 questions on this topic of d- and f-Block Elements (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise d- and f-Block Elements freeAll d- and f-Block Elements questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. Which of the following configurations represents a transition metal ion that is attracted most strongly into an applied magnetic field?

  1. Sc³⁺ (aq)
  2. Ti⁴⁺ (aq)
  3. Zn²⁺ (aq)
  4. Mn²⁺ (aq)
Show answer and explanation

Answer: (D) Mn²⁺ (aq)

Paramagnetic substances are attracted into an applied magnetic field; the greater the number of unpaired electrons, the stronger the attraction. Mn²⁺ has a 3d⁵ configuration with 5 unpaired electrons (paramagnetic), whereas Sc³⁺ (3d⁰), Ti⁴⁺ (3d⁰), and Zn²⁺ (3d¹⁰) have no unpaired electrons and are diamagnetic.

medium

2. A solution containing hydrated Ti³⁺ ions is purple, but it turns colourless when heated strongly to drive off all water molecules. What is the fundamental reason for this loss of colour?

  1. Ti³⁺ gets oxidised to Ti⁴⁺ in the absence of water.
  2. In the absence of ligands, crystal field splitting disappears and d-d transitions cannot occur.
  3. The coordination number of Ti³⁺ increases from 6 to 8 upon heating.
  4. The 3d electron drops into the 3p subshell when water is removed.
Show answer and explanation

Answer: (B) In the absence of ligands, crystal field splitting disappears and d-d transitions cannot occur.

The colour of [Ti(H₂O)₆]³⁺ arises from the excitation of an electron from the t₂g to the eg level (a d-d transition) induced by the crystal field of water ligands. In the absence of water ligands (anhydrous state), there is no crystal field splitting; the five d-orbitals remain degenerate, preventing d-d transitions and rendering the substance colourless.

hard

3. Given below are two statements regarding the transition elements: Statement I: The highest oxidation state of manganese in its fluorides is +4 (MnF₄), whereas in its oxides it reaches +7 (Mn₂O₇). Statement II: Oxygen possesses a higher capability than fluorine to stabilize higher oxidation states because of its ability to form multiple pπ-dπ bonds with transition metals. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Oxygen stabilizes the highest oxidation state (+7 in Mn₂O₇) superiorly compared to fluorine (+4 in MnF₄) because oxygen can form multiple bonds (pπ-dπ overlap) with transition metal atoms, whereas fluorine can only form single covalent bonds. Thus, both statements are correct and Statement II properly explains Statement I.

medium

4. Given below are two statements: Statement I: Transition metals readily form interstitial compounds with small non-metal atoms like H, C, and N. Statement II: Interstitial compounds are chemically more reactive than the parent transition metals and possess lower melting points. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because small atoms fit into the interstitial spaces of the metal lattice. Statement II is incorrect because interstitial compounds are chemically inert and have higher melting points than pure metals.

easy

5. Which of the following compounds exhibits colour primarily due to ligand-to-metal charge transfer (LMCT) rather than d-d transition?

  1. [Cu(H₂O)₄]²⁺
  2. [Fe(H₂O)₆]³⁺
  3. [Ni(H₂O)₆]²⁺
  4. K₂Cr₂O₇
Show answer and explanation

Answer: (D) K₂Cr₂O₇

In K₂Cr₂O₇, chromium is in the +6 oxidation state with a 3d⁰ electronic configuration, meaning no d-electrons are available for d-d transitions. The intense orange colour is due to charge transfer from the oxygen 2p orbitals (ligand) to empty chromium 3d orbitals (metal).

hard

6. Consider the transition series monoxides: TiO, VO, CrO, and MnO. Which of the following trends in character is correct as the atomic number increases from Ti to Mn?

  1. Basic character increases and covalent character increases.
  2. Metallic bonding strength increases leading to sharply increasing melting points.
  3. Reducing character increases while standard electrode potential E°(M²⁺/M) becomes more negative.
  4. Basic character decreases while acidic character increases.
Show answer and explanation

Answer: (D) Basic character decreases while acidic character increases.

For transition metal oxides in a given oxidation state (+2), as effective nuclear charge increases across the series (Ti to Mn), ionic character gradually decreases and basicity decreases (moving towards more acidic/amphoteric nature in higher oxides, and less basic monoxides). TiO is more basic and strongly reducing, whereas MnO is least basic among these and weakly basic.

medium

7. According to the intermediate compound formation theory of catalysis, the catalytic action of vanadium(V) oxide in the oxidation of SO₂ to SO₃ involves which sequence of redox changes of vanadium?

  1. V⁵⁺ is first oxidised to V⁶⁺ by O₂, then reduced to V⁵⁺ by SO₂
  2. V⁵⁺ is first reduced to V⁴⁺ by SO₂, then re-oxidised to V⁵⁺ by O₂
  3. V⁵⁺ acts solely as an electron-pair acceptor without changing oxidation state
  4. V⁵⁺ is reduced to metallic vanadium which adsorbs SO₂ on its surface
Show answer and explanation

Answer: (B) V⁵⁺ is first reduced to V⁴⁺ by SO₂, then re-oxidised to V⁵⁺ by O₂

In the Contact process, V₂O₅ oxidises SO₂ to SO₃ while vanadium is reduced from +5 to +4 (forming V₂O₄): V₂O₅ + SO₂ → V₂O₄ + SO₃. The V₂O₄ is subsequently re-oxidised back to V₂O₅ by molecular oxygen: 2V₂O₄ + O₂ → 2V₂O₅.

medium

8. Which of the following pairs of transition metal ions both have a calculated spin-only magnetic moment equal to 3.87 BM?

  1. Cr³⁺ and Co²⁺
  2. V³⁺ and Fe³⁺
  3. Ti³⁺ and Cu²⁺
  4. Mn²⁺ and Fe³⁺
Show answer and explanation

Answer: (A) Cr³⁺ and Co²⁺

A spin-only magnetic moment of 3.87 BM corresponds to √(3(3+2)) = √15 ≈ 3.87 BM, which means n = 3 unpaired electrons. Cr³⁺ is 3d³ (3 unpaired electrons) and Co²⁺ is 3d⁷ (t₂g⁵ eg² in weak field, giving 3 unpaired electrons). Both possess 3 unpaired electrons.

medium

9. Consider the following statements regarding the physical and magnetic properties of first-row transition metal oxides: Statement I: MnO is basic, while Mn₂O₇ is an acidic oxide that dissolves in water to form HMnO₄. Statement II: As the oxidation state of a transition metal increases, the covalent character of its oxide increases due to higher polarising power of the metal cation. In light of the above statements, choose the correct option:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true, but Statement II is false.
  4. Statement I is false, but Statement II is true.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are true.

According to NCERT, in lower oxidation states, transition metal oxides like MnO are basic, ionic, and have low covalent character. In higher oxidation states, such as Mn₂O₇ (+7), high polarising power of the metal cation (Fajans' rules) leads to predominant covalent character, making Mn₂O₇ a covalent, green oil that is strongly acidic and forms HMnO₄ in water.

hard

10. Match List-I (Catalyst / Process) with List-II (Role / Intermediate Oxidation States): List-I: (A) Fe³⁺ in I⁻ and S₂O₈²⁻ reaction (B) V₂O₅ in Contact process (C) TiCl₄ + Al(CH₃)₃ (D) Finely divided Ni List-II: (I) Ziegler-Natta polymerisation (II) Alternates between V(V) and V(IV) (III) Alternates between Fe(III) and Fe(II) (IV) Heterogeneous catalytic hydrogenation Choose the correct match from the options given below:

  1. A-II, B-III, C-I, D-IV
  2. A-IV, B-II, C-I, D-III
  3. A-III, B-I, C-II, D-IV
  4. A-III, B-II, C-I, D-IV
Show answer and explanation

Answer: (D) A-III, B-II, C-I, D-IV

Fe³⁺ oxidises I⁻ to I₂ and gets reduced to Fe²⁺, which is then re-oxidised by S₂O₈²⁻ (A-III). V₂O₅ is reduced to VO₂ (V⁴⁺) by SO₂ and re-oxidised by O₂ back to V₂O₅ (B-II). TiCl₄ with Al(CH₃)₃ is the Ziegler-Natta catalyst used for alkene polymerisation (C-I). Finely divided Ni acts as a heterogeneous catalyst for hydrogenation (D-IV).

medium

11. The spin-only magnetic moment of the free Fe³⁺ ion is:

  1. 5.92 BM
  2. 1.73 BM
  3. 3.87 BM
  4. 2.83 BM
Show answer and explanation

Answer: (A) 5.92 BM

Fe³⁺ has 5 unpaired electrons; μ = √(n(n + 2)) = √(35) = 5.92 BM.

medium

12. Match List-I (Property/Feature) with List-II (Reason/Consequence) for transition elements: List-I: (A) High enthalpies of atomisation (B) Catalytic property (C) Tendency to form interstitial compounds (D) Complex formation ability List-II: (I) Small size, high nuclear charge, and availability of vacant d-orbitals (II) Large number of unpaired electrons involved in metallic bonding (III) Variable oxidation states and ability to form unstable intermediates (IV) Trapping of small non-metal atoms in crystal lattice voids Choose the correct match from the options given below:

  1. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  2. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  3. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
  4. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Show answer and explanation

Answer: (A) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

High enthalpies of atomisation are due to extensive interatomic bonding by unpaired d electrons (A-II). Catalysis relies on variable oxidation states and intermediate formation (B-III). Interstitial compounds form by trapping small atoms like H, C, or N in voids (C-IV). Complex formation is facilitated by small cation size, high charge density, and vacant d-orbitals (D-I).

Keep going with d- and f-Block Elements. All 499 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in d- and f-Block Elements