Properties of Transition Elements Transition Elements – General introduction: NEET MCQs with Answers

15 free questions on Properties of Transition Elements Transition Elements – General introduction for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: General introduction, electronic configuration, occurrence and characteristics, general trends in properties of the first-row transition elements - physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation,

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Question 1

Previous-year question
Identify the INCORRECT statement. [NEET 2020]
  1. Transition metals and their compounds are known for catalytic activity because of their multiple oxidation states and ability to form complexes
  2. Interstitial compounds are formed when small atoms such as H, C or N are trapped inside the crystal lattices of metals
  3. The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same
  4. Cr²⁺ (d⁴) is a stronger reducing agent than Fe²⁺ (d⁶) in water
Show answer
Answer: (C) The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same
Explanation: In CrO₄²⁻, x + 4(–2) = –2 gives x = +6; in Cr₂O₇²⁻, 2x + 7(–2) = –2 gives x = +6. Chromium is +6 in both ions, so the claim that the states differ is false.

Question 2

Previous-year question
Match the metal ions in Column I with their spin-only magnetic moments in Column II. Column I: (A) Co³⁺ (B) Cr³⁺ (C) Fe³⁺ (D) Ni²⁺ Column II: (i) √8 BM (ii) √35 BM (iii) √3 BM (iv) √24 BM (v) √15 BM [NEET 2018]
  1. A–(iv), B–(v), C–(ii), D–(i)
  2. A–(i), B–(ii), C–(iii), D–(iv)
  3. A–(iii), B–(v), C–(i), D–(ii)
  4. A–(iv), B–(i), C–(ii), D–(iii)
Show answer
Answer: (A) A–(iv), B–(v), C–(ii), D–(i)
Explanation: Co³⁺ is d⁶ with 4 unpaired electrons, so μ = √24; Cr³⁺ is d³ with 3, so √15; Fe³⁺ is d⁵ with 5, so √35; Ni²⁺ is d⁸ with 2, so √8.

Question 3

Previous-year question
Because of lanthanoid contraction, which of the following pairs of elements have nearly the same atomic radii? (Atomic numbers are given in brackets.) [AIPMT 2015]
  1. Zr (40) and Nb (41)
  2. Zr (40) and Hf (72)
  3. Zr (40) and Ta (73)
  4. Ti (22) and Zr (40)
Show answer
Answer: (B) Zr (40) and Hf (72)
Explanation: Zr and Hf lie one below the other in group 4. The 14 lanthanoids intervene before Hf and their poor 4f shielding contracts the size, so the expected increase from Zr to Hf is wiped out and both have radii of about 159–160 pm.

Question 4

Previous-year question
Which is the correct order of increasing energy of the listed orbitals in the titanium atom? [AIPMT 2015]
  1. 3s 4s 3p 3d
  2. 4s 3s 3p 3d
  3. 3s 3p 3d 4s
  4. 3s 3p 4s 3d
Show answer
Answer: (D) 3s 3p 4s 3d
Explanation: By the (n + l) rule: 3s = 3, 3p = 4, 4s = 4 and 3d = 5. Between 3p and 4s, which tie at 4, the lower n comes first, so the increasing order is 3s < 3p < 4s < 3d.

Question 5

Previous-year question
The number of d-electrons in Fe²⁺ (Z = 26) is NOT equal to the number of electrons in which one of the following? [AIPMT 2015]
  1. p-electrons in Cl (Z = 17)
  2. d-electrons in Fe (Z = 26)
  3. p-electrons in Ne (Z = 10)
  4. s-electrons in Mg (Z = 12)
Show answer
Answer: (A) p-electrons in Cl (Z = 17)
Explanation: Fe²⁺ is 3d⁶, i.e. 6 d-electrons. Chlorine has 2p⁶ 3p⁵ = 11 p-electrons, which does not match. Fe atom has 6 d-electrons, Ne has 6 p-electrons and Mg has 1s² 2s² 3s² = 6 s-electrons, all equal to 6.

Question 6

Previous-year question
A magnetic moment of 2.84 BM is shown by (At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27) [AIPMT 2015]
  1. Ti³⁺
  2. Cr²⁺
  3. Co²⁺
  4. Ni²⁺
Show answer
Answer: (D) Ni²⁺
Explanation: μ = 2.84 BM corresponds to √8, i.e. n = 2 unpaired electrons. Ni²⁺ is 3d⁸ and has exactly two unpaired electrons. Ti³⁺ (d¹) gives 1.73, Cr²⁺ (d⁴) 4.90 and Co²⁺ (d⁷) 3.87 BM.

Question 7

Previous-year question
The pair of compounds that can exist together in solution is [AIPMT 2014]
  1. FeCl₃, SnCl₂
  2. HgCl₂, SnCl₂
  3. FeCl₂, SnCl₂
  4. FeCl₃, KI
Show answer
Answer: (C) FeCl₂, SnCl₂
Explanation: FeCl₂ and SnCl₂ are both reducing agents, so no redox reaction occurs between them and they coexist. In the other pairs the oxidising species (Fe³⁺ or Hg²⁺) is reduced by Sn²⁺ or oxidises I⁻, so the two cannot remain together.
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Question 8

Previous-year question
Cu²⁺ salts are more stable than Cu⁺ salts in aqueous solution because of [NEET 2023 · NCERT (New) Pg. 100]
  1. enthalpy of atomisation
  2. hydration energy
  3. second ionisation enthalpy
  4. first ionisation enthalpy
Show answer
Answer: (B) hydration energy
Explanation: In aqueous medium Cu⁺ undergoes disproportionation: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s) Although the second ionisation enthalpy of copper is large, the hydration enthalpy of the smaller, doubly charged Cu²⁺ ion (≈ −2100 kJ mol⁻¹) is far more negative than that of Cu⁺ (≈ −580 kJ mol⁻¹). This large excess of hydration energy more than compensates for the second ionisation enthalpy, so Cu²⁺ is the stable species in water.

Question 9

Previous-year question
The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺. This is because the reduction of Mn³⁺ involves a change of [NEET 2024]
  1. d⁵ to d⁴ configuration
  2. d⁵ to d² configuration
  3. d⁴ to d⁵ configuration
  4. d³ to d⁵ configuration
Show answer
Answer: (C) d⁴ to d⁵ configuration
Explanation: Mn³⁺ is [Ar]3d⁴. On accepting one electron it becomes Mn²⁺ = [Ar]3d⁵, an exactly half-filled subshell with maximum exchange energy. This strong thermodynamic driving force makes Mn³⁺ a powerful oxidising agent, so E°(Mn³⁺/Mn²⁺) = +1.57 V. By contrast, reducing Cr³⁺ (d³, itself a stable half-filled t₂g set) to Cr²⁺ (d⁴) destroys stability, giving E° = −0.41 V; and reducing Fe³⁺ (d⁵) to Fe²⁺ (d⁶) destroys a half-filled set, giving a moderate +0.77 V.

Question 10

Previous-year question
The transition elements have the general electronic configuration [AIPMT 1991]
  1. ns²np⁶nd¹⁻¹⁰
  2. (n−1)d¹⁻¹⁰, ns⁰⁻², np⁰⁻⁶
  3. (n−1)d¹⁻¹⁰ ns¹⁻²
  4. nd¹⁻¹⁰ ns²
Show answer
Answer: (C) (n−1)d¹⁻¹⁰ ns¹⁻²
Explanation: In the transition elements the d-orbitals of the penultimate shell are filled progressively while the outermost s-orbital holds one or two electrons. The general configuration is therefore (n−1)d¹⁻¹⁰ ns¹⁻². Here (n−1) refers to the inner shell whose d-orbitals may hold one to ten electrons, and the outermost ns orbital may hold one or two. The ns¹ cases in the 3d series are chromium ([Ar]3d⁵4s¹) and copper ([Ar]3d¹⁰4s¹), where the half-filled and fully filled d subshells provide extra stability.

Question 11

Previous-year question
The common oxidation states of titanium are [AIPMT 1994]
  1. +2, +3
  2. +3, +4
  3. +3, −4
  4. +2, +3, +4
Show answer
Answer: (B) +3, +4
Explanation: Titanium is [Ar]3d²4s². The +4 state is by far the most common and stable, since losing all four valence electrons gives the noble-gas core of argon (as in TiO₂ and TiCl₄). The +3 state is the next most familiar (TiCl₃, the violet Ti³⁺ ion). The +2 state does exist (TiO, TiCl₂) but Ti²⁺ is very strongly reducing and is not encountered in ordinary chemistry, so +3 and +4 are described as the common states.

Question 12

Previous-year question
Mercury is the only metal which is liquid at 0 °C. This is due to its [AIPMT 1995]
  1. high vapour pressure
  2. weak metallic bond
  3. high ionisation energy
  4. both (2) and (3)
Show answer
Answer: (D) both (2) and (3)
Explanation: Mercury has a very high ionisation enthalpy, so its 6s electrons take little part in metallic bonding. The metallic bond is therefore very weak (enthalpy of atomisation about 61 kJ mol⁻¹), and mercury remains liquid at 0 °C, for both these reasons.

Question 13

Previous-year question
Which of the following does not represent the correct order of the property indicated? [AIPMT 1997]
  1. Ni²⁺ > Cr²⁺ > Fe²⁺ > Mn²⁺ (ionic size)
  2. Sc > Ti > Cr > Mn (atomic size)
  3. Mn²⁺ > Fe²⁺ > Co²⁺ > Ni²⁺ (number of unpaired electrons)
  4. Fe²⁺ > Co²⁺ > Ni²⁺ > Cu²⁺ (number of unpaired electrons)
Show answer
Answer: (A) Ni²⁺ > Cr²⁺ > Fe²⁺ > Mn²⁺ (ionic size)
Explanation: M²⁺ radii generally decrease across the series as effective nuclear charge rises, so Ni²⁺ is the smallest of these ions and cannot head an order of ionic size. The atomic size order Sc > Ti > Cr > Mn and the unpaired electron orders (Mn²⁺ 5, Fe²⁺ 4, Co²⁺ 3, Ni²⁺ 2, Cu²⁺ 1) are correct.

Question 14

Previous-year question
Which of the following elements is responsible for the oxidation of water to O₂ in biological processes? [AIPMT 1997]
  1. Fe
  2. Cu
  3. Mn
  4. Mo
Show answer
Answer: (C) Mn
Explanation: The oxygen-evolving complex of photosystem II in green plants contains a Mn₄CaO₅ cluster which carries out the four-electron oxidation of water to dioxygen during photosynthesis: 2H₂O → O₂ + 4H⁺ + 4e⁻ For comparison, iron occurs in haemoglobin and the cytochromes, copper in plastocyanin and cytochrome c oxidase, and molybdenum in nitrogenase and nitrate reductase.

Question 15

Previous-year question
Which one of the following ionic species will impart colour to an aqueous solution? [AIPMT 1998]
  1. Ti⁴⁺
  2. Cu⁺
  3. Zn²⁺
  4. Cr³⁺
Show answer
Answer: (D) Cr³⁺
Explanation: Colour arises from d–d transitions, which require a partly filled d subshell. Cr³⁺ = [Ar]3d³ → 3 unpaired electrons, coloured ✔ Ti⁴⁺ = [Ar]3d⁰ → colourless Cu⁺ = [Ar]3d¹⁰ → colourless Zn²⁺ = [Ar]3d¹⁰ → colourless Only Cr³⁺ has an incompletely filled d subshell, so only it is coloured.

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