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Class XII · Chemistry

d- and f-Block Elements: NEET MCQ practice

293 practice questions on d- and f-Block Elements, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

293

Practice questions

143

Easy

144

Medium

6

Hard

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What d- and f-Block Elements covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Transition elements- general introduction, electronic configuration, occurrence, characteristics
  • General trends- physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloy formation
  • Preparation, properties and uses of K2Cr2O7 and KMnO4
  • Lanthanoids- electronic configuration, oxidation states, lanthanoid contraction
  • Actinoids- electronic configuration and oxidation states

Sample questions on d- and f-Block Elements

Try each one, then open the answer and explanation.

Question 1 · Practice question

Malachite is an ore of:

  1. A.Copper
  2. B.Magnesium
  3. C.Calcium
  4. D.Both magnesium and calcium
Show answer

Answer: A. Copper

Malachite is a green mineral with the formula CuCO3·Cu(OH)2, and it is a common ore of copper. Magnesium ores include magnesite and dolomite, while calcium ores include limestone and gypsum. The option 'Both magnesium and calcium' is incorrect because malachite does not contain those metals.

Question 2 · Practice question

Among the following series of transition metal ions, the one where all metal ions have 3d 2 electronic configuration is (Atomic number, Ti = 22, V = 23, Cr = 24, Mn = 25) [AIEEE 2004]

  1. A.Ti 3+, V 2+, Cr 3+, Mn 4+
  2. B.Ti +, V 4+, Cr 6+, Mn 7+
  3. C.Ti 4+, V 3+, Cr 2+, Mn 3+
  4. D.Ti 2+, V 3+, Cr 4+, Mn 5+
Show answer

Answer: D. Ti 2+, V 3+, Cr 4+, Mn 5+

The correct answer is D because for each ion: Ti²⁺ (3d²), V³⁺ (3d²), Cr⁴⁺ (3d²), and Mn⁵⁺ (3d²) — all have exactly two electrons in the 3d subshell. Common mistakes occur if you incorrectly remove electrons from 3d before 4s or misremember the anomalous configuration of chromium.

Question 3 · Practice question

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of + 2 oxidation state will be there in which of the following order?

  1. A.Mn > Cr > Fe > Co
  2. B.Fe > Mn > Co > Cr
  3. C.Co > Mn > Fe > Cr
  4. D.Cr > Mn > Co > Fe
Show answer

Answer: A. Mn > Cr > Fe > Co

The stability of the +2 oxidation state relative to the +3 state is determined by the standard reduction potential E° for M³⁺/M²⁺ couples – the higher the E°, the more stable the +2 state. The known order is Co³⁺/Co²⁺ (+1.97 V) > Mn³⁺/Mn²⁺ (+1.51 V) > Fe³⁺/Fe²⁺ (+0.77 V) > Cr³⁺/Cr²⁺ (-0.41 V), giving exactly Co > Mn > Fe > Cr. Option (a) wrong because Cr²⁺ is least stable, not second; option (b) incorrectly places Fe first, while option (d) places Cr first.

Question 4 · Practice question

The catalytic activity of transition metals and their compounds is mainly due to

  1. A.their magnetic behaviour
  2. B.their ful-filled d-orbitals
  3. C.their ability to adopt variable oxidation states
  4. D.their chemical reactivity
Show answer

Answer: C. their ability to adopt variable oxidation states

Transition metals can change oxidation states easily, which allows them to form intermediate complexes during a reaction and then revert back, speeding up the process. While they do show magnetic behaviour and reactivity, these are not the primary reasons for their catalytic action; the key is their variable oxidation states, a well-established fact in the NCERT syllabus.

Question 5 · Practice question

Many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as 2Cu + → Cu + Cu 2+. This is due to

  1. A.free energy change of reaction is positive
  2. B.free energy of reaction is negative
  3. C.second ionisation energy of Cu is higher than first ionisation energy
  4. D.hydration energy of Cu + is less than Cu 2+
Show answer

Answer: B. free energy of reaction is negative

Cu⁺ disproportionates because Cu²⁺ gets significantly more stabilized in water due to its higher charge/size ratio, leading to much more negative hydration enthalpy than Cu⁺, making the overall free energy change negative. Option (b) is merely a consequence; option (d) is the actual cause given in textbooks.

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