d- and f-Block Elements — NEET Chemistry MCQs with answers

d- and f-Block Elements is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 593 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Transition element general properties · KMnO4 and K2Cr2O7 chemistry · Lanthanoids and actinoids · Colour and magnetism of d-block species · General introduction to d-block series · General properties across the three transition series · Oxidation states across a group · Crystal field effects across the transition series · Applications of transition metals · Oxidation states of transition metals · Magnetic and structural properties across the three transition series · Oxidation states across the three transition series · Standard Electrode Potentials · Coloured Ions · Catalytic Properties · KMnO4 · K2Cr2O7 · f-Block · Alloys and Interstitial Compounds · Oxides · Applications · Complex Formation · Important compounds · Lanthanoids · Actinoids · d-Block Elements · f-Block Elements · Transition Elements · Coordination Compounds · Coordination Compounds (brief overlap concept) · Coordination Compounds (transition metal application) · Variable oxidation states · Color of transition metal ions · Lanthanoid contraction · Catalytic properties · Oxidizing property of KMnO4 · Lanthanoids vs Actinoids · Magnetic properties · Standard electrode potentials · Interstitial compounds

8 free d- and f-Block Elements practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Colour and magnetism of d-block species

   The colour of most transition metal ions in aqueous solution arises mainly due to which phenomenon?

   • A.d-d electronic transitions within partially filled d-orbitals split by the ligand field
   • B.Charge transfer from the metal nucleus to the surrounding water molecules
   • C.Absorption of radiation by completely filled d-orbitals undergoing s-p transitions
   • D.Excitation of electrons from the 4s orbital to the 4p orbital
   Show answer and explanation

   Answer: A. d-d electronic transitions within partially filled d-orbitals split by the ligand field

   In transition metal ions, the ligand field splits degenerate d-orbitals into sets of different energy. Electrons absorb visible light to jump between these split d-levels (d-d transition), and the transmitted complementary colour is observed.

   Colour in transition metal complexes/ions is explained by crystal field theory: surrounding ligands split the five degenerate d-orbitals into lower and higher energy sets (e.g. t2g and eg in octahedral fields). If the metal ion has a partially filled d-subshell (d1-d9), an electron can absorb a photon of visible light and be promoted from the lower to higher set. The wavelength absorbed corresponds to the crystal field splitting energy, and the complementary wavelength is transmitted, giving the observed colour. Ions with d0 or d10 configuration have no such transition available and are typically colourless.

   Common mistake: Attributing colour to vague 'electron excitation' without specifying it is specifically a d-d transition made possible by ligand-field splitting

2. Question 2 · easy · Lanthanoids and actinoids

   The steady decrease in atomic and ionic radii of the lanthanoids from La to Lu is primarily caused by:

   • A.imperfect shielding of one 4f electron by another 4f electron, so effective nuclear charge on outer electrons rises steadily
   • B.complete shielding by the 5d electrons, which cancels the increase in nuclear charge
   • C.a sudden jump in oxidation state from +2 to +4 across the series
   • D.successive addition of electrons to the 6s orbital, which contracts faster than 4f orbitals expand
   Show answer and explanation

   Answer: A. imperfect shielding of one 4f electron by another 4f electron, so effective nuclear charge on outer electrons rises steadily

   As atomic number increases across La to Lu, electrons are added to the inner 4f subshell, which shields the outer electrons from the increasing nuclear charge poorly and imperfectly (worse than d or p electrons). This causes a steady, cumulative decrease in atomic/ionic size called lanthanoid contraction.

   In the lanthanoid series, each added electron enters the 4f orbital, which is buried beneath the 5s and 5p subshells. The 4f orbitals have poor radial shielding of one 4f electron by another, so each successive proton added to the nucleus is not fully compensated by the extra 4f electron's shielding. Effective nuclear charge felt by the outer 6s (and 5d) electrons therefore rises steadily along the series, pulling the electron cloud inward. Since this poor shielding operates cumulatively over all 14 elements, the total contraction (about 20 pm from La3+ to Lu3+) is larger than the contraction seen across a normal d-block row despite each individual step being small.

   Common mistake: Attributing the contraction to d-electron shielding or oxidation state changes instead of poor 4f shielding.

3. Question 3 · easy · Lanthanoids and actinoids

   Zirconium (Zr) and hafnium (Hf), belonging to group 4, show almost identical atomic radii and very similar chemical properties. This is best explained by:

   • A.both elements having the same number of valence electrons only
   • B.hafnium being a synthetic radioactive element with no independent chemistry
   • C.the radius increase expected from adding a new principal shell (period 5 to 6) being almost exactly cancelled by lanthanoid contraction preceding Hf
   • D.Zr and Hf being isotopes of the same element
   Show answer and explanation

   Answer: C. the radius increase expected from adding a new principal shell (period 5 to 6) being almost exactly cancelled by lanthanoid contraction preceding Hf

   Hafnium follows the lanthanoids (La to Lu) in period 6, so its expected radius increase over Zr (from an added principal quantum shell) is almost exactly cancelled by the cumulative lanthanoid contraction, leaving Zr and Hf with nearly equal radii and near-identical chemistry.

   Moving down a group normally increases atomic radius due to an added principal shell. Zr (period 5, after Y) would normally be smaller than Hf (period 6, after La and the lanthanoids) by this logic, but Hf's radius would actually be larger without any correction. However, because the 14 lanthanoids intervene between La and Hf, the lanthanoid contraction shrinks the effective size of Hf by an amount that almost exactly offsets the expected increase from the extra shell. The near-identical radii (Zr ~160 pm, Hf ~159 pm) make their ions difficult to separate chemically — the same is true for the Nb/Ta and Mo/W pairs in later groups.

   Common mistake: Thinking radius similarity is coincidental or attributing it to isotopy rather than the intervening f-block contraction.

4. Question 4 · medium · Lanthanoids and actinoids

   Consider the following statements comparing lanthanoids and actinoids: (I) Actinoids show a much wider range of oxidation states than lanthanoids because 5f, 6d and 7s orbitals lie close in energy and all can participate in bonding. (II) All known actinoids are radioactive, whereas only promethium among the lanthanoids is radioactive. (III) 5f orbitals are shielded by outer electrons more effectively than 4f orbitals are, making actinoid compounds purely ionic like lanthanoid compounds. (IV) Actinoid contraction per element is slightly greater than lanthanoid contraction per element due to poorer shielding by 5f electrons. Which statements are correct?

   • A.I, II and III only
   • B.I, II and IV only
   • C.II, III and IV only
   • D.I, III and IV only
   Show answer and explanation

   Answer: B. I, II and IV only

   Statements I, II and IV are established facts: actinoids show a wider oxidation-state range (5f/6d/7s near-degeneracy), all actinoids are radioactive (only Pm is radioactive among lanthanoids), and actinoid contraction per element exceeds lanthanoid contraction. Statement III is false — 5f orbitals are shielded LESS effectively than 4f orbitals, giving actinoid bonding greater covalent character, not purely ionic character.

   (I) True: because 5f, 6d and 7s in actinoids are comparable in energy, more electrons can be involved in bonding, giving oxidation states from +3 up to +7 (e.g., Np, Pu, Am), unlike lanthanoids which are overwhelmingly +3. (II) True: all actinoids (Th to Lr) are radioactive; among lanthanoids only promethium (Pm, Z=61) is radioactive, the rest are stable. (III) False: 5f orbitals extend further from the nucleus and are shielded by the intervening 6s/6p electrons less effectively than 4f orbitals are shielded, so actinoid ions show greater covalent character in bonding than lanthanoid ions, not purely ionic behaviour. (IV) True: because of poorer 5f shielding, the radius decrease per element (actinoid contraction) is somewhat larger than the analogous decrease per element in lanthanoid contraction. Hence only I, II, IV are correct — option B.

   Common mistake: Accepting statement III as true by assuming 5f behaves like 4f in shielding and bonding character.

5. Question 5 · easy · Lanthanoids and actinoids

   As the lanthanoid series is traversed from La(OH)3 to Lu(OH)3, the basic strength of the hydroxides:

   • A.increases steadily, because nuclear charge increases along the series
   • B.remains exactly constant, since all lanthanoid ions carry the same +3 charge
   • C.shows no regular trend and varies randomly from element to element
   • D.decreases steadily, because the decreasing ionic radius (lanthanoid contraction) increases charge density and covalent character of the M-OH bond
   Show answer and explanation

   Answer: D. decreases steadily, because the decreasing ionic radius (lanthanoid contraction) increases charge density and covalent character of the M-OH bond

   As Ln3+ ionic radius shrinks across the series due to lanthanoid contraction, charge density on the smaller cation increases, giving the M-OH bond more covalent character and making the hydroxide less basic. Hence La(OH)3 is the most basic and Lu(OH)3 the least basic lanthanoid hydroxide.

   Base strength of an ionic hydroxide correlates inversely with the covalent character of the metal-oxygen bond, which in turn increases as the cation becomes smaller and more polarising (Fajans' rules intuition, applicable qualitatively here). Across La3+ to Lu3+, the charge remains fixed at +3 but ionic radius steadily decreases (lanthanoid contraction), so charge density (charge/radius) rises steadily. Higher charge density polarises the O-H bond region more, giving increasingly covalent, less ionic M-OH character, which reduces the tendency to release OH- ions in solution. Consequently basicity decreases smoothly from La(OH)3 (most basic, largest ion) to Lu(OH)3 (least basic, smallest ion) — this is a standard NCERT-level consequence of lanthanoid contraction.

   Common mistake: Believing basicity should increase with nuclear charge, ignoring that ionic size and charge density are the operative factors here.

6. Question 6 · medium · Lanthanoids and actinoids

   Chemically separating the individual lanthanoid elements from one another (e.g., by ion-exchange chromatography) is notoriously difficult mainly because:

   • A.all lanthanoids are radioactive and decay before separation can be completed
   • B.the lanthanoid contraction causes only a very small, gradual decrease in ionic radius across 14 chemically similar +3 ions, so their chemical behaviour (solubility, complex formation) is nearly identical
   • C.lanthanoids do not form stable complexes with any chelating agents, preventing ion-exchange based separation
   • D.each lanthanoid has a completely different, unpredictable oxidation state, making a common separation method impossible
   Show answer and explanation

   Answer: B. the lanthanoid contraction causes only a very small, gradual decrease in ionic radius across 14 chemically similar +3 ions, so their chemical behaviour (solubility, complex formation) is nearly identical

   Because lanthanoid contraction changes ionic radius only very gradually across 14 elements while the oxidation state stays uniformly +3, the lanthanoids are extremely close in chemical behaviour. This near-identical behaviour (not radioactivity or lack of complexation) is what makes separating them by ordinary chemical means so difficult.

   Separation methods rely on differences in chemical/physical behaviour. Within the lanthanoid series, all elements adopt the same characteristic +3 oxidation state and their ionic radii change only slightly and gradually from element to element (roughly 1 pm per step) due to lanthanoid contraction. This means solubility products, complex stability constants, and ion-exchange affinities are all extremely close for neighbouring lanthanoids, so classical precipitation or single-pass separation fails. Modern methods (ion-exchange chromatography with citrate or similar complexing eluents, solvent extraction) exploit the very small but real systematic radius-dependent differences in complex stability to achieve separation only through many repeated equilibration stages, historically requiring thousands of steps before ion-exchange resins made it practical.

   Common mistake: Attributing separation difficulty to radioactivity or absence of complex formation rather than the small, gradual radius change causing near-identical chemistry.

7. Question 7 · medium · Lanthanoids and actinoids

   Unlike the lanthanoids, several actinoids (e.g., uranium, neptunium, plutonium) readily show oxidation states well above +4, such as +5, +6, and even +7. The chemical reason for this difference is that:

   • A.actinoids have larger atomic radii than lanthanoids, and larger atoms always show more variable oxidation states
   • B.in actinoids, the 5f, 6d and 7s orbitals are comparable in energy, so electrons from all three can be lost in bonding, unlike lanthanoids where 4f electrons are too deeply buried (energetically well below 5d/6s) to participate readily
   • C.actinoids are transition metals while lanthanoids are inner transition metals, and only transition metals can exceed +4
   • D.lanthanoids are non-metals whereas actinoids are metals, and only metals can show oxidation states above +4
   Show answer and explanation

   Answer: B. in actinoids, the 5f, 6d and 7s orbitals are comparable in energy, so electrons from all three can be lost in bonding, unlike lanthanoids where 4f electrons are too deeply buried (energetically well below 5d/6s) to participate readily

   In actinoids, the 5f, 6d and 7s orbitals lie close enough in energy that electrons from all three can participate in bonding, allowing higher oxidation states (+5, +6, +7 in U, Np, Pu). In lanthanoids, the 4f orbitals are much lower in energy than 5d/6s and are too contracted/buried to readily participate, restricting most lanthanoids to +3.

   The key structural difference is orbital energy separation. For lanthanoids, once the 4f subshell begins filling, its orbitals drop to noticeably lower energy than 5d and 6s and become spatially contracted (shielded within the atom), so 4f electrons are essentially chemically inert core-like electrons for most of the series — only the outer 5d/6s (and occasionally one f electron under special f0/f7/f14 driving forces) are lost in bonding, capping most lanthanoids at +3. In actinoids, 5f, 6d and 7s remain much closer in relative energy (partly due to relativistic effects and weaker actinide 5f binding), so more electrons across these three subshells are energetically accessible for bonding. This allows early actinoids especially (Th through Am) to reach higher oxidation states — e.g., UF6 (U +6), NpO2+ (Np +5/+6), PuO2^2+ (Pu +6) — a pattern not seen for lanthanoid analogues.

   Common mistake: Attributing the oxidation-state range difference to size, metallic character, or classification labels instead of orbital energy proximity.

8. Question 8 · medium · Lanthanoids and actinoids

   A student reasons: "Since actinoids have partially filled 5f orbitals just as lanthanoids have partially filled 4f orbitals, actinoid ions should show weak, pale f-f transition colours exactly like lanthanoid ions, and their compounds should be similarly poor at forming covalent bonds to organic/carbonyl ligands." What is wrong with this reasoning?

   • A.Nothing is wrong; actinoid and lanthanoid ions behave identically in colour intensity and covalent bonding ability because both have partially filled f orbitals
   • B.Actinoid ions are colourless because 5f-5f transitions are strictly forbidden by symmetry, unlike 4f-4f transitions in lanthanoids
   • C.The reasoning wrongly assumes 5f orbitals behave like 4f orbitals in extent and shielding; because 5f orbitals are more spatially extended and less shielded, they participate more in covalent bonding (e.g., with carbonate, oxalate, cyclopentadienyl-type ligands) and often give more intense colours/charge-transfer bands than the largely non-bonding 4f orbitals of lanthanoids
   • D.The reasoning is wrong only because lanthanoids, not actinoids, are radioactive, which is unrelated to colour or bonding at all
   Show answer and explanation

   Answer: C. The reasoning wrongly assumes 5f orbitals behave like 4f orbitals in extent and shielding; because 5f orbitals are more spatially extended and less shielded, they participate more in covalent bonding (e.g., with carbonate, oxalate, cyclopentadienyl-type ligands) and often give more intense colours/charge-transfer bands than the largely non-bonding 4f orbitals of lanthanoids

   The student incorrectly assumes 5f orbitals behave exactly like 4f orbitals. In reality 5f orbitals are more radially extended and less effectively shielded than 4f, so they participate more significantly in covalent bonding and often produce more intense colours/charge-transfer character in actinoid complexes than the largely core-like, weakly interacting 4f orbitals of lanthanoids.

   Both f-block series have partially filled f orbitals in many of their common ions, so both can show f-f transition colours in principle. However, the degree of orbital involvement in bonding differs sharply: lanthanoid 4f orbitals are contracted and shielded by filled 5s/5p subshells, making them largely non-bonding/core-like, so lanthanoid coordination chemistry is dominated by ionic, electrostatic interactions with only weak, pale f-f transition colours. Actinoid 5f orbitals, by contrast, are more radially extended and less well shielded by the 6s/6p subshells, allowing greater orbital overlap with ligand orbitals; this gives actinoid complexes noticeably more covalent character (evident in organometallic/coordination compounds of U, Np, Pu) and often more intense colours, partly from enhanced f-f transition intensity and partly from charge-transfer bands not significant in lanthanoid chemistry. The student's error is treating '5f is partially filled just like 4f' as implying identical orbital behaviour, when the spatial extent/shielding difference is the crux of the actinoid-lanthanoid distinction.

   Common mistake: Assuming lanthanoids and actinoids behave identically simply because both are f-block series with partially filled f orbitals.

Questions about d- and f-Block Elements for NEET

More Chemistry chapters

← Basic Principles of Organic ChemistryAll Chemistry chaptersp-Block Elements →

Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (593 active practice questions in this chapter today).