The d- and f-Block Elements NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every The d- and f-Block Elements question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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The d- and f-Block Elements previous year questions
40 questions. Tap an option to check your answer.
Q1NEET 2021
Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of :
Not quite — the answer is C.
Between Zr (4d) and Hf (5d) the fourteen 4f elements are filled; poor shielding by 4f electrons causes the lanthanoid contraction, cancelling the expected size increase. Being in the same group alone would predict Hf to be larger.
Q2NEET 2021
The incorrect statement among the following is :
Not quite — the answer is B.
Most trivalent lanthanoid ions are coloured in the solid state and in solution because of f–f transitions (La³⁺ and Lu³⁺ are exceptions), so calling them colourless is the incorrect statement. Actinoid contraction from element to element is indeed greater, because 5f electrons shield poorly.
Q3NEET 2020 · Q103
The calculated spin only magnetic moment of Cr²⁺ ion is :
Not quite — the answer is B.
Cr²⁺ is 3d⁴ with 4 unpaired electrons, so μ = √(n(n+2)) = √24 ≈ 4.90 BM. The value 3.87 BM corresponds to 3 unpaired electrons (Cr³⁺) and 5.92 BM to 5 unpaired electrons.
Q4NEET 2020 · Q126
Identify the incorrect statement.
Not quite — the answer is D.
In CrO₄²⁻, x − 8 = −2 gives +6, and in Cr₂O₇²⁻, 2x − 14 = −2 gives +6, so chromium has the same oxidation state in both; the statement saying they differ is incorrect. Cr²⁺ does reduce more strongly than Fe²⁺ (Cr²⁺ → Cr³⁺ reaches stable t₂g³), and the statements on catalysis and interstitial compounds are true.
Q5NEET 2019 · Q80
The manganate and permanganate ions are tetrahedral, due to:
Not quite — the answer is A.
In MnO₄²⁻ and MnO₄⁻ the Mn is tetrahedrally σ-bonded to four O atoms, and π-bonding arises from overlap of filled oxygen p-orbitals with manganese d-orbitals (NCERT). Oxygen has no d-orbitals in its valence shell, so d–d overlap is impossible.
Q6NEET 2018 · Q69
Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code: Column I: a. Co³⁺ b. Cr³⁺ c. Fe³⁺ d. Ni²⁺; Column II: i. √8 B.M. ii. √35 B.M. iii. √3 B.M. iv. √24 B.M. v. √15 B.M.
Not quite — the answer is C.
μ = √(n(n+2)): Co³⁺ (3d⁶, 4 unpaired) √24; Cr³⁺ (3d³, 3 unpaired) √15; Fe³⁺ (3d⁵, 5 unpaired) √35; Ni²⁺ (3d⁸, 2 unpaired) √8. Matching Ni²⁺ with √3 would assume only one unpaired electron.
Q7NEET 2018 · Q70
Which one of the following ions exhibits d-d transition and paramagnetism as well?
Not quite — the answer is D.
In MnO₄²⁻ manganese is +6 (3d¹): one unpaired electron makes it paramagnetic and allows d-d transitions. MnO₄⁻ (Mn⁺⁷), Cr₂O₇²⁻ and CrO₄²⁻ (Cr⁺⁶) are d⁰ and diamagnetic; their colour is due to charge transfer.
Q8NEET 2017 · Q1
HgCl₂ and I₂ both when dissolved in water containing I⁻ ions the pair of species formed is:
Not quite — the answer is C.
In excess iodide, HgCl₂ first gives HgI₂, which dissolves in more I⁻ as the complex HgI₄²⁻ (as in Nessler's reagent), while I₂ combines with I⁻ to form the triiodide ion I₃⁻. HgI₂ cannot remain as such because excess I⁻ is present.
Q9NEET 2017 · Q35
The reason for greater range of oxidation states in actinoids is attributed to:
Not quite — the answer is C.
In actinoids the 5f, 6d and 7s orbitals are close in energy, so varying numbers of electrons from all three can be used in bonding, giving many oxidation states. 4f and 5d levels belong to the lanthanoids, and radioactivity or contraction does not create extra oxidation states.
Q10NEET 2017 (Gujarati paper) · Q21
Which of the following lanthanoids shows +4 oxidation state to acquire noble gas configuration? (At. nos.: La = 57, Ce = 58, Eu = 63 and Yb = 70)
Not quite — the answer is A.
Ce is [Xe]4f¹5d¹6s²; losing four electrons gives Ce⁴⁺ with the noble gas configuration [Xe]. Eu and Yb show +2 to reach stable 4f⁷ and 4f¹⁴, and La shows only +3, which already gives [Xe].
Q11NEET 2016 Phase 1 · Q59
Which one of the following statements is correct when SO₂ is passed through acidified K₂Cr₂O₇ solution?
Not quite — the answer is D.
SO₂ acts as a reducing agent: it is oxidised to sulphate while orange dichromate (Cr +6) is reduced to Cr³⁺, giving green Cr₂(SO₄)₃. The solution turns green rather than colourless, and SO₂ is oxidised, not reduced.
Q12NEET 2016 Phase 1 · Q60
The electronic configurations of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:
Not quite — the answer is D.
Eu (63) has a half-filled 4f⁷ shell: [Xe]4f⁷6s². In Gd (64) the extra electron goes to 5d to keep the stable 4f⁷: [Xe]4f⁷5d¹6s². Tb (65) has [Xe]4f⁹6s². Writing Gd as 4f⁸6s² ignores the extra stability of the half-filled f subshell.
Q13AIPMT 2015
Magnetic moment 2.84 B.M. is given by : (At. Nos., Ni = 28, Ti = 22, Cr = 24, Co = 27)
Not quite — the answer is B.
μ = √(n(n + 2)) = 2.84 BM gives n = 2. Ni²⁺ (3d⁸) has 2 unpaired electrons, whereas Ti³⁺ (3d¹) has 1, Co²⁺ (3d⁷) has 3 and Cr²⁺ (3d⁴) has 4.
Q14AIPMT 2015
Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii ? (Numbers in the parenthesis are atomic numbers).
Not quite — the answer is D.
The 14 lanthanoids between La and Hf cause a contraction that cancels the expected size increase from the 4d to the 5d series. So Zr (160 pm) and Hf (159 pm) have almost the same radius. Ti and Zr differ normally because no f-block intervenes.
Q15AIPMT 2014
The reaction of aqueous KMnO₄ with H₂O₂ in acidic conditions gives:
Not quite — the answer is A.
In acid, MnO₄⁻ is reduced to Mn²⁺ while H₂O₂ is oxidised to O₂: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. Ozone is never formed, and Mn⁴⁺/MnO₂ forms only in neutral or weakly basic medium.
Q16AIPMT 2014
In acidic medium, H₂O₂ changes Cr₂O₇²⁻ to CrO₅ which has two (–O–O–) bonds. Oxidation state of Cr in CrO₅ is:
Not quite — the answer is B.
CrO₅ has one oxide oxygen (−2) and two peroxide O–O groups (four O at −1). x − 2 − 4 = 0 gives x = +6; treating all five O as −2 would give the impossible +10.
Q17AIPMT 2014
Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28)
Not quite — the answer is A.
μ = √(n(n + 2)) = 2.83 BM gives n = 2 unpaired electrons. Ni²⁺ (3d⁸) has 2, while Cr³⁺ (3d³) has 3, Mn²⁺ (3d⁵) has 5 and Ti³⁺ (3d¹) has 1.
Q18AIPMT 2014
Reason of lanthanoid contraction is:
Not quite — the answer is D.
4f orbitals are diffuse and shield the outer electrons poorly, so as nuclear charge rises across the series the effective nuclear charge increases and radii shrink. Increasing nuclear charge alone would not explain it; the poor f-shielding is the cause.
Q19NEET 2013
Which of the following does not give oxygen on heating ?
Not quite — the answer is D.
(NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O on heating, releasing nitrogen, not oxygen. KClO₃ and Zn(ClO₃)₂ release O₂, and K₂Cr₂O₇ on strong heating gives K₂CrO₄, Cr₂O₃ and O₂.
Q20NEET 2013
Which of the following lanthanoid ions is diamagnetic ? (At. No. Ce = 58, Sm = 62, Eu = 63, Yb = 70)
Not quite — the answer is D.
Yb (Z = 70) is [Xe]4f¹⁴6s², so Yb²⁺ is 4f¹⁴ with every electron paired: diamagnetic. Ce²⁺ (4f²), Sm²⁺ (4f⁶) and Eu²⁺ (4f⁷) all have unpaired f electrons.
Q21NEET 2013
KMnO₄ can be prepared from K₂MnO₄ as per the reaction: 3MnO₄²⁻ + 2H₂O ⇌ 2MnO₄⁻ + MnO₂ + 4OH⁻. The reaction can go to completion by removing OH⁻ ions by adding
Not quite — the answer is C.
CO₂ is weakly acidic and neutralises OH⁻ (forming carbonate) without reacting with the permanganate, pushing the equilibrium to the right. HCl and SO₂ are reducing agents that would be oxidised by MnO₄⁻, and KOH would add OH⁻ and push the reaction backwards.
Q22NEET 2013
Which of the following statements about the interstitial compounds is incorrect?
Not quite — the answer is B.
Interstitial compounds (small atoms like H, C, N trapped in metal lattices) keep metallic conductivity, are very hard and have high melting points, but they are chemically inert. So calling them chemically reactive is incorrect.
Q23AIPMT 2012 Mains
Which one of the following does not correctly represent the correct order of the property indicated against it?
Not quite — the answer is C.
Manganese has a much lower melting point (about 1246 °C) than chromium (about 1907 °C) because its stable 3d⁵4s² configuration weakens metallic bonding, so Ti < V < Cr < Mn for melting point is wrong. The other orders are right: Cr has the highest 2nd ionisation enthalpy (removing an electron from stable 3d⁵), and unpaired electrons rise from d¹ to d⁴ across Ti³⁺ to Mn³⁺.
Q24AIPMT 2012 Mains
Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential E°(M²⁺/M) value has a positive sign?
Not quite — the answer is C.
Copper is the only one of these with a positive E°(M²⁺/M), about +0.34 V, because its high atomisation enthalpy and low hydration enthalpy are not compensated. Fe, Co and Ni all have negative values.
Q25AIPMT 2012 Mains
Which of the following exhibits only +3 oxidation state?
Not quite — the answer is C.
Actinium (5f⁰) shows only the +3 oxidation state. Thorium is mainly +4, protactinium up to +5 and uranium shows +3 to +6.
Q26AIPMT 2012 Prelims
Which of the statements is not true?
Not quite — the answer is C.
Na₂Cr₂O₇ is hygroscopic, so K₂Cr₂O₇, which is not, is preferred as a primary standard in volumetric analysis; the statement favouring sodium dichromate is false. The others are true: raising pH converts orange Cr₂O₇²⁻ to yellow CrO₄²⁻, and H₂S is oxidised to colloidal sulphur, giving a milky appearance.
Q27AIPMT 2011 Prelims
Acidified K₂Cr₂O₇ solution turns green when Na₂SO₃ is added to it. This is due to the formation of
Not quite — the answer is B.
Sulphite reduces dichromate in acid: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. The green colour is due to Cr³⁺, i.e. Cr₂(SO₄)₃. Chromate is yellow and Cr would not be reduced to +2.
Q28AIPMT 2010 Mains · Q114
Which of the following oxidation states is the most common among the lanthanoids?
Not quite — the answer is D.
All lanthanoids show the +3 oxidation state, which is the characteristic and most common one. +2 (Eu, Yb) and +4 (Ce) occur only for a few elements with stable f⁰, f⁷ or f¹⁴ configurations.
Q29AIPMT 2010 Prelims · Q54
Which of the following pairs has the same size?
Not quite — the answer is D.
Because of the lanthanoid contraction, Hf (5d) is almost the same size as Zr (4d), so Zr⁴⁺ and Hf⁴⁺ have nearly identical radii. Ti⁴⁺ is a 3d ion and clearly smaller than Zr⁴⁺.
Q30AIPMT 2010 Prelims · Q65
Which one of the following ions has electronic configuration [Ar]3d⁶? (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
Not quite — the answer is A.
Co (Z = 27) is [Ar]3d⁷4s²; removing 3 electrons gives Co³⁺ = [Ar]3d⁶. Fe³⁺ is 3d⁵, Mn³⁺ is 3d⁴ and Ni³⁺ is 3d⁷.
Q31AIPMT 2010 Prelims · Q68
Which of the following ions will exhibit colour in aqueous solutions?
Not quite — the answer is C.
Ti³⁺ is 3d¹ with one unpaired electron, so a d-d transition gives colour (violet). Sc³⁺ (3d⁰), La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) have no unpaired d/f electrons and are colourless.
Q32AIPMT 2009 Prelims · Q156
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
Not quite — the answer is B.
3d⁵4s² (manganese) has seven electrons that can take part in bonding and shows oxidation states from +2 up to +7, the widest range here. 3d⁵4s¹ (Cr) goes up to +6, and the others to +4 and +5.
Q33AIPMT 2008 Prelims · Q71
The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is
Not quite — the answer is B.
The second electron is removed from M⁺. Cr⁺ is 3d⁵, a stable half-filled shell, so removing a 3d electron costs the most. Mn⁺ (3d⁵4s¹) loses its 4s electron more easily, and V⁺ and Ti⁺ follow in order of nuclear charge. Observed values (≈1590, 1510, 1410, 1310 kJ mol⁻¹) give Cr > Mn > V > Ti. Mn > Cr would hold for the third, not the second, ionisation enthalpy.
Q34AIPMT 2007 Prelims · Q84
Which one of the following ions is the most stable in aqueous solution? (At. No. Ti = 22, V = 23, Cr = 24, Mn = 25)
Not quite — the answer is D.
Cr³⁺ (3d³) has a half-filled t₂g set and a very high hydration energy, making it the most stable in water. Ti³⁺ and V³⁺ are oxidised easily, and Mn³⁺ (d⁴) disproportionates.
Q35AIPMT 2007 Prelims · Q100
Identify the incorrect statement among the following:
Not quite — the answer is B.
Because of lanthanoid contraction, 4d and 5d elements of the same group have nearly identical radii (e.g. Zr and Hf) and very similar properties. Saying they have no similarities is therefore the incorrect statement; the others are true.
Q36AIPMT 2006 Prelims · Q59
More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is :
Not quite — the answer is B.
In actinoids the 5f, 6d and 7s orbitals are very close in energy, closer than 4f, 5d and 6s in lanthanoids, and 5f electrons are less tightly held. So more electrons can take part in bonding, giving many more oxidation states. A larger 5f–6d gap would have the opposite effect.
Q37AIPMT 2006 Prelims · Q62
In which of the following pairs are both the ions coloured in aqueous solution ? (At. no. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27)
Not quite — the answer is A.
Colour needs partly filled d orbitals for d–d transitions. Ni²⁺ (3d⁸) and Ti³⁺ (3d¹) are both coloured. Sc³⁺ is 3d⁰ and Cu⁺ is 3d¹⁰, so each pair containing one of them has a colourless ion.
Q38AIPMT 2005 Prelims · Q63
The aqueous solution containing which one of the following ions will be colourless ? (Atomic no. : Sc = 21, Fe = 26, Ti = 22, Mn = 25)
Not quite — the answer is A.
Sc³⁺ is 3d⁰, so no d–d transition is possible and its solution is colourless. Ti³⁺ (3d¹), Mn²⁺ (3d⁵) and Fe²⁺ (3d⁶) have unpaired d electrons and are coloured.
Q39AIPMT 2005 Prelims · Q64
Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy ?
Not quite — the answer is D.
The third electron of Mn is removed from Mn²⁺ (3d⁵), a stable half-filled configuration, so Mn has the highest third ionisation enthalpy. For Fe, removal from 3d⁶ gives the stable 3d⁵, so it is easier.
Q40AIPMT 2005 Prelims · Q94
The main reason for larger number of oxidation states exhibited by the actinides than the corresponding lanthanides, is :
Not quite — the answer is A.
In actinides the 5f, 6d and 7s levels are closer in energy than 4f, 5d and 6s in lanthanides, so more electrons can take part in bonding, giving more oxidation states. A larger energy gap would do the opposite.
How often The d- and f-Block Elements came up
Questions from this chapter in each paper on this site.
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