CBSE Class 12 Chemistry The d and f Block Elements NEET JEE Questions Set 03

Single Correct Answer

Question 1. Which one of the following is the correct increasing order of the magnitude of ionic radii of Ce³⁺, La³⁺, Pm³⁺ and Yb³⁺? (DPMT 2010)
(a) Yb³⁺ < Pm³⁺ < La³⁺ < Ce³⁺
(b) Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺
(c) Pm³⁺ < La³⁺ < Ce³⁺ < Yb³⁺
(d) Ce³⁺ < Yb³⁺ < Pm³⁺ < La³⁺
Answer:
Ans. (b)
La³⁺ (106 pm) > Ce³⁺ (103 pm) > Pm³⁺ (98 pm) > Yb³⁺ (87 pm)
Ionic size decreases from La³⁺ to Lu³⁺ due to lanthanide contraction. Thus, Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺

Teacher's Note:
1. For Ln³⁺ ions just arrange them by atomic number: higher Z, smaller ion.

Question 2. Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect? (AIEEE 2009)
(a) The ionic sizes of Ln (III) decrease in general with increasing atomic number.
(b) Ln (III) compounds are generally colourless.
(c) Ln (III) hydroxides are mainly basic in character
(d) Because of the large size of the Ln (III) ions, the bonding in its compounds is predominantly ionic in character.
Answer:
Ans. (b)
Ln (III) compounds are generally coloured due to partly filled f-orbitals which permit f-f transition.

Teacher's Note:
1. Only La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) and a few others are colourless; most Ln³⁺ ions are coloured.

Question 3. Which is not correct statement about the chemistry of 3d and 4f series elements? (DCE 2009)
(a) 3d-elements show more oxidation states than 4f-series elements
(b) The energy difference between 3d and 4s orbitals is very little
(c) Europium (II) is more stable than Ce (II)
(d) The paramagnetic character in 3d-series elements increases from scandium to copper
Answer:
Ans. (d)
The paramagnetic character in 3d series elements first increases from Sc to Mn as the number of unpaired electrons increases and then decreases as the number of unpaired electrons decreases from Fe to Cu.

Teacher's Note:
1. Paramagnetism tracks unpaired electrons, which peak at the half-filled d⁵ stage, not at the end of the series.

Question 4. The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because (AIEEE 2007)
(a) the 5f orbitals extend farther from the nucleus than the 4f orbitals
(b) the 5f orbitals are more buried than the 4f orbitals
(c) there is a similarity between 4f and 5f orbitals in their angular part of the wave function
(d) the actinoids are more reactive than the lanthanoids
Answer:
Ans. (a)
As 5f orbitals extend farther from the nucleus than 4f orbitals, the attraction on the 5f electrons in actinoid series decreases. Hence, they can easily participate in bonding and, therefore, exhibit more number of oxidation states.

Teacher's Note:
1. 5f electrons are less shielded inside the atom than 4f electrons, so they are available for bonding (U shows up to +6, Np/Pu up to +7).

Question 5. The reason for greater range of oxidation states in actinoids is attributed to (NEET 2017)
(a) the radioactive nature of actinoids
(b) actinoid contraction
(c) 5f, 6d and 7s levels having comparable energies
(d) 4f and 5d levels being close in energies
Answer:
Ans. (c)
The greater range of oxidation states shown by actinoids is due to 5f, 6d and 7s levels having comparable energies and thus all their electrons can take part in bond formation.

Teacher's Note:
1. Option (d) is the lanthanoid set of orbitals; for actinoids the relevant levels are 5f, 6d and 7s.

Question 6. Reason of lanthanoid contraction is (AIPMT 2014)
(a) negligible screening effect of ‘f’-orbitals
(b) increasing nuclear charge
(c) decreasing nuclear charge
(d) decreasing screening effect
Answer:
Ans. (a)
Due to poor screening or shielding effect of 4f-orbitals, nucleus will exert a stronger attraction and size of atom or ion goes on decreasing as we move in the series with increase in atomic number.

Teacher's Note:
1. Nuclear charge rises by one each step while the added 4f electron shields poorly, so effective nuclear charge rises and size falls.

Question 7. Which one of the following statements related to lanthanons is incorrect? (NEET Phase II 2016)
(a) All the lanthanons are much more reactive than aluminium
(b) Ce (+4) solutions are widely used as oxidizing agent in volumetric analysis
(c) Europium shows + 2 oxidation state
(d) The basicity decreases as the ionic radius decreases from Pr to Lu.
Answer:
Ans. (a)
(a) Lanthanons are less reactive than aluminium due to high IP (Lanthanoid contraction)
(b) Ce⁺⁴ is good oxidizing agent and easily converted into Ce⁺³
(c) Eu (63) = 4f⁷ 5d⁰ 6s², Eu⁺² = 4f⁷
(d) In lanthanoids series ‘Ce’ to ‘Lu’ ionic radius regularly decreases and covalent character increases and hence basic character of hydroxides decreases.

Teacher's Note:
1. The earlier lanthanoids behave much like calcium; their reactivity falls along the series, so ‘all much more reactive than Al’ is too sweeping.

Question 8. Identify the incorrect statement from the following : (NEET 2020)
(a) The overall decrease in atomic and ionic radii from lanthanum to lutetium is called lanthanoid contraction
(b) Zirconium and hafnium have identical radii of 160 pm and 159 pm, respectively as a consequence of lanthanoid contraction
(c) Lanthanoids reveal only + 3 oxidation state
(d) The lanthanoid ions other than the f⁰ type and the f¹⁴ type are all paramagnetic
Answer:
Ans. (c)
Most common oxidation state of lanthanoids is +3 but these can show +2 and +4 oxidation states in solution or in solid compounds as well.

Teacher's Note:
1. Remember the exceptions: Eu²⁺, Yb²⁺ (+2) and Ce⁴⁺, Tb⁴⁺ (+4) are stabilised by f⁰, f⁷ or f¹⁴ configurations.

Question 9. The incorrect statement among the following is : (NEET 2021)
(a) Actinoid contraction is greater for element to element than lanthanoid contraction
(b) Most of the trivalent lanthanoid ions are colourless in the solid state
(c) Lanthanoids are good conductors of heat and electricity
(d) Actinoids are highly reactive metals especially when finely divided
Answer:
Ans. (b)
Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solution.

Teacher's Note:
1. Actinoid contraction is larger from element to element because 5f electrons shield even more poorly than 4f electrons.

Question 10. Misch metal is an alloy consisting mainly of (JEE Main 2020)
(a) lanthanoid and actinoid metals
(b) actinoid and transition metals
(c) lanthanoid metals
(d) actinoid metals
Answer:
Ans. (c)
Misch metal is an alloy consisting mainly of lanthanoid metals (about 95%) with iron (about 5%) and traces of S, C, Ca and Al.

Teacher's Note:
1. Misch metal is used in Mg-based alloys for bullets, shells and lighter flints; it contains no actinoids.

Question 11. Gadolinium has a low value of third ionization enthalpy because of (NEET 2022)
(a) small size
(b) high exchange enthalpy
(c) high electronegativity
(d) high basic character
Answer:
Ans. (b)
Electronic configuration of Gadolinium (Gd) is Gd = [Xe] 4f⁷ 5d¹ 6s².
In case of third ionization enthalpy, electron will be removed from 5d-orbital giving a more stable configuration [Xe] 4f⁷ which will have high exchange energy. Therefore, less energy will be required to remove third electron.

Teacher's Note:
1. Whenever removing an electron leaves a half-filled or completely filled subshell, the corresponding ionization enthalpy is lower than expected.

Question 12. All Cu (II) halides are known except iodide. The reason for this is that (J & K CET 2011)
(a) iodide ion is a bulky ion
(b) Cu²⁺ oxidizes iodide to iodine
(c) Cu²⁺ (aq) has much more negative hydration enthalpy
(d) Cu²⁺ ion has smaller size.
Answer:
Ans. (b)
Iodide (I⁻) ions are stronger reducing agent. They reduce Cu²⁺ to Cu⁺ (2CuI₂ → Cu₂I₂ + I₂).
In other words, Cu²⁺ oxidizes I⁻ ions to I₂ (2Cu²⁺ + 4I⁻ → Cu₂I₂(s) + I₂)

Teacher's Note:
1. This same reaction is used in iodometric estimation of copper: liberated I₂ is titrated with thiosulphate.

Question 13. In nitroprusside ion, the iron and NO exist as Fe^II and NO⁺ rather than Fe^III and NO. These forms can be differentiated by
(a) estimating the concentration of iron
(b) measuring the concentration of CN⁻
(c) measuring the solid state magnetic moment
(d) thermally decomposing the compound.
Answer:
Ans. (c)
Magnetic moment helps to find the number of unpaired electrons. Low spin Fe^II (d⁶) with NO⁺ gives a diamagnetic ion, whereas Fe^III (d⁵) with NO (one unpaired electron each) would be paramagnetic.

Teacher's Note:
1. When two descriptions differ only in how electrons are distributed, a magnetic measurement is the tool that can tell them apart.

Question 14. A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is
(a) HgI₂
(b) HgO
(c) Pb₃O₄
(d) (NH₄)₂ Cr₂O₇
Answer:
Ans. (a)
HgI₂ (red ppt.) + 2KI → K₂[HgI₄] (soluble)
HgI₂ —Δ→ Hg + I₂ (violet vapours)

Teacher's Note:
1. Violet fumes point to iodine and metal droplets on the cool part point to mercury; K₂[HgI₄] is Nessler’s reagent.

Question 15. [Fe(H₂O)₅NO]²⁺ is a complex formed in the brown ring test for NO₃⁻ ion. In this complex
(a) NO transfers its electron to Fe²⁺ so that we have iron as Fe(I) and NO as NO⁺
(b) There are three unpaired electrons so that its magnetic moment is 3.87 B.M.
(c) The colour is due to charge transfer
(d) All the above statements are correct
Answer:
Ans. (d)
NO → NO⁺ + e⁻, Fe²⁺ (3d⁶, 4 unpaired e⁻) + e⁻ → Fe⁺ (3d⁷, 3 unpaired e⁻)
\( \mu \) for Fe⁺ \( = \sqrt{n(n+2)} = \sqrt{3\times 5} \) BM \( = \sqrt{15} \) BM = 3.87 BM

Teacher's Note:
1. In the brown ring complex the formal charges are Fe⁺ and NO⁺; the brown colour arises from charge transfer between them.

Question 16. When Hg₂Cl₂ ionizes, the ions produced and the unpaired electrons present on the cation respectively are
(a) 2 Hg⁺ and 2 Cl⁻, two
(b) Hg₂²⁺ and 2 Cl⁻, two
(c) Hg₂²⁺ and 2 Cl⁻, one
(d) Hg₂²⁺ and 2 Cl⁻, zero
Answer:
Ans. (d)
Hg₂Cl₂ → Hg₂²⁺ + 2Cl⁻. Each Hg⁺ has one unpaired electron. Hence, Hg₂²⁺ (with an Hg–Hg bond) has no unpaired electron.

Teacher's Note:
1. Mercury(I) always exists as the dimer [Hg–Hg]²⁺; the two odd electrons pair up to form the metal–metal bond, so mercurous salts are diamagnetic.

Question 17. Which one of the following ions exhibits d-d transition and paramagnetism as well? (NEET 2018)
(a) MnO₄⁻
(b) Cr₂O₇²⁻
(c) CrO₄²⁻
(d) MnO₄²⁻
Answer:
Ans. (d)
MnO₄⁻ = Mn⁷⁺ = [Ar] 3d⁰, unpaired e⁻, n = 0, diamagnetic
MnO₄²⁻ = Mn⁶⁺ = [Ar] 3d¹, unpaired e⁻, n = 1, paramagnetic
Cr₂O₇²⁻ = Cr⁶⁺ = [Ar] 3d⁰, unpaired e⁻, n = 0, diamagnetic
CrO₄²⁻ = Cr⁶⁺ = [Ar] 3d⁰, unpaired e⁻, n = 0, diamagnetic
Thus, only MnO₄²⁻ (3d¹) can show d-d transition and is paramagnetic, while the other ions (3d⁰) are diamagnetic and owe their colour to charge transfer.

Teacher's Note:
1. A d–d transition needs at least one d electron, so every d⁰ species is ruled out at once.

Question 18. Which of the following reactions are disproportionation reactions?
(i) 2Cu⁺ → Cu²⁺ + Cu⁰
(ii) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
(iii) 2KMnO₄ —Δ→ K₂MnO₄ + MnO₂ + O₂
(iv) 2MnO₄⁻ + 3Mn²⁺ + 2H₂O → 5MnO₂ + 4H⁺
Select the correct option from the following : (NEET 2019)
(a) (i) and (iv) only
(b) (i) and (ii) only
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Answer:
Ans. (b)
Disproportionation reactions are those in which the same reactant is oxidized as well as reduced simultaneously
2Cu⁺ → Cu²⁺ + Cu⁰
3MnO₄²⁻ (+6) + 4H⁺ → 2MnO₄⁻ (+7) + MnO₂ (+4) + 2H₂O
In (iii) Mn is only reduced (oxygen is oxidised) and (iv) is a comproportionation (Mn +7 and +2 both give +4).

Teacher's Note:
1. In (iii) it is O²⁻, not Mn, that is oxidised; and (iv) is the reverse of disproportionation (two oxidation states meet in one).

Question 19. Identify the incorrect statement (NEET 2020)
(a) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes
(b) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals
(c) The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same
(d) Cr²⁺ (d⁴) is a stronger reducing agent than Fe²⁺ (d⁶) in water
Answer:
Ans. (c)
Chromium is in +6 oxidation state in both CrO₄²⁻ and Cr₂O₇²⁻; their interconversion depends only on pH. Hence (c) is incorrect.
(d) is correct: Cr²⁺ → Cr³⁺ gives stable t_2g³ (E°_{Cr³⁺/Cr²⁺} = −0.41 V), while Fe²⁺ → Fe³⁺ has E° = +0.77 V.

Teacher's Note:
1. Chromate–dichromate interconversion (2CrO₄²⁻ + 2H⁺ ⇄ Cr₂O₇²⁻ + H₂O) is an acid-base change, not a redox change.

Question 20. The composition of gun metal is (JEE Main 2021)
(a) Cu, Zn, Sn
(b) Al, Mg, Mn, Cu
(c) Cu, Ni, Fe
(d) Cu, Sn, Fe
Answer:
Ans. (a)
Gun metal is an alloy of copper (about 88%), tin (about 10%) and zinc (about 2%).

Teacher's Note:
1. Al, Mg, Mn, Cu is duralumin; remember gun metal as a bronze (Cu + Sn) with a little zinc.

One or More Correct Answers

Question 21. The correct statement(s) about Cr²⁺ and Mn³⁺ is (are) [Atomic numbers of Cr = 24 and Mn = 25] (JEE Advanced 2015)
(a) Cr²⁺ is a reducing agent
(b) Mn³⁺ is an oxidizing agent
(c) Both Cr²⁺ and Mn³⁺ exhibit d⁴ electronic configuration
(d) When Cr²⁺ is used as a reducing agent, the chromium ion attains d⁵ configuration
Answer:
Ans. (a, b, c)
(a) Cr²⁺ is a reducing agent as it gets oxidized to Cr³⁺ (3d³ or t_2g³ which is a stable half-filled configuration).
(b) Mn³⁺ is an oxidizing agent as it gets reduced to Mn²⁺ (3d⁵ which is more stable half-filled configuration).
(c) Cr (24) = 3d⁵ 4s¹ ∴ Cr²⁺ = 3d⁴; Mn (25) = 3d⁵ 4s² ∴ Mn³⁺ = 3d⁴. Thus, both Cr²⁺ and Mn³⁺ have d⁴ electronic configuration.
(d) When Cr²⁺ is used as a reducing agent, it is oxidized to Cr³⁺ which has d³ and not d⁵ configuration.

Teacher's Note:
1. Same d⁴ count, opposite behaviour: Cr²⁺ moves to t_2g³ by losing an electron, Mn³⁺ moves to d⁵ by gaining one.

Question 22. Fusion of MnO₂ with KOH in presence of O₂ produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt X. The manganese containing ions present in W and X respectively are Y and Z. Correct statement(s) is (are). (JEE Advanced 2019)
(a) In both Y and Z, π-bonding occurs between p-orbitals of oxygen and d-orbitals of manganese
(b) both Y and Z are coloured and have tetrahedral shape
(c) In aqueous acidic solution, Y undergoes disproportionation reaction to give Z and MnO₂
(d) Y is diamagnetic in nature while Z is paramagnetic
Answer:
Ans. (a, b, c)
2MnO₂ + 4KOH + O₂ —Fusion→ 2K₂MnO₄ (W) + 2H₂O
On electrolysis of K₂MnO₄, MnO₄²⁻ gets oxidized to MnO₄⁻: 2K₂MnO₄ + 2H₂O —Electrolysis→ 2KMnO₄ (X) + 2KOH + H₂
In W, Mn containing ion is MnO₄²⁻ (Y). In X, Mn containing ion is MnO₄⁻ (Z).
Both Y and Z are coloured (Y is green while Z is purple). Y (MnO₄²⁻ or K₂MnO₄) undergoes disproportionation reaction in aqueous acidic solution as 3MnO₄²⁻ (+6) + 4H⁺ → 2MnO₄⁻ (+7, Z) + MnO₂ (+4) + 2H₂O
Both MnO₄⁻ and MnO₄²⁻ are tetrahedral in shape involving pπ–dπ bonding.
Y (MnO₄²⁻, Mn +6) is paramagnetic while Z (MnO₄⁻, Mn +7) is diamagnetic in nature.

Teacher's Note:
1. Manganate (d¹) is paramagnetic and green; permanganate (d⁰) is diamagnetic and purple; option (d) swaps them.

Question 23. In an experiment, m grams of a compound X (gas/liquid/solid) taken in a container is loaded in a balance as shown in the figure I below. In the presence of a magnetic field, the pan with X is either deflected upwards (figure II) or deflected downwards (figure III) depending on the compound X. Identify the correct statement(s) (JEE Advanced 2020)
(a) If X is H₂O (l), deflection of the pan is upwards
(b) If X is K₄[Fe(CN)₆] (s), deflection of the pan is upwards
(c) If X is O₂ (g), deflection of the pan is downward
(d) If X is C₆H₆ (l), deflection of the pan is downward

Answer:
Ans. (a, b, c)
Paramagnetic substances are attracted by the magnetic field, so magnetic balance shows downward deflection while diamagnetic substances are repelled by the magnetic field and hence show upward deflection.
H₂O, K₄[Fe(CN)₆] and C₆H₆ are diamagnetic while O₂ (g) is paramagnetic.

Teacher's Note:
1. K₄[Fe(CN)₆] contains low-spin Fe²⁺ (t_2g⁶), so it is diamagnetic despite being a transition metal complex.

Passage-Based Questions

Question 24. Comprehension 2: Potassium permanganate is prepared from the mineral pyrolusite, MnO₂. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium, it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of H₂SO₄. However, before using it as a titrant, it is first standardized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration, 13.4 g of dry pure sodium oxalate (molar mass = 134 g mol⁻¹) was dissolved in 100 mL of distilled water and then 100 mL of 2M H₂SO₄ were added. The solution was cooled to 25-30°C. Now to this solution, 0.1 M KMnO₄ solution was added till a very faint pink colour persisted.
Mohr salt, FeSO₄.(NH₄)₂SO₄.6H₂O, is preferred over FeSO₄.7H₂O for standardization of KMnO₄ solution because
(a) Mohr salt is a double salt while ferrous sulphate is a single salt.
(b) Mohr salt is not hygroscopic but FeSO₄.7H₂O is hygroscopic
(c) Mohr salt contains only ferrous ions whereas ferrous sulphate contains some ferric ions.
(d) Mohr salt solution can be titrated even in the absence of H₂SO₄
Answer:
Ans. (c)
FeSO₄.7H₂O contains some ferric ions due to aerial oxidation, whereas Mohr salt is stable and contains only ferrous ions.

Teacher's Note:
1. A primary standard must have an exactly known composition; partial oxidation to Fe³⁺ would make the FeSO₄ reading wrong.

Question 25. Comprehension 2: Potassium permanganate is prepared from the mineral pyrolusite, MnO₂. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium, it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of H₂SO₄. However, before using it as a titrant, it is first standardized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration, 13.4 g of dry pure sodium oxalate (molar mass = 134 g mol⁻¹) was dissolved in 100 mL of distilled water and then 100 mL of 2M H₂SO₄ were added. The solution was cooled to 25-30°C. Now to this solution, 0.1 M KMnO₄ solution was added till a very faint pink colour persisted.
If instead of H₂SO₄, HCl or HNO₃ of suitable concentration were used, the volume of KMnO₄ solution used would have been
(a) less in case of HCl but more in case of HNO₃
(b) more in case of HCl but less in case of HNO₃
(c) more in both cases
(d) less in both cases.
Answer:
Ans. (b)
If HCl were used, the oxygen produced from KMnO₄ will be partly used up for oxidation of HCl. Hence, KMnO₄ required will be more. If HNO₃ were used, it itself acts as oxidizing agent and will oxidize the reductant. Hence, KMnO₄ solution used will be less.

Teacher's Note:
1. H₂SO₄ is chosen because dilute sulphuric acid is neither oxidised by permanganate nor an oxidant itself.

Question 26. Comprehension 2: Potassium permanganate is prepared from the mineral pyrolusite, MnO₂. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium, it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of H₂SO₄. However, before using it as a titrant, it is first standardized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration, 13.4 g of dry pure sodium oxalate (molar mass = 134 g mol⁻¹) was dissolved in 100 mL of distilled water and then 100 mL of 2M H₂SO₄ were added. The solution was cooled to 25-30°C. Now to this solution, 0.1 M KMnO₄ solution was added till a very faint pink colour persisted.
If of 0.1 M K₂Cr₂O₇ solution were used in place 0.1 M KMnO₄ in presence of H₂SO₄ in each case, the volume of 0.1 M K₂Cr₂O₇ solution used would be
(a) 200 mL
(b) 400 mL
(c) 333.3 mL
(d) 666.6 mL
Answer:
Ans. (c)
If K₂Cr₂O₇ were used, the reaction would have been
Cr₂O₇²⁻ + 14H⁺ + 3C₂O₄²⁻ → 2Cr³⁺ + 7H₂O + 6CO₂
Moles of K₂Cr₂O₇ required for 0.1 mol of sodium oxalate = \( \frac{1}{3}\times 0.1 \) = 0.03333. Hence, volume of 0.1 M K₂Cr₂O₇ solution required = 333.3 mL.

Teacher's Note:
1. Dichromate accepts 6 electrons per ion against 5 for permanganate, so less volume is needed: 400 × 5/6 = 333.3 mL.

Matrix Match Questions

Teacher's Note:
1. Mn₂O₇ from cold conc. H₂SO₄ is an explosive oil; never add conc. H₂SO₄ to solid KMnO₄ carelessly.

Teacher's Note:
1. ‘Rare earths’ are La and the lanthanoids (with Sc and Y); Th and U are actinoids, not rare earths.

Integer and Numerical Answer Questions

Question 29. Manganese (VI) has the ability to disproportionate in acidic solution. The difference in oxidation states of two species of Mn formed in acidic solution is .......... (JEE Main 2022)
Answer:
Ans. 3
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
[MnO₄]⁻ gives x − 8 = −1 or x = +7
MnO₂ gives x − 4 = 0 or x = +4
Difference = 3

Teacher's Note:
1. In a disproportionation the products lie on either side of the starting state: +6 goes to +7 and +4.

Question 30. The number of correct statement(s) from the following for copper (at. no. 29) is/are .........
(A) Cu (II) complexes are always paramagnetic.
(B) Cu (I) complexes are generally colourless.
(C) Cu (I) is easily oxidized.
(D) In Fehling solution, the active reagent has Cu (I). (JEE Main 2022)
Answer:
Ans. 3
(A) is correct because Cu (II) has d⁹ configuration which has one unpaired electron.
(B) is correct because Cu (I) has d¹⁰ configuration.
(C) is correct because Cu (I) is oxidized to Cu (II) in aqueous solution.
(D) is not correct because in Fehling solution, active reagent is Cu (II) which is reduced to Cu (I) on reaction with aldehydes.
Number of correct statements = 3

Teacher's Note:
1. A d⁹ ion always has one unpaired electron whatever the ligand, so every Cu(II) complex is paramagnetic.

Question 31. The number of terminal oxygen atoms present in the product B obtained from the following reaction is ..........
FeCr₂O₄ + Na₂CO₃ + O₂ → A + Fe₂O₃ + CO₂
A + H⁺ → B + H₂O + Na⁺ (JEE Main 2022)
Answer:
Ans. 6
FeCr₂O₄ + Na₂CO₃ + O₂ → Fe₂O₃ + CO₂ + Na₂CrO₄ (A)
Na₂CrO₄ + H⁺ → Cr₂O₇²⁻ (B) + H₂O + Na⁺
Structure of B: two CrO₄ tetrahedra sharing one O; each Cr carries two Cr=O and one Cr–O⁻ terminal oxygens, with one bridging Cr–O–Cr oxygen.
Terminal O-atoms = 6

Teacher's Note:
1. Of the 7 oxygens in Cr₂O₇²⁻, only the bridging one is non-terminal.

Question 32. An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after addition of 1 mL of 3% H₂O₂, a blue alcohol layer was obtained. The blue colour is due to the formation of a chromium (VI) compound ‘X’. What is the number of oxygen atoms bonded to chromium through only single bonds in a molecule of X? (JEE Advanced 2020)
Answer:
Ans. 4
K₂CrO₄ —H⁺/H₂O₂→ CrO₅ (Deep blue solution)
CrO₅ has one Cr=O and two peroxo (O–O) groups, each peroxo O bonded to Cr by a single bond.
O-atoms bonded to Cr through single bonds = 4

Teacher's Note:
1. CrO₅ is unstable in water, which is why it is extracted into amyl alcohol (or ether) to see the blue colour.

Question 33. To measure the quantity of MnCl₂ dissolved in aqueous solution, it was completely converted to KMnO₄ using the reaction.
MnCl₂ + K₂S₂O₈ + H₂O → KMnO₄ + H₂SO₄ + HCl (equation not balanced)
Few drops of concentrated HCl were added to this solution and gently warmed. Further oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl₂ (in mg) present in the initial solution is.............. .
(Atomic weights in g mol⁻¹ : Mn = 55, Cl = 35.5) (JEE Advanced 2018)
Answer:
Ans. 126
2MnCl₂ + 5K₂S₂O₈ + 8H₂O → 2KMnO₄ + 4K₂SO₄ + 6H₂SO₄ + 4HCl ...(i)
2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂ ...(ii)
Mass of oxalic acid added = 225 mg. Millimoles of oxalic acid added = \( \frac{225}{90} \) = 2.5
From eqn. (ii), 5 millimoles of oxalic acid react with KMnO₄ = 2 millimoles
∴ 2.5 millimoles of oxalic acid will react with KMnO₄ = 1 millimole
From eqn. (i), 2 millimoles of KMnO₄ react with MnCl₂ = 2 millimoles ∴ 1 millimole will react with MnCl₂ = 1 millimole
Thus, MnCl₂ present in the initial solution = 1 millimole = 1 × 126 mg = 126 mg

Teacher's Note:
1. Molar mass of MnCl₂ = 55 + 2 × 35.5 = 126 g mol⁻¹; Mn is conserved, so mmol MnCl₂ = mmol MnO₄⁻.

Assertion–Reason Questions

Question 34. Assertion: Pd shows diamagnetic nature.
Reason: Aufbau rule is violated in writing electronic configuration of Pd.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Pd has 4d¹⁰ 5s⁰ configuration rather than 4d⁸ 5s² configuration and hence is diamagnetic in nature.

Teacher's Note:
1. Pd is the only element with no electron in its outermost (5s) shell in the ground state.

Question 35. Assertion: In Cr₂O₇²⁻ ion, all the Cr—O bond lengths are equal.
Reason: In Cr₂O₇²⁻ ion, all the O—Cr—O bond angles are equal.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Two Cr—O bonds have different bond lengths than the remaining six Cr—O bonds.
Correct R. Cr—O—Cr bond in the middle is different from other O—Cr—O bond angles.

Teacher's Note:
1. The two bridging Cr–O bonds (179 pm) are longer than the six terminal ones (163 pm).

Question 36. Assertion: K₂Cr₂O₇ is used as primary standard in volumetric analysis.
Reason: It has a good solubility in water. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. K₂Cr₂O₇ is not much soluble in cold water. However, it is obtained in pure state and is not hygroscopic in nature.

Teacher's Note:
1. A primary standard must be available pure, be non-hygroscopic and have a known stable composition; high solubility is not the criterion.

Question 37. Assertion: Change in colour of acidic solution of potassium dichromate by breath is used to test drunken drivers.
Reason: Change in colour is due to the complexation of alcohol with potassium dichromate. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. Acidified K₂Cr₂O₇ solution oxidizes alcohol and itself is reduced to Cr₂(SO₄)₃ which is green in colour.

Teacher's Note:
1. The orange → green change is a redox change (Cr +6 → +3), with ethanol oxidised to ethanal/ethanoic acid.

Question 38. Assertion: The purple colour of KMnO₄ is due to the charge transfer transition.
Reason: The intense colour of most of the transition metal complexes is due to d-d transition. (AIIMS 2014)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. The intense colour of KMnO₄ is not due to d-d transition but due to charge transfer from O to M (Ligand to Metal) reducing the oxidation state of Mn from +7 to +6 temporarily.

Teacher's Note:
1. Both statements are true, but R cannot explain A because Mn(VII) has no d electrons.

Question 39. Assertion: Atomic size of silver is almost the same as that of gold.
Reason: d-subshell has low penetration power and produces poor shielding. (AIIMS 2011)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. Atomic size of silver is almost equal to that of gold due to lanthanoid contraction.

Teacher's Note:
1. The similar size of 4d and 5d elements comes from the poor shielding of the 4f electrons (lanthanoid contraction).

Question 40. Assertion: The size of Bk³⁺ ion is less than that of Np³⁺ ion.
Reason: The above is a consequence of lanthanoid contraction. (JEE Main 2021)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. ₉₃Np³⁺ > ₉₇Bk³⁺ in size because as atomic number increases, ionic size decreases. It is due to actinoid contraction.

Teacher's Note:
1. Np and Bk are actinoids, so the relevant effect is actinoid contraction (poor shielding by 5f electrons).