CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Why is Ce⁴⁺ in aqueous solution a good oxidising agent?
Answer:
Because Ce is most stable in Ce³⁺ state in aqueous solution, so Ce⁴⁺ readily gains an electron to change into Ce³⁺.

Teacher's Note:
1. E°(Ce⁴⁺/Ce³⁺) is about +1.74 V, which shows its strong oxidising power.

Question 2. Why do Zr and Hf exhibit similar properties?
Answer:
Due to lanthanide contraction.

Teacher's Note:
1. Lanthanoid contraction makes Hf almost the same size as Zr (about 160 pm), so their chemistry is nearly identical.

Question 3. What is lanthanoid contraction?
Answer:
The regular slow decrease in the atomic or ionic radii of lanthanoids with increasing atomic number.

Teacher's Note:
1. It arises from the poor shielding of 4f electrons; mention this cause if the question asks "why".

Question 4. Why is Cu (Z = 29) considered a transition metal?
Answer:
Due to its partially filled d-orbital (3d⁹) in Cu²⁺ state.

Teacher's Note:
1. An element is a transition element if its atom or any of its common ions has an incompletely filled d-subshell.

Question 5. Why is KMnO₄ or MnO₄⁻ ion coloured?
Answer:
Due to charge transfer (from oxygen to manganese).

Teacher's Note:
1. Mn⁷⁺ has no d electrons, so the colour cannot be from d–d transitions.

Question 6. Why generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements. [CBSE 2010]
Answer:
Due to decrease in size and increase in mass.

Teacher's Note:
1. Density = mass/volume: atomic mass rises while the atomic volume falls along the series.

Question 7. A substance is found to have a magnetic moment of 3.9 B.M. How many unpaired electrons does it contain?
Answer:
Using the formula, μ = \( \sqrt{n(n+2)} \) B.M., n = 3.

Teacher's Note:
1. Check: √(3 × 5) = √15 = 3.87 ≈ 3.9 BM.

Question 8. Why in permanganate ion, there is a covalency between Mn and oxygen?
Answer:
In MnO₄⁻ ion, Mn is in highest oxidation state, viz., +7. In high oxidation state, transition metals form covalent bonds (according to Fajan's rules as oxidation state increases, ionic character decreases).

Teacher's Note:
1. A small, highly charged "Mn⁷⁺" would polarise O²⁻ strongly, so the bonds are covalent.

Question 9. Why are transition elements so named ?
Answer:
Transition elements are so named because their properties are inbetween those of s-block and p-block.

Teacher's Note:
1. The d-block sits between the s-block and p-block in the periodic table, so its properties show a transition from the electropositive s-block to the more electronegative p-block.

Question 10. Why are transition elements known as d-block elements ?
Answer:
Transition elements are called d-block element because the last electron enters (n − 1) d-orbital, i.e., d-orbital of the penultimate shell.

Teacher's Note:
1. Remember that the differentiating electron goes to the (n − 1)d subshell, not the nd subshell, e.g., 3d for the fourth period.

Question 11. The electronic configuration of a transition element in +3 oxidation state is [Ar] 3d⁷. Find out its atomic number. (CBSE Sample Paper 2017)
Answer:
M³⁺ = [Ar]¹⁸ 3d⁷, i.e., it has 25 electrons.
It has been formed by loss of 3 e⁻ from metal M.
Hence, no. of electrons in neutral metal atom = 25 + 3 = 28 e⁻.
∴ No. of protons = Atomic no. = 28

Teacher's Note:
1. Always add back the electrons lost: for an ion Mⁿ⁺, atomic number = electrons in the ion + n. Here the element is nickel.

Short Answer Questions (2 Marks)

Question 12. Write electronic configuration of Cu²⁺ and Co²⁺.
Answer:
Cu²⁺ = [Ar] 3d⁹ 4s⁰
Co²⁺ = [Ar] 3d⁷

Teacher's Note:
1. Transition metals lose 4s electrons first, before 3d electrons.

Question 13. Balance the following equations:
(a) MnO₄⁻ + Fe²⁺ + H⁺ →
(b) Cr₂O₇²⁻ + Sn²⁺ + H⁺ →
Answer:
(a) MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(b) Cr₂O₇²⁻ + 3Sn²⁺ + 14H⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Teacher's Note:
1. Check a balanced ionic equation by charge as well as atoms: in (a) both sides carry +17.

Question 14. Briefly explain why electronic configurations of lanthanoids are not known with certainty?
Answer:
4f/5d subshells are very close in energy. Hence electrons can jump from 4f to 5d or vice-versa.

Teacher's Note:
1. Because the energies are so close, only Ce, Gd and Lu have a 5d¹ electron in the ground state; the rest are 4fⁿ6s².

Question 15. Why are Zn, Cd, Hg soft and have low melting point?
Answer:
Due to weak interatomic attraction/absence of unpaired electrons.

Teacher's Note:
1. These metals have a full (n − 1)d¹⁰ subshell, so d electrons do not take part in metallic bonding.

Question 16. What is the effect of decreasing pH on K₂CrO₄ solution?
Answer:
K₂CrO₄ solution changes into K₂Cr₂O₇/yellow colour changes into orange colour.
2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

Teacher's Note:
1. The chromate–dichromate change is an acid–base equilibrium, not a redox reaction: Cr is +6 in both.

Question 17. Which of the following is/are transition element and why?
Zn, Cd, Ag, Fe, Ni
Answer:
Fe, Ni, Ag
Because they have incompletely filled d-orbitals in the atom (Fe, Ni) or in a common oxidation state (Ag²⁺: 4d⁹). Zn and Cd have completely filled d¹⁰ orbitals in the atom as well as in their ions.

Teacher's Note:
1. Ag is a transition element even though the atom is 4d¹⁰5s¹, because in its +2 state it is 4d⁹.

Question 18. What are interstitial compounds? Give example.
Answer:
When small atoms like C, H, B and N occupy interstitial sites in the lattice of transition metals, the compounds formed are called interstitial compounds.
Example: TiC, Fe₃H.

Teacher's Note:
1. Interstitial compounds are usually non-stoichiometric and are neither typically ionic nor covalent.

Question 19. Why are Zn, Cd, Hg volatile metals? Explain.
Answer:
Due to weak interatomic attraction/low boiling point.

Teacher's Note:
1. Weak metallic bonding (no unpaired d electrons) means little energy is needed to vaporise them.

Question 20. Why is first ionisation energy of 5d elements higher than those of 3d and 4d elements?
Answer:
In 5d elements, the 4f electrons (filled before 5d) shield the nucleus poorly. Hence, the effective nuclear charge acting on the outer electrons increases and they are held more firmly, giving higher first ionisation energy.

Teacher's Note:
1. This is a consequence of lanthanoid contraction: poor 4f shielding raises the effective nuclear charge in the 5d series.

Question 21. Explain ‘Misch metal’ and write its use.
Answer:
It is an alloy of 95% lanthanoid and 5% iron and traces of S, C, Ca and Al. Used in lighter flint, bullet tips etc.

Teacher's Note:
1. Its Mg-based alloy is used to make bullets, shells and lighter flints.

Question 22. The following two reactions of HNO₃ with Zn are given:
(a) Zn + conc. HNO₃ → Zn(NO₃)₂ + X + H₂O
(b) Zn + dil. HNO₃ → Zn(NO₃)₂ + Y + H₂O
Identify X and Y.
Answer:
X = NO₂
Y = N₂O

Teacher's Note:
1. The more dilute the acid, the more the nitrogen is reduced: conc. gives NO₂ (+4), dilute gives N₂O (+1).

Short Answer Questions (3 Marks)

Question 23. Explain the following observations giving an appropriate reason for each :
(i) The enthalpies of atomization of transition elements are quite high.
(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e., 3^rd series).
(iii) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation. [CBSE Delhi 2012]
Answer:
(i) The transition elements exhibit high enthalpies of atomization because they have large number of unpaired electron in their atoms. Due to which they have stronger interatomic interaction and hence stronger bonding between atoms.
(ii) 4d and 5d transition elements (2nd and 3rd series) are larger in size than the corresponding 3d elements. Hence, the valence electrons are less tightly held and form metal-metal bond more frequently.
(iii) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation.
Mn²⁺ → 3d⁵
Fe²⁺ → 3d⁶
As Mn²⁺ has stable configuration hence it is resistant towards oxidation. While in Fe²⁺, electronic configuration is 3d⁶, so it can lose one electron to give stable configuration 3d⁵.

Teacher's Note:
1. In (iii) the key word is half-filled: oxidising Mn²⁺ would destroy a stable 3d⁵ set, while oxidising Fe²⁺ creates one.

Question 24. A mixed oxide of iron and chromium is fused with sodium carbonate in free access of air to form a yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B), which is a strong oxidizing agent. Identify compound (A) and (B). Write chemical reactions involved. [CBSE Comptt. OD 2017]
Answer:
A : Na₂CrO₄   B : Na₂Cr₂O₇
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O

Teacher's Note:
1. The 'mixed oxide of iron and chromium' is chromite ore, FeCr₂O₄ (FeO·Cr₂O₃); write both balanced equations, as unbalanced ones lose marks.

Question 25. What is lanthanoid contraction ? What are its two consequences ? [CBSE Comptt. Delhi 2013]
Answer:
The steady decrease in the size of lanthanoid ions (M³⁺) with the increase in atomic number is called lanthanoid contraction.
Consequences of lanthanoid contraction :
(i) Separation of lanthanoids is difficult due to lanthanoid contraction because due to this contraction their ionic radii is very small.
(ii) The basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃. Due to lanthanoid contraction, the size of M³⁺ ions decreases and there is increase in the covalent character in M – OH bond.

Teacher's Note:
1. State the cause too: the 4f electrons shield each other poorly, so the effective nuclear charge rises steadily across the series.
2. Another consequence worth remembering: Zr and Hf (and Nb/Ta, Mo/W) have almost identical radii.

Question 26. (i) How would you account for the following :
(a) Actinoid contraction is greater than lanthanoid contraction.
(b) Transition metals form coloured compounds.
(ii) Complete the following equation : 2MnO₄⁻ + 6H⁺ + 5NO₂⁻ → [CBSE Delhi 2015]
Answer:
(i) (a) Actinoid contraction is greater than lanthanoid contraction due to ineffective shielding by intervening 5f-electrons.
(b) Transition elements generally forms coloured compounds on account of d-d transition. When the visible light falls on the compounds, they absorb certain radiations and reflect others. The colour observed corresponds to the complementary colour of the light absorbed.
(ii) 2MnO₄⁻ + 6H⁺ + 5NO₂⁻ → 5NO₃⁻ + 2Mn²⁺ + 3H₂O

Teacher's Note:
1. 5f orbitals are more diffuse than 4f orbitals, so they shield even more poorly – hence the larger contraction from element to element.

Question 27. How would you account for the following :
(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
(ii) The E°_M²⁺/M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behaviour.
(iii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series. [CBSE OD 2012]
Answer:
(i) Lanthanoids exhibit occasionally +2, +4 ions because of extra stability of empty, half-filled and completely filled 4f-subshell respectively i.e., 4f⁰, 4f⁷, 4f¹⁴ configuration.
(ii) The E° (M²⁺/M) for copper is positive. This is because high energy is required to transform Cu to Cu²⁺ which is not balanced by its hydration enthalpy.
(iii) In moving down a group, there is increase in number of shells leading to increase in size. So, size of elements of 4d series is larger than those of 3d series. The similar atomic radii of second and third transition series is due to phenomenon called lanthanoid contraction, associated with intervention of 4f-orbitals which must be filled before the 5d series of elements begin.

Teacher's Note:
1. Examples for (i): Ce⁴⁺ (4f⁰) and Eu²⁺ (4f⁷), Yb²⁺ (4f¹⁴).

Question 28. How would you account for the following :
(i) With the same d-orbital configuration (d⁴) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidizing agent ?
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoids series ?
(iii) Most of the transition metal ions exhibit characteristic colours in aqueous solutions ? [CBSE Delhi 2012]
Answer:
(i) Cr (24) = 3d⁵4s¹
Configuration of Cr²⁺ = 3d⁴
As Cr³⁺ has stable t_2g³ configuration so, Cr²⁺ has a tendency to change to Cr³⁺. Hence, it is a reducing agent.
Mn (25) = 3d⁵4s²
Mn³⁺ = 3d⁴
Mn³⁺ changes to Mn²⁺ (3d⁵) which is half filled state configuration. Hence, Mn³⁺ acts as oxidising agent.
(ii) Actinoids exhibit a large number of oxidation states because 5f, 6d and 7s levels have almost comparable energies (very small energy difference between sub shells) than the energy difference between 4f and 5d orbitals in case of lanthanoid.
(iii) Most of the transition metal ions exhibit characteristic colours in aqueous solutions because of d-d transitions, as they have maximum number of unpaired electrons.

Teacher's Note:
1. Same d⁴, opposite behaviour: Cr²⁺ loses an electron to reach d³, while Mn³⁺ gains one to reach d⁵ – both move towards a more stable configuration.

Question 29. Explain the following observations :
(i) Many of the transition elements are known to form interstitial compounds.
(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
(iii) The numbers of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. [CBSE Delhi 2012]
Answer:
(i) Many of the transition elements are known to form interstitial compounds because of unpaired electrons in the d-orbital. Transition elements have vacant interstitial sites and are able to trap small atoms like H, C or N to form such compounds.
(ii) As the atomic radii decreases moving across from titanium to Cu, its volume will decrease and density is expected to increase.
(iii) Actinoids exhibit a large number of oxidation states because 5f, 6d and 7s levels have almost comparable energies (very small energy difference between sub-shells than the energy difference between 4f and 5d orbitals) in case of lanthanoids.

Teacher's Note:
1. For (ii) mention both factors: atomic mass increases while the metallic radius decreases, so mass per unit volume rises from Ti to Cu.

Question 30. (a) Deduce the number of 3d electrons in the following ions: Cu²⁺, Sc³⁺
(b) Why do transition metals form alloys?
(c) Why are Zn²⁺ salts white?
Answer:
(a) Cu²⁺ : 9 electrons; Sc³⁺ : 0 electron
(b) Transition metals have similar atomic radii.
(c) Absence of unpaired electron.

Teacher's Note:
1. Zn²⁺ is 3d¹⁰: with a full d subshell no d–d transition is possible, so its salts are white.

Question 31. Complete and balance the following equations:
(a) MnO₄²⁻ + H⁺ →
(b) KMnO₄ \( \xrightarrow{\text{heat}} \)
(c) MnO₄⁻ + FeC₂O₄ \( \xrightarrow{\text{H}^+} \)
Answer:
(a) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
(b) 2KMnO₄ \( \xrightarrow{\text{heat}} \) K₂MnO₄ + MnO₂ + O₂
(c) 3MnO₄⁻ + 5FeC₂O₄ + 24H⁺ → 3Mn²⁺ + 5Fe³⁺ + 10CO₂ + 12H₂O

Teacher's Note:
1. In (c) FeC₂O₄ loses 3 electrons (1 from Fe²⁺ and 2 from C₂O₄²⁻), so 3 MnO₄⁻ (5 e⁻ each) react with 5 FeC₂O₄.

Question 32. Describe the oxidising action of K₂Cr₂O₇ with the following. Write ionic equations for its reaction (acidic medium) with:
(a) I⁻ (b) Iron(II) (c) H₂S
Answer:
(a) It liberates I₂ from I⁻.
Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
(b) It oxidises Fe²⁺ to Fe³⁺.
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
(c) It oxidises H₂S to sulphur.
Cr₂O₇²⁻ + 3H₂S + 8H⁺ → 2Cr³⁺ + 7H₂O + 3S

Teacher's Note:
1. All three use the same half-reaction Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O; only the reductant part changes.

Teacher's Note:
1. Most actinoid differences come from the 5f, 6d and 7s levels being close in energy, which allows more electrons to take part in bonding.

Long Answer Questions (5 Marks)

Question 34. (i) Account for the following:
(a) Transition metals form large number of complex compounds. [CBSE 2019 (56/2/1)]
(b) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.
(c) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compare to Cr³⁺/Cr²⁺.
(ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [CBSE Delhi 2017]
Answer:
(i) (a) The tendency to form complex compounds is due to: (1) Small size and high charge on metal ion. (2) The availability of d orbitals for accommodating electrons donated by the ligand.
(b) In low oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. On the other hand, in higher oxidation state of the metal, valence electrons are involved in bonding and are not available. Instead effective nuclear charge is high and hence it can accept electrons and behave as an acid.
(c) Much large third ionisation energy of Mn (where the required change is stable half filled d⁵ to d⁴) is mainly responsible for this.
(ii) Similarities (Any one)
1. Both show mainly an oxidation state of +3.
2. Actinoids show actinoid contraction like lanthanoid contraction is shown by lanthanoids.
3. Both are electropositive and very reactive.
Differences (Any one)
1. Except promethium (Pm) lanthanoids are non-radioactive whereas actinoids are radioactive.
2. Lanthanoids do not form oxocation whereas actinoids form oxocation.
3. Lanthanoids have less tendency towards complex formation whereas actinoids have greater tendency towards complex formation.

Teacher's Note:
1. For (c), compare Cr³⁺ (d³, stable half-filled t_2g), whose reduction to Cr²⁺ is unfavourable, giving a negative E° (−0.41 V).

Question 35. Assign reasons for the following.
(i) The enthalpies of atomisation of transition elements are high.
(ii) The transition metals and many of their compounds act as good catalysts.
(iii) From element to element, the actinoid contraction is greater than the lanthanoid contraction.
(iv) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that of Cr³⁺/Cr²⁺.
(v) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as a transition element. [CBSE 2019 (56/2/3)]
Answer:
(i) This is because transition metals have strong metallic bonds as they have a large number of unpaired electrons.
(ii) The catalytic activity of transition metals is attributed to the following reasons:
(a) Because of their variable oxidation states, transition metals form unstable intermediate compounds and provide a new path with lower activation energy for the reaction.
(b) In some cases, the transition metal provides a suitable large surface area with free valencies on which reactants are adsorbed.
(iii) This is due to poorer shielding by 5f-electrons in actinoids than that by 4f-electrons in the lanthanoids.
(iv) This is due to much larger third ionisation energy of Mn as Mn²⁺ is very stable on account of stable d⁵ configuration.
(v) This is because scandium has partially filled d-orbitals in the ground state (3d¹4s²).

Teacher's Note:
1. Sc shows only +3 (Sc³⁺ is 3d⁰), but the incomplete 3d¹ in the atom is enough to make it a transition element.

Question 36. (i) Following are the transition metal ions of 3d series:
Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic number : Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following :
(a) Which ion is most stable in an aqueous solution and why?
(b) Which ion is a strong oxidizing agent and why?
(c) Which ion is colourless and why?
(ii) Complete the following equations:
(a) 2MnO₄⁻ + 16H⁺ + 5S²⁻ →
(b) KMnO₄ —heat→ [CBSE OD 2017]
Answer:
(i) (a) Cr³⁺, half filled t_2g³
(b) Mn³⁺, due to stable d⁵ configuration in Mn²⁺
(c) Ti⁴⁺, No unpaired electrons
(ii) (a) 2MnO₄⁻ + 16H⁺ + 5S²⁻ → 5S + 2Mn²⁺ + 8H₂O
(b) 2KMnO₄ —Δ→ K₂MnO₄ + MnO₂ + O₂

Teacher's Note:
1. Answer the 'why' part of each sub-question with the configuration: Cr³⁺ 3d³, Mn³⁺ 3d⁴ → Mn²⁺ 3d⁵, Ti⁴⁺ 3d⁰.
2. On heating (513 K), KMnO₄ decomposes into manganate, MnO₂ and oxygen; write the coefficient 2.

Question 37. Describe the preparation of potassium permanganate from pyrolusite ore. Write balanced chemical equation for one reaction to show oxidizing nature of potassium permanganate. [CBSE Comptt. OD 2013]
Answer:
Potassium permanganate is prepared from pyrolusite ore with KOH in the presence of oxidising agent like KNO₃. The dark green potassium manganate undergoes electrolytic oxidation to produce potassium permanganate.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Oxidising nature of potassium permanganate can be shown as :
Ferrous to ferric
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
[Fe²⁺ → Fe³⁺ + e⁻] × 5
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Teacher's Note:
1. Pyrolusite is MnO₂; the green manganate (Mn +6) is converted to purple permanganate (Mn +7) either by electrolytic oxidation or by disproportionation in acid.
2. Show the oxidising action with half-reactions and multiply to equalise electrons, as done here for Fe²⁺.

Teacher's Note:
1. For the atomic-radii trend, mention the reason at each stage: increasing nuclear charge, poor shielding by d electrons, and electron–electron repulsion near the end of the series.
2. Mn shows the maximum number of oxidation states (+2 to +7) in the 3d series – a favourite one-mark question.