Class 12 Chemistry

Chapter 4 Solutions — The d- and f-Block Elements

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Overview

Step-by-step NCERT solutions for The d- and f-Block Elements (Chapter 4, CBSE Class 12 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the The d- and f-Block Elements textbook chapter.

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What these solutions cover

All 38 questions in The d- and f-Block Elements are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

1.
Write down the electronic configuration of:
- (i) Cr³⁺
- (ii) Pm³⁺
- (iii) Cu⁺
- (iv) Ce⁴⁺
- (v) Co²⁺
- (vi) Lu²⁺
- (vii) Mn²⁺
- (viii) Th⁴⁺
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d₃, 3d₅, 3d₈ and 3d₄?
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
In what way is the electronic configuration of the transition elements different from that of the non transition elements?
What are the different oxidation states exhibited by the lanthanoids?
11.
Explain giving reasons:
- (i) Transition metals and many of their compounds show paramagnetic behaviour.
- (ii) The enthalpies of atomisation of the transition metals are high.
- (iii) The transition metals generally form coloured compounds.
- (iv) Transition metals and their many compounds act as good catalyst.
What are interstitial compounds? Why are such compounds well known for transition metals?
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
- (i) iodide
- (ii) iron(II) solution and
- (iii) H₂S
16.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with
- (i) iron(II) ions
- (ii) SO₂ and
- (iii) oxalic acid? Write the ionic equations for the reactions.
17.
For M²⁺/M and M³⁺/M²⁺ systems the E° values for some metals are as follows: Cr²⁺/Cr = -0.9V; Cr³⁺/Cr²⁺ = -0.4V; Mn²⁺/Mn = -1.2V; Mn³⁺/Mn²⁺ = +1.5V; Fe²⁺/Fe = -0.4V; Fe³⁺/Fe²⁺ = +0.8V. Use this data to comment upon:
- (i) the stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and
- (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium…
Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu²⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.
Compare the stability of +2 oxidation state for the elements of the first transition series.
20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
- (i) electronic configuration
- (ii) atomic and ionic sizes and
- (iii) oxidation state
- (iv) chemical reactivity.
21.
How would you account for the following:
- (i) Of the d₄ species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.
- (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
- (iii) The d₁ configuration is very unstable in ions.
What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?
25.
Give examples and suggest reasons for the following features of the transition metal chemistry:
- (i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
- (ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
- (iii) The highest oxidation state is exhibited in oxoanions of a metal.
26.
Indicate the steps in the preparation of:
- (i) K₂Cr₂O₇ from chromite ore.
- (ii) KMnO₄ from pyrolusite ore.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Use Hund's rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
- (i) electronic configuration
- (ii) oxidation states and
- (iii) chemical reactivity.
Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
35.
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
- (i) electronic configurations
- (ii) oxidation states
- (iii) ionisation enthalpies and
- (iv) atomic sizes.
Write down the number of 3d electrons in each of the following ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
What can be inferred from the magnetic moment values of the following complex species? K₄[Mn(CN)₆] (mu = 2.2 BM); [Fe(H₂O)₆]²⁺ (mu = 5.3 BM); K₂[MnCl₄] (mu = 5.9 BM)

Sample

Worked examples

3 questions from The d- and f-Block Elements, solved step by step.

Exercises

What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Definition:

Disproportionation is the process in which an element in a single oxidation state is simultaneously oxidised (to a higher oxidation state) and reduced (to a lower oxidation state) in the same reaction. In other words, when a particular oxidation state becomes less stable relative to the states one higher and one lower, it undergoes disproportionation.

Example 1 — Manganese(VI) in acidic solution:

MnO₄²⁻ (manganate, Mn = +6) disproportionates in acid to give permanganate Mn(+7) and MnO₂ Mn(+4):

3 MnO₄²⁻ + 4 H⁺ → 2 MnO₄⁻ + MnO₂ + 2 H₂O

(Mn goes from +6 → +7 (oxidised) and +6 → +4 (reduced))

Example 2 — Copper(I) in aqueous solution:

Cu⁺ (cuprous, +1) is unstable in aqueous solution and disproportionates into Cu²⁺ (+2) and Cu⁰ (0):

2 Cu⁺(aq) → Cu²⁺(aq) + Cu(s)

(Cu goes from +1 → +2 (oxidised) and +1 → 0 (reduced))

The E°(Cu⁺/Cu) = +0.52V and E°(Cu²⁺/Cu⁺) = +0.16V; the cell E° for the disproportionation is +0.52 - 0.16 = +0.36V > 0, so the reaction is spontaneous.

Answer: Disproportionation occurs when a species in one oxidation state simultaneously oxidises and reduces itself to give products in higher and lower oxidation states. Example 1: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (Mn⁶⁺ → Mn⁷⁺ and Mn⁴⁺). Example 2: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s) (Cu⁺ → Cu²⁺ and Cu⁰).

Exercises

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

The last element in the actinoid series is Lawrencium (Lr), atomic number Z = 103.

Electronic configuration:

Lr: [Rn] 5f₁₄ 6d₁ 7s₂

(The 5f subshell is completely filled with 14 electrons, one electron in 6d, and two in 7s. Some sources also write [Rn] 5f₁₄ 7s₂ 7p₁ depending on quantum chemical calculations, but NCERT uses [Rn] 5f₁₄ 6d₁ 7s₂.)

Possible oxidation state:

Since Lr has completely filled 5f₁₄ subshell (analogous to Lu in the lanthanoid series with 4f₁₄), the most stable oxidation state is +3.

In the +3 state, Lr³⁺ has configuration [Rn] 5f₁₄ — the completely filled and stable 5f₁₄ configuration is retained.

Loss of the 6d₁ and 7s₂ electrons (3 electrons total) gives the +3 ion.

The +3 state is consistent with the trend observed for heavier actinoids (Fm, Md, No, Lr) which all predominantly show +3, behaving similarly to lanthanoids.

No is unusual in showing +2 (as No²⁺ has the stable 5f₁₄ configuration like Yb²⁺ with 4f₁₄); Lr returns to +3.

Answer: Last actinoid: Lawrencium (Lr), Z=103. Electronic configuration: [Rn] 5f₁₄ 6d₁ 7s₂. The most stable and expected oxidation state is +3. In Lr³⁺, the stable completely filled 5f₁₄ subshell is retained (by losing 6d₁ and 7s₂ electrons). +3 is consistent with the trend of heavier actinoids.

Exercises

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Z = 61 — Promethium (Pm), a lanthanoid (Z=57-71):

Core: [Xe] (Z=54), then 4f electrons fill from Ce (Z=58).

Pm: [Xe] 4f₅ 6s₂

Z = 91 — Protactinium (Pa), an actinoid (Z=89-103):

Core: [Rn] (Z=86)

Pa: [Rn] 5f₂ 6d₁ 7s₂

(Pa has one 6d electron because 5f and 6d are close in energy at the beginning of actinoid series)

Z = 101 — Mendelevium (Md), an actinoid:

Core: [Rn] (Z=86), then fill 5f from Th (Z=90).

Md: [Rn] 5f₁₃ 7s₂

Z = 109 — Meitnerium (Mt), a 6d transition metal (4th series):

Core: [Rn] (Z=86), then 5f₁₄ (fills from Th to Lr), then 6d series starts at Rf (Z=104).

Configuration after [Rn]: 5f₁₄ + filling 6d from Rf(104):

Rf: 6d₂, Db: 6d₃, Sg: 6d₄, Bh: 6d₅, Hs: 6d₆, Mt: 6d₇

Mt: [Rn] 5f₁₄ 6d₇ 7s₂

Answer: Z=61 (Pm): [Xe] 4f₅ 6s₂. Z=91 (Pa): [Rn] 5f₂ 6d₁ 7s₂. Z=101 (Md): [Rn] 5f₁₃ 7s₂. Z=109 (Mt): [Rn] 5f₁₄ 6d₇ 7s₂.

All 38 questions in The d- and f-Block Elements are solved step by step in the Solutions PDF above.

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Chapter 4 Solutions — The d- and f-Block Elements

Overview

What these solutions cover

Exercises

Worked examples

What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

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