CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. How many elements are present in the d-block of the periodic table ?
Answer:
40.

Teacher's Note:
1. Four series of ten elements each (3d, 4d, 5d and 6d) give 4 × 10 = 40.

Question 2. How many transition series of elements are there in the periodic table ? Name them.
Answer:
There are four main transition series of elements corresponding to filling of 3d, 4d, 5d and 6d sublevels in 4th, 5th, 6th and 7th periods. They are known as first transition series or 3d series, second transition or 4d series, third transition or 5d series and fourth transition or 6d series.

Teacher's Note:
1. Note the mismatch between the series name and the period: the 3d series lies in the 4th period because 3d fills after 4s.

Question 3. In the transition series, starting from lanthanum (₅₇La), the next element hafnium (₇₂Hf) has an atomic number of 72. Why do we observe this jump in atomic number ?
Answer:
This is because after ₅₇La, filling of 4 f-orbital starts which is complete at atomic number 71. These 14 elements, therefore, belong to f-block and placed separately at the bottom of the periodic table.

Teacher's Note:
1. The 14 lanthanoids (Ce to Lu, Z = 58 to 71) fit in between La and Hf, which is why the jump is 72 − 57 = 15.

Question 4. Which out of the following is/are transition element/s and why ?
Zn, Ag, Cd, Au.
Answer:
Ag and Au are transition elements because they have incompletely filled d-subshell.

Teacher's Note:
1. Ag and Au have d¹⁰ in the atom but d⁹ in the common +2/+3 states (Ag²⁺, Au³⁺), whereas Zn and Cd stay d¹⁰ in all their states.

Question 5. Both Cu and Zn have completely filled 3d atomic orbitals. Cu is considered as transition element but Zn is not. Explain. (Assam Board 2012)
Answer:
This is because Cu in one of the oxidation states viz. + 2, has incompletely filled 3d subshell (3d⁹) while Zn has completely filled 3d¹⁰ only.

Teacher's Note:
1. The definition counts an element as transition if the atom OR any of its common ions has a partly filled d-subshell.

Question 6. The atomic sizes of Fe, Co and Ni are nearly same. Explain with reason. (Raj. Board 2011)
Or
Atomic size of 3d series elements from chromium to copper is almost the same. Give reason. (Karnataka Board 2012)
Answer:
As we move from left to right along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the d-subshell increases the screening effect which counterbalances the effect of increased nuclear charge.

Teacher's Note:
1. Name both opposing effects in your answer: increasing nuclear charge (shrinks) and increasing shielding by (n − 1)d electrons (expands).

Question 7. How does the ionic/covalent character of the compounds of a transition metal vary with its oxidation state ?
Answer:
As the oxidation state increases, more and more valence shell electrons are involved in bonding. The atomic core becomes less shielded resulting in the increased attraction on the electrons. Because of this, ionic character of bonds decreases with the increase in oxidation state.

Teacher's Note:
1. Example to quote: MnO is ionic and basic, while Mn₂O₇ is covalent and acidic.

Question 8. What is the common oxidation state of Cu, Ag and Au ?
Answer:
All these three metals show a common oxidation state of +1.

Teacher's Note:
1. All three have the outer configuration (n − 1)d¹⁰ ns¹, so losing the single s-electron gives +1.

Question 9. Which metal in the first transition series (3d series) exhibits + 1 oxidation state most frequently and why? (CBSE 2013)
Answer:
Copper in the first transition series exhibits + 1 oxidation state most frequently. This is because Cu is 3d¹⁰ 4s¹. When one electron is lost, the configuration becomes most stable due to fully filled d¹⁰ configuration.

Teacher's Note:
1. Even so, in aqueous solution Cu⁺ disproportionates, so Cu²⁺ is the more stable ion in water.

Question 10. Give an example of disproportionation reaction.
Answer:
Copper (I) salts decompose in water. Half the Cu (I) changes to copper metal and the other half to copper (II), i.e., 2Cu⁺ → Cu + Cu²⁺. This is known as disproportionation.

Teacher's Note:
1. In disproportionation the same species is both oxidised and reduced: Cu(+1) goes to Cu(0) and Cu(+2).

Short Answer Questions (2 Marks)

Question 11. Complete the equations:
(a) KMnO₄ \( \xrightarrow{\Delta} \)
(b) 3K₂MnO₄ \( \xrightarrow[\text{acidic medium}]{\text{on standing for long time}} \)
Answer:
(a) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
(b) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Teacher's Note:
1. In (b) manganate disproportionates: Mn(+6) goes to Mn(+7) in MnO₄⁻ and Mn(+4) in MnO₂.

Question 12. Out of Fe and Cu, which one would exhibit higher melting point?
Answer:
Fe, due to large number of unpaired d-electrons/more interatomic attraction.

Teacher's Note:
1. Fe (3d⁶4s²) melts at about 1811 K, Cu (3d¹⁰4s¹) at about 1358 K.

Question 13. Sc, the first member of first transition series does not exhibit variable oxidation state. Why?
Answer:
Due to noble gas electronic configuration in +3 oxidation state no other oxidation state is stable.

Teacher's Note:
1. Sc³⁺ is [Ar] 3d⁰, so all three valence electrons are lost together.

Question 14. Though copper, silver and gold have completely filled sets of d-orbitals yet they are considered as transition metals. Why?
Answer:
These metals in their common oxidation states have incompletely filled d-orbitals, e.g., Cu²⁺ has 3d⁹ and Au³⁺ has 5d⁸ configuration.

Teacher's Note:
1. The definition allows an incomplete d subshell either in the atom or in a common oxidation state.

Question 15. Name the elements which are not really transition elements but are discussed with them. Why is it so?
Answer:
Zinc, cadmium and mercury have completely filled sets of d-orbitals, and therefore, are not really transition metals. However, they are often discussed with them because of the similarities of some of their properties to those of transition metals. Moreover, they are the end members of these series.

Teacher's Note:
1. Zn, Cd and Hg keep d¹⁰ in the atom and in their common +2 ions, so they never have a partly filled d subshell.

Question 16. Explain why Fe is a transition metal but Na is not?
Answer:
Fe contains incompletely filled 3d subshell (3d⁶ 4s²). Hence, it is a d-block element, i.e., a transition element.
Na has no d-subshell. Last electron in it enters 3s orbital (1s² 2s² 2p⁶ 3s¹). Hence, it belongs to s-block.

Teacher's Note:
1. The block is decided by the subshell that receives the last (differentiating) electron.

Question 17. Why is there striking similarities (horizontal and vertical) in successive members of the transition series?
Answer:
This is because of the fact that along a horizontal row, electrons enter an incomplete inner shell in building up an atom, while the outer level remains almost unchanged.
In vertical columns, similarities are due to similar electronic configurations even in the d-subshells.

Teacher's Note:
1. Since the outer 4s configuration hardly changes along the row, neighbouring transition metals behave alike – unlike main-group elements.

Question 18. Why the properties of third transition series are very similar to second transition series?
Or
Why the second and third members in each group of the transition elements have very similar atomic radii? [CBSE 2012]
Answer:
In the third transition series after lanthanum, there is lanthanoid contraction. Due to this contraction, the size of any atom of the third transition series is almost the same as that of the element lying just above in the second transition series. This leads to similarity in their properties.

Teacher's Note:
1. Name the cause clearly in such answers: "lanthanoid contraction" is the key phrase examiners look for.

Question 19. The melting and boiling points of Zn, Cd and Hg are low. Why?
Or
Why Zn, Cd and Hg are soft and have low m.pt.?
Answer:
In Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, the metallic bonds present in them are weak. That is why they have low melting and boiling points.

Teacher's Note:
1. With d¹⁰ configuration only the ns electrons take part in metallic bonding.

Question 20. K₂PtCl₆ is a well known compound whereas corresponding Ni compound is not known. State a reason for it.
Answer:
This is because Pt⁴⁺ is more stable than Ni⁴⁺ as the sum of four ionization enthalpies of Pt is less than that of Ni.

Teacher's Note:
1. Lower total ionisation enthalpy makes a high oxidation state easier to reach and more stable.

Question 21. Why do transition elements show variable oxidation states? [HP Board 2013, CBSE 2013]
Answer:
In the transition elements, the energies of (n−1) d-orbitals and ns orbitals are very close. Hence, electrons from both can participate in bonding.

Teacher's Note:
1. Note that their oxidation states usually differ by one unit (Fe²⁺/Fe³⁺), unlike p-block elements (differ by two).

Question 22. Giving reasons indicate which one of the following would be coloured?
Cu⁺, VO²⁺, Sc³⁺, Ni²⁺ (At. Nos. Cu = 29, V = 23, Sc = 21, Ni = 28)
Answer:
VO²⁺ (3d¹) and Ni²⁺ (3d⁸) are coloured due to incompletely filled d-orbitals (d-d transition is possible).
Cu⁺ (3d¹⁰) and Sc³⁺ (3d⁰) are colourless.

Teacher's Note:
1. In VO²⁺ vanadium is +4 (3d¹); always work out the oxidation state of the metal in an oxo-ion first.

Short Answer Questions (3 Marks)

Question 23. (a) Why is separation of lanthanoid elements difficult?
(b) Transition metals exhibit higher enthalpies of atomisation. Explain why?
(c) Why do the transition metals have high enthalpy of hydration?
Answer:
(a) Due to lanthanide contraction, the size of these elements is nearly same.
(b) Transition metals contain large number of unpaired electrons, and they have strong interatomic attractions.
(c) Due to their small size and large nuclear charge.

Teacher's Note:
1. Similar size means similar chemical properties; lanthanoids are separated by ion-exchange methods that exploit tiny size differences.

Question 24. (a) Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment.
(b) Is lanthanum a f-block element?
Answer:
(a) ₅₈Ce = [Xe] 4f¹ 5d¹ 6s²
Ce³⁺ = [Xe] 4f¹, one unpaired electron
μ = \( \sqrt{n(n+2)} \) = \( \sqrt{1(1+2)} \) = 1.73 BM
(b) No, it is a d-block element.

Teacher's Note:
1. For lanthanoid ions the spin-only formula is an approximation because orbital contribution is large, but for Ce³⁺ (one f electron) it is the expected exam answer.

Question 25. Account for the following:
(a) Silver chloride dissolves in excess of NH₃.
(b) Cuprous chloride is diamagnetic while cupric chloride is paramagnetic.
(c) In CrO₄²⁻ ion, all the Cr–O bond lengths are equal.
Answer:
(a) AgCl forms a soluble complex with NH₃.
AgCl + 2NH₃ → [Ag(NH₃)₂]Cl
(b) Cu⁺ : 3d¹⁰ 4s⁰ – All electrons are paired.
Cu²⁺ : 3d⁹ – Here, one unpaired electron is present.
(c) Due to resonance.

Teacher's Note:
1. Paramagnetism needs at least one unpaired electron; always write the ion’s configuration to justify it.

Question 26. The E° values in respect of electrodes of Cr, Mn and iron are:
Cr³⁺/Cr²⁺ = −0.4 V
Mn³⁺/Mn²⁺ = +1.5 V
Fe³⁺/Fe²⁺ = +0.8 V
Compare the feasibilities of further oxidation of these ions.
Answer:
Cr³⁺ is more stable than Cr²⁺ (negative E°), so Cr²⁺ is oxidised most easily.
Mn²⁺ is more stable than Mn³⁺ (large positive E°), so Mn²⁺ is oxidised with great difficulty.
Fe²⁺ is oxidised to Fe³⁺ less easily than Cr²⁺ but more easily than Mn²⁺.
Order of feasibility of further oxidation: Cr²⁺ > Fe²⁺ > Mn²⁺
Order of stability of +2 oxidation state: Mn²⁺ > Fe²⁺ > Cr²⁺

Teacher's Note:
1. The more positive the M³⁺/M²⁺ potential, the harder it is to oxidise M²⁺ to M³⁺.

Question 27. Write any three properties of interstitial compounds.
Answer:
(a) They are chemically inert.
(b) They retain metallic conductivity.
(c) They have higher melting points than their pure metals.
(d) These are harder and more corrosion resistant.

Teacher's Note:
1. Steel and cast iron are hard because of interstitial carbon in the iron lattice, a good everyday example.

Question 28. Account for the following:
(a) All scandium salts are white.
(b) The first ionisation energies of the 5d series are higher than 3d and 4d transition elements in respective groups.
(c) Ce³⁺ can be easily oxidised to Ce⁴⁺.
Answer:
(a) Sc has only +3 oxidation state (3d⁰), there is no unpaired electron.
(b) Due to lanthanide contraction (poor shielding by 4f electrons), effective nuclear charge increases.
(c) Due to gain of noble gas electronic configuration (Ce⁴⁺ = [Xe] 4f⁰).

Teacher's Note:
1. Ce⁴⁺ is still a strong oxidant in water because it tends to return to the very stable +3 state.

Question 29. A green chromium compound (A) on fusion with alkali gives a yellow compound (B) which on acidification gives an orange coloured compound (C). Identify A, B, C. Write equations for reactions.
Answer:
A : Cr₂O₃ B : Na₂CrO₄ C : Na₂Cr₂O₇
2Cr₂O₃ (A) + 4Na₂CO₃ + 3O₂ → 4Na₂CrO₄ (B) + 4CO₂
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ (C) + Na₂SO₄ + H₂O

Teacher's Note:
1. The colour sequence green (Cr³⁺) → yellow (CrO₄²⁻) → orange (Cr₂O₇²⁻) is the usual clue in chromium identification questions.

Question 30. When an oxide of Mn (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). Identify A, B, C.
Answer:
A : MnO₂ B : K₂MnO₄ C : KMnO₄
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Teacher's Note:
1. Green manganate (Mn +6) disproportionates to purple permanganate (Mn +7) and MnO₂ (Mn +4).

Question 31. The E° values in respect of the electrodes of chromium (Z = 24), manganese (Z = 25) and iron (Z = 26) are: Cr³⁺/Cr²⁺ = −0.4 V, Mn³⁺/Mn²⁺ = +1.5 V, Fe³⁺/Fe²⁺ = +0.8 V. On the basis of the above information compare the feasibilities of further oxidation of their +2 oxidation states.
Answer:
−ve value of E° for Cr³⁺/Cr²⁺ shows that Cr²⁺ is least stable. Greater +ve value for Mn³⁺/Mn²⁺ than that for Fe³⁺/Fe²⁺ shows that Mn²⁺ is more stable than Fe²⁺.
Hence, stability of +2 oxidation state is in the order: Cr²⁺ < Fe²⁺ < Mn²⁺ or the oxidation of their +2 state to +3 state is in the order: Cr²⁺ > Fe²⁺ > Mn²⁺

Teacher's Note:
1. The more positive E°(M³⁺/M²⁺), the more stable M²⁺ and the harder it is to oxidise.

Question 32. How would you account for the fact that the transition metals and their compounds are found to be good catalysts in many processes? [CBSE 2007]
Answer:
According to activated complex theory (intermediate compound formation theory), due to variable oxidation states, transition metal combines with the reactant/s to form an intermediate (an unstable species) called activated complex which finally decomposes to form the products regenerating the catalyst.
A + C (catalyst) → [AC] (Activated complex/Intermediate) → B (Product) + C (Catalyst)
The reaction proceeds through a new path with lower activation energy and hence is faster.
According to adsorption theory, transition metal provides a large surface area with free valencies on which the reactants are adsorbed. As a result, the concentration of the reactants on the surface increases and the rate of the reaction increases.

Teacher's Note:
1. Two key reasons to quote: variable oxidation states (intermediate formation) and the ability to provide a surface (free valencies), e.g. Fe in Haber process, V₂O₅ in Contact process.

Question 33. Decide giving reason which one of the following pairs exhibits the property indicated:
(i) Sc³⁺ or Cr³⁺ exhibits paramagnetism
(ii) V or Mn exhibits more number of oxidation states
(Atomic numbers: Sc = 21, Cr = 24, V = 23, Mn = 25)
Answer:
(i) ₂₁Sc = [Ar] 3d¹ 4s², Sc³⁺ = [Ar] —No unpaired electron
₂₄Cr = [Ar] 3d⁵ 4s¹, Cr³⁺ = [Ar] 3d³ —Three unpaired electron
Hence, Cr³⁺ exhibits paramagnetism.
(ii) Mn exhibits more number of oxidation states.
Reason. ₂₃V = 3d³ 4s². Its ox. states can be +2, +3, +4, +5
₂₅Mn = 3d⁵ 4s². Its ox. states can be +2, +3, +4, +5, +6, +7.

Teacher's Note:
1. Paramagnetism needs at least one unpaired electron; d⁰ and d¹⁰ ions are diamagnetic.

Question 34. Describe giving reason which one of the following pairs has the property indicated?
(i) Fe or Cu has higher melting point.
(ii) Co²⁺ or Ni²⁺ has lower magnetic moment.
Answer:
(i) Fe has higher melting point than Cu. This is because Fe has four unpaired electrons in 3d-subshell while Cu has only one electron in the 4s-subshell. Hence, metallic bonds in Fe are much stronger than those in Cu.
(ii) ₂₇Co = [Ar] 3d⁷ 4s², Co²⁺ = [Ar] 3d⁷ (3 unpaired electron)
₂₈Ni = [Ar] 3d⁸ 4s², Ni²⁺ = [Ar] 3d⁸ (2 unpaired electrons)
Hence, Ni has lower magnetic moment than Co.

Teacher's Note:
1. Fewer unpaired electrons → lower spin-only moment: Co²⁺ 3.87 BM, Ni²⁺ 2.83 BM.

Long Answer Questions (5 Marks)

Question 35. When chromite ore is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (A) is obtained. On acidifying the yellow solution with sulphuric acid, compound (B) is crystallized out. When compound (B) is treated with KCl, orange crystals of compound (C) crystallize out. Identify (A), (B) and (C) and write the reactions involved. [CBSE Comptt. OD 2017]
Answer:
A : Na₂CrO₄   B : Na₂Cr₂O₇   C : K₂Cr₂O₇
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Teacher's Note:
1. The last step is a simple double decomposition (no redox); potassium dichromate separates because it is much less soluble than sodium dichromate.

Question 36. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. [NCERT]
Answer:
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.
(iii) The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.
(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding (M−M bonding).
(v) The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Teacher's Note:
1. A good example for (ii): Fe and Mn favour +2/+3, but Ru, Os and Re readily show +6 to +8 (e.g. OsO₄).

Teacher's Note:
1. Present comparisons as a table with matching rows – examiners award marks per correct pair.
2. Actinoids are also radioactive, which makes their chemistry harder to study; this is a useful extra point.

Teacher's Note:
1. Paramagnetism depends on the number of unpaired electrons: more unpaired electrons means a larger magnetic moment (Mn³⁺ 4.90 BM vs Cr³⁺ 3.87 BM).

Question 39. (i) Account for the following:
(a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(b) Zirconium and Hafnium exhibit similar properties.
(c) Transition metals act as a catalysts.
(ii) Complete the following equations:
(a) 2MnO₂ + 4KOH + O₂ —Δ→
(b) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ →
(Atomic nos. : Mn = 25, Cr = 24) [CBSE OD 2016]
Answer:
(i) (a) Oxygen has the ability to form multiple bonds (pπ–dπ) with the metal atom, so Mn shows its highest oxidation state of +7 in Mn₂O₇. Fluorine can form only single bonds, so the highest oxidation state of Mn with fluorine is +4 (MnF₄).
(b) Zirconium (4d series) and hafnium (5d series) have almost identical atomic radii because of lanthanoid contraction. The 4f electrons filled before the 5d series shield poorly, so the expected increase in size on adding a new shell is cancelled. Hence Zr and Hf have similar size and similar properties.
(c) Transition metals act as good catalysts because of their ability to adopt multiple oxidation states and to form complexes (unstable intermediates), and because they provide a large surface area for the reactants to be adsorbed.
(ii) (a) 2MnO₂ + 4KOH + O₂ —Δ→ 2K₂MnO₄ + 2H₂O
(b) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O

Teacher's Note:
1. In (ii)(a) MnO₂ (Mn +4) is oxidised by air to green potassium manganate (Mn +6); this is the first step in making KMnO₄.