CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Write all possible oxidation states of an element having atomic number 25.
Answer:
E.C. of ₂₅Mn = [Ar]¹⁸ 3d⁵ 4s². It shows oxidation states of + 2, + 3, + 4, + 5, + 6 and + 7.

Teacher's Note:
1. The highest oxidation state of Mn (+7) equals the total number of 3d and 4s electrons (5 + 2).

Question 2. Why does vanadium pentoxide act as a catalyst ?
Answer:
Vanadium exists in multiple oxidation states.

Teacher's Note:
1. V₂O₅ catalyses the contact process (2SO₂ + O₂ → 2SO₃) by switching between V(+5) and V(+4).

Question 3. Out of [Sc(H₂O)₆]³⁺ and [Ti(H₂O)₆]³⁺ ions which is coloured and why ? Give reason. (HP Board 2011, Assam Board 2013)
Answer:
[Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless. This is because Ti³⁺ has one electron in the d-orbital (3d¹) which can absorb energy corresponding to yellow wavelength and undergo d-d transition (from t_2g to e_g). But Sc³⁺ is 3d⁰, i.e., it has no electron in 3d subshell.

Teacher's Note:
1. A d⁰ or d¹⁰ ion cannot show d–d transitions, so it is colourless; that is the quick test for questions of this kind.

Question 4. Which of the following cations are coloured in aqueous solution and why?
Sc³⁺, V³⁺, Ti⁴⁺, Mn²⁺
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (CBSE 2013)
Answer:
Sc³⁺ = 3d⁰ 4s⁰ (No unpaired electron), V³⁺ = 3d² 4s⁰ (2 unpaired electrons), Ti⁴⁺ = 3d⁰ 4s⁰ (No unpaired electron), Mn²⁺ = 3d⁵ 4s⁰ (5 unpaired electrons). Thus, only V³⁺ and Mn²⁺ are coloured because they have unpaired electrons.

Teacher's Note:
1. Write the configuration of each ion first; remember the 4s electrons are removed before the 3d electrons.

Question 5. Copper (I) compounds are white and diamagnetic while copper (II) compounds are paramagnetic and form coloured compounds. Explain. (Hr. Board 2012, MP Board 2013)
Answer:
In copper (I) ion, all orbitals are completely filled (1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰) and, therefore, it is diamagnetic and forms white compounds. Copper (II) ion has electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ (has one unpaired electron) and, therefore, it is paramagnetic and forms coloured compounds.

Teacher's Note:
1. One unpaired electron explains both properties of Cu²⁺: paramagnetism and the d–d transition that gives colour.

Question 6. Which of the two : V (IV) or V (V) is paramagnetic ? Give reasons. At No of V = 23.
Answer:
₂₃V = [Ar] 3d³ 4s², V (IV) = [Ar] 3d¹,
V (V) = [Ar] 3d⁰
Thus, V (IV) has one unpaired electron while V (V) has no unpaired electron. Hence, V (IV) is paramagnetic.

Teacher's Note:
1. Paramagnetism needs at least one unpaired electron; V(V) is d⁰ and therefore diamagnetic.

Question 7. What do you mean by 18 carat gold ?
Answer:
18 carat gold contains 18 parts by weight of pure gold in 24 parts by weight of the alloy, i.e. 75% gold and 25% copper.

Teacher's Note:
1. Pure gold is 24 carat, so percentage of gold = (carat/24) × 100; for 22 carat this gives 91.7%.

Question 8. Which elements are called ferrous metals ?
Answer:
The elements of the first triad of Groups 8, 9 and 10, i.e., iron, cobalt and nickel are called ferrous metals.

Teacher's Note:
1. Fe, Co and Ni are also the three ferromagnetic elements of the 3d series, which is an easy way to remember the triad.

Question 9. Which elements are called platinum metals ?
Answer:
The elements of the second and third triad of Groups 8, 9 and 10, i.e., Ru, Rh, Pd ; Os, Ir and Pt are called platinum metals.

Teacher's Note:
1. Six elements in all: the 4d triad (Ru, Rh, Pd) and the 5d triad (Os, Ir, Pt).

Question 10. What are coinage metals ? Why are they so called ?
Answer:
Copper, silver and gold are called coinage metals because in the early times, they were used in making coins.

Teacher's Note:
1. The coinage metals form Group 11 and all have the outer configuration (n − 1)d¹⁰ ns¹.

Question 11. Write the general electronic configuration of transition elements or d-block elements. (Karnataka Board 2012)
Answer:
(n − 1) d¹⁻¹⁰ ns⁰⁻²

Teacher's Note:
1. The ns⁰ case is palladium (4d¹⁰ 5s⁰); most elements have ns¹ or ns².

Short Answer Questions (2 Marks)

Question 12. Scandium forms no coloured ions, yet it is regarded as a transition element. Explain why?
Answer:
Scandium in the ground state has one electron in the 3d- subshell. Hence, it is regarded as a transition element.
However, in its common oxidation state +3, it has no electron in 3d subshell (3d⁰). Hence, it does not form coloured ions.

Teacher's Note:
1. Sc meets the definition through its atom (3d¹ 4s²), not through its ion.

Question 13. Why are the ionization energies of 5d elements greater than 3d elements?
Answer:
In the 5d series, after lanthanum (Z = 57), there is lanthanoid contraction. In each group, the size of 5d element is smaller while nuclear charge is greater than 3d element. Hence, ionization energies of 5d elements are greater than 3d elements.

Teacher's Note:
1. The 32 extra protons of a 5d element are poorly shielded by the 4f electrons, so the outer electrons are held more tightly.

Question 14. The 4d and 5d series of transition metals have more frequent metal-metal bonding in their compounds than do the 3d metals. Explain.
Answer:
In the same group of d-block elements, the 4d and 5d transition elements have larger size than that of 3d element. Thus, the valence electrons are less tightly held and hence can form metal-metal bond more frequently.
(That is why melting points of 4d and 5d series as well as enthalpies of atomisation are higher than those of 3d series)

Teacher's Note:
1. The 4d and 5d orbitals are more extended, so they overlap better between neighbouring metal atoms.

Question 15. Why is first ionization enthalpy of Cu higher than that of Na? [Assam Board 2013]
Answer:
Cu = [Ar]¹⁸ 3d¹⁰ 4s¹, Na = [Ne]¹⁰ 3s¹
Though electron to be removed in both cases is from outermost s-orbital but nuclear charge of Cu is much higher than that of Na. Hence, attraction on 4s electron in Cu is much greater than on 3s electron in Na. Moreover, removal of 1e⁻ from 3s in Na gives a stable noble gas configuration whereas in case of Cu, it gives pseudo noble gas configuration.

Teacher's Note:
1. The 3d¹⁰ electrons of Cu shield the 4s electron poorly, so it feels a much larger effective nuclear charge.

Question 16. Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state
E°_{Cr³⁺/Cr²⁺} = −0.4 V, E°_{Mn³⁺/Mn²⁺} = +1.5 V, E°_{Fe³⁺/Fe²⁺} = +0.8 V. [CBSE Sample Paper 2018, CBSE 2018 (C)]
Answer:
Higher the standard reduction potential, more easily is M³⁺ reduced to M²⁺, i.e., greater is the stability of +2 oxidation state. Thus, increasing order of stability of +2 oxidation state will be Mn²⁺ > Fe²⁺ > Cr²⁺ or increasing order is Cr²⁺ < Fe²⁺ < Mn²⁺.

Teacher's Note:
1. Read the question carefully: "increasing order" must be written with < signs from least to most stable.

Question 17. On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti⁴⁺ species may be used to oxidise Fe(II) to Fe(III)
Ti⁴⁺ + e⁻ → Ti³⁺, E° = +0.01 V
Fe³⁺ + e⁻ → Fe²⁺, E° = +0.77 V [CBSE Sample Paper 2018]
Answer:
We have to predict the spontaneity of the reaction
Ti⁴⁺ + Fe²⁺ → Ti³⁺ + Fe³⁺
E°_cell = +0.01 + (−0.77 V) = −0.76 V
As E°_cell is −ve, the above reaction is non-spontaneous, i.e., Ti⁴⁺ cannot oxidize Fe(II) to Fe(III).

Teacher's Note:
1. E°_cell = E°(species reduced) − E°(species oxidised) = 0.01 − 0.77; a negative value means the reaction does not occur.

Question 18. What is the effect of pH on the solution of K₂Cr₂O₇?
Answer:
In aqueous solution:
Cr₂O₇²⁻ (Dichromate ion, orange red) + H₂O ⇄ 2CrO₄²⁻ (Chromate ion, yellow) + 2H⁺
In acidic medium (i.e., decreasing pH), equilibrium shifts backward and the colour is orange red. In basic medium (i.e., increasing pH), equilibrium will shift forward and the solution is yellow.

Teacher's Note:
1. Use Le Chatelier's principle: added H⁺ pushes the equilibrium towards dichromate; added OH⁻ removes H⁺ and favours chromate.

Question 19. Give the relationship between the equivalent weight and molecular weight of KMnO₄ in (a) acidic (b) alkaline and (c) neutral media.
Answer:
(a) E = M/5
(b) In strongly alkaline medium, E = M. In weakly alkaline medium E = M/3
(c) E = M/3.

Teacher's Note:
1. Equivalent weight = M / (electrons gained): acidic 5 (Mn²⁺), strongly alkaline 1 (MnO₄²⁻), neutral or weakly alkaline 3 (MnO₂).

Question 20. Why is it not advisable to dissolve KMnO₄ in conc. H₂SO₄?
Answer:
With conc. H₂SO₄, KMnO₄ reacts to form Mn₂O₇ which on warming decomposes to MnO₂:
2KMnO₄ + 2H₂SO₄ (conc.) → Mn₂O₇ + 2KHSO₄ + H₂O
2Mn₂O₇ → 4MnO₂ + 3O₂ (decomposes explosively on warming)

Teacher's Note:
1. Mn₂O₇ is an oily, highly unstable oxide; this is a safety point often asked in exams.

Question 21. The +3 oxidation states of lanthanum (Z = 57), gadolinium (Z = 64) and lutetium (Z = 71) are especially stable. Why?
Answer:
This is because in the +3 oxidation state, they have empty, half-filled and completely filled 4f sub-shell respectively.

Teacher's Note:
1. La³⁺ = 4f⁰, Gd³⁺ = 4f⁷, Lu³⁺ = 4f¹⁴.

Question 22. Why Zr and Hf or Nb and Ta exhibit similar properties?
Or
Zirconium (atomic number 40) and hafnium (atomic number 72) occur together in minerals and they exhibit similar properties. Give reasons.
Answer:
Due to lanthanoid contraction, Hf (Z = 72) has size similar to that of Zr (Z = 40). Hence, their properties are similar and therefore, occur together. For the same reason, Nb and Ta have similar size and hence similar properties.

Teacher's Note:
1. Atomic radii: Zr 160 pm, Hf 159 pm – nearly equal despite 32 more electrons in Hf.

Short Answer Questions (3 Marks)

Question 23. Give reasons for each of the following:
(i) Transition metal fluorides are ionic in nature whereas bromides and chlorides are usually covalent in nature.
(ii) Chemistry of all the lanthanoids is quite similar.
Answer:
(i) As electronegativity of halogens decreases in the order F > Cl > Br, the ionic character of transition metal halides decreases in the order M—F > M—Cl > M—Br. Hence, fluorides are ionic whereas chlorides and bromides are covalent.
(ii) The change in the size of the lanthanoids due to lanthanoid contraction is very small as we proceed from La (Z = 57) to Lu (Z = 71). Hence, their chemical properties are similar. Moreover, their valence shell configuration remains the same because the electrons are added into the inner 4 f-subshell. Hence, they show similar chemical properties.

Teacher's Note:
1. Larger electronegativity difference between metal and halogen → more ionic bond.

Question 24. Assign reason for each of the following statements:
(i) The largest number of oxidation states are exhibited by the elements in the middle of the first row transition elements.
(ii) The atomic radii decrease in size with the increasing atomic number in the lanthanoid series.
Answer:
(i) The maximum number of oxidation states are shown by that element in a transition series which has maximum number of unpaired electrons. This is so in the middle of the series.
(ii) This is due to lanthanoid contraction. As the atomic number increases, the nuclear charge increases by one unit at each step and the extra electron enters the 4f subshell. Because of the diffused shape of f-orbitals, the shielding of one 4f electron by another is imperfect, so the effective nuclear charge increases and the atomic radii decrease steadily.

Teacher's Note:
1. Fewer oxidation states at the start (too few electrons) and at the end (too many paired electrons) of the series.

Question 25. Assign reasons for each of the following:
(i) Manganese exhibits the highest oxidation state of +7. [Raj. Board 2012]
(ii) Unlike Cr³⁺, Mn²⁺, Fe³⁺ and the subsequent M³⁺ ions of 3d series of elements, the 4d and 5d series metals generally do not form stable cationic species. [CBSE 2011]
Answer:
(i) Electronic configuration of ₂₅Mn is 3d⁵ 4s². As all the 7 electrons of 3d and 4s can participate in bonding, it exhibits highest oxidation state of +7.
(ii) In the 4d and 5d series higher oxidation states are more stable (because difference of energy between 4d and 5s or 5d and 6s is smaller than between 3d and 4s). In the higher oxidation states, bonds formed are mostly covalent. Hence, they do not form cationic species.

Teacher's Note:
1. High oxidation states of a metal are usually found in oxo-anions or covalent compounds (MnO₄⁻, OsO₄), not as free cations.

Question 26. How is the variability of oxidation states of the transition elements different from that of the non-transition elements? Illustrate with examples.
Answer:
The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., Fe²⁺ and Fe³⁺, Cu⁺ and Cu²⁺ etc.
In case of non-transition elements, the oxidation states differ by units of two, e.g., Pb²⁺ and Pb⁴⁺, Sn²⁺ and Sn⁴⁺ etc. Moreover, in transition elements, the higher oxidation states are more stable for heavier elements in a group. For example, in group 6, Mo (VI) and W (VI) are more stable than Cr (VI). In p-block elements, the lower oxidation states are more stable for heavier members due to inert pair effect, e.g., in group 14, Pb (II) is more stable than Pb (IV).

Teacher's Note:
1. Two differences to state: steps of one versus two units, and the opposite trend down a group.

Teacher's Note:
1. Zn (−0.76 V) is also more negative than expected because Zn²⁺ has the stable full 3d¹⁰ configuration.

Teacher's Note:
1. Higher the oxidation state of the metal in an oxide, more covalent and more acidic the oxide.

Question 29. The outer electronic configurations of two members of the lanthanoid series are as follows: 4f¹ 5d¹ 6s² and 4f⁷ 5d⁰ 6s². What are their atomic numbers? Predict the oxidation states exhibited by these elements in their compounds.
Answer:
Complete E.C. of 1st lanthanoid = [Xe]⁵⁴ 4f¹ 5d¹ 6s². Atomic No. = 58 (Ce)
Complete E.C. of 2nd lanthanoid = [Xe]⁵⁴ 4f⁷ 5d⁰ 6s². Atomic No. = 63 (Eu)
Oxidation states of 1st lanthanoid = +2 (4f²), +3 (4f¹), +4 (4f⁰)
Oxidation states of 2nd lanthanoid = +2 (4f⁷) and +3 (4f⁶).

Teacher's Note:
1. Atomic number = 54 (Xe core) + all the electrons written after it.

Question 30. The electronic configurations of actinide elements are not known with certainty. Explain.
Answer:
In actinides, 5f and 6d subshells are close in energy. The outermost 7s orbital remains filled with 2 electrons (7s²). The electron can easily jump from 5f to 6d or vice versa. Further, irregularities in electronic configurations are also related to the stabilities of f⁰, f⁷ and f¹⁴ occupancy of the 5f orbitals. Hence, they show a large number of oxidation states (Moreover, they are radioactive with short half-lives. Hence, their properties cannot be studied easily.)

Teacher's Note:
1. Two reasons to mention: closeness of 5f/6d energies and the radioactivity (short half-lives) of most actinoids.

Question 31. Account for the following:
(i) Oxidizing power in the series: VO₂⁺ < Cr₂O₇²⁻ < MnO₄⁻
(ii) Actinoid contraction is greater from element to element than lanthanoid contraction.
(iii) Oxoanions of a metal show higher oxidation states.
Answer:
(i) This is due to increasing stability of the lower species to which they are reduced.
(ii) This is due to poor shielding effect of 5f electrons of actinoids than 4f electrons of lanthanoids.
(iii) This is due to high electronegativity and multiple bond formation with metal by oxygen.

Teacher's Note:
1. In (i) the metal is in its highest state in each case (V +5, Cr +6, Mn +7); the oxidising power increases along the series.

Question 32. Account for the following:
(a) Europium (II) is more stable than cerium (II).
(b) Transition metals have high enthalpies of atomisation.
(c) Actinoid ions are generally coloured.
Answer:
(a) Europium has stable electronic configuration, i.e., [Xe] 4f⁷ 5d⁰ 6s⁰.
(b) Due to large number of unpaired electrons in their atoms, there is larger and stronger metallic bonding.
(c) Unpaired electrons are present in their ions which undergo f – f transition.

Teacher's Note:
1. Eu²⁺ has the half-filled 4f⁷ configuration; Ce²⁺ (4f²) has no such stability.

Question 33. What are transition elements ? Write electronic configurations of ₂₄Cr and ₂₉Cu and also mention their valencies and cations. (Jharkhand Board 2011)
or
What are transition elements ? Why are they called d-block elements ? Write the electronic configuration of the first and the last member of 3d transition series. (J & K Board 2012)
or
What are transition elements ? Write two characteristics of the transition elements. (CBSE 2015)
Answer:
Transition elements are the elements which have incompletely filled d-orbitals in their ground state or in any one of their oxidation states. They are called d-block elements because the last electron enters the (n − 1) d-orbital.
₂₄Cr = [Ar] 3d⁵ 4s¹ ; valencies +2, +3 (also +6) ; cations Cr²⁺, Cr³⁺
₂₉Cu = [Ar] 3d¹⁰ 4s¹ ; valencies +1, +2 ; cations Cu⁺, Cu²⁺
or First member of 3d series : ₂₁Sc = [Ar] 3d¹ 4s² ; last member : ₃₀Zn = [Ar] 3d¹⁰ 4s²
or Two characteristics : (i) They show variable oxidation states. (ii) They form coloured ions/compounds (they also form complexes, act as catalysts and are paramagnetic).

Teacher's Note:
1. Cr and Cu are the two exceptions in the 3d series: a half-filled d⁵ or full d¹⁰ subshell is more stable, so one 4s electron moves into 3d.

Long Answer Questions (5 Marks)

Question 34. (i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so ?
(ii) Which transition metal of 3d series has positive E°(M²⁺/M) value and why ?
(iii) Out of Cr³⁺ and Mn³⁺, which is a stronger oxidizing agent and why ?
(iv) Name a member of the lanthanoid series which is well known to exhibit + 2 oxidation state.
(v) Complete the following equation : MnO₄⁻ + 8H⁺ + 5e⁻ → [CBSE Delhi 2014]
Answer:
(i) Mn, because of presence of 5 unpaired electrons in 3d subshell.
(ii) Cu, because enthalpy of atomization and ionisation enthalpy is not compensated by hydration enthalpy.
(iii) Mn³⁺ because Mn²⁺ is more stable due to its half filled (3d⁵) configuration.
(iv) Eu⁺² (Eu).
(v) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Teacher's Note:
1. For (i), Mn can use its two 4s and five 3d electrons, giving every state from +2 up to +7.

Question 35. On the basis of Lanthanoid contraction, explain the following :
(a) Nature of bonding in La₂O₃ and Lu₂O₃.
(b) Trends in the stability of oxo salts of lanthanoids from La to Lu.
(c) Stability of the complexes of lanthanoids.
(d) Radii of 4d and 5d block elements.
(e) Trends in acidic character of lanthanoid oxides. [NCERT Exemplar]
Answer:
(a) La₂O₃ is ionic because La has largest atomic size and lowest ionization enthalpy, whereas Lu₂O₃ is covalent because Lu has smallest atomic size and has highest ionization enthalpy.
(b) As the size decreases from La to Lu, therefore, the stability of oxo cations also decreases.
(c) The stability of complexes increases as the size of lanthanoid decreases.
(d) Radii of 4d and 5d elements are almost same due to lanthanoid contraction.
(e) Acidic character of oxides of lanthanoid increases because metallic character decreases due to decrease in atomic size and increase in ionization enthalpy.

Teacher's Note:
1. Every part follows from one idea: Ln³⁺ becomes smaller from La to Lu, so its polarising power and covalent character rise.
2. Smaller, more highly charged ions bind ligands more strongly, which is why complex stability increases towards Lu.

Question 36. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of KNO₃ to give compound (B). On heating compound (C) with conc. H₂SO₄ and NaCl, Cl₂ gas is liberated and compound (D) of manganese is formed. Identify A, B, C, D along with reactions involved.
Answer:
A : KMnO₄ B : K₂MnO₄ C : MnO₂ D : MnCl₂
2KMnO₄ (A) \( \xrightarrow{\Delta} \) K₂MnO₄ (B) + MnO₂ (C) + O₂
MnO₂ (C) + 2KOH + KNO₃ → K₂MnO₄ (B) + KNO₂ + H₂O
MnO₂ (C) + 4NaCl + 4H₂SO₄ (conc.) → MnCl₂ (D) + 4NaHSO₄ + 2H₂O + Cl₂

Teacher's Note:
1. In "identify A, B, C, D" questions, balanced equations carry marks; check Mn oxidation states +7 → +6/+4 → +2 along the chain.

Question 37. (a) What is meant by disproportionation of an oxidation state? Give one example.
(b) Draw the structures of Cr₂O₇²⁻, CrO₄²⁻, MnO₄⁻.
(c) What is the effect of lanthanoid contraction beyond lanthanoids?
Answer:
(a) When any atom or ion undergoes oxidation and reduction simultaneously it is called disproportionation.
2Cu⁺ → Cu²⁺ + Cu
(b)
(c) Size of respective 4d and 5d series elements becomes comparable from fourth group onwards (e.g., Zr and Hf).

Teacher's Note:
1. Dichromate is two tetrahedra sharing one oxygen (one Cr–O–Cr bond); chromate and permanganate are single tetrahedra.

Question 38. What is a transition element ? Give reasons for the following :
(i) Transition elements have high enthalpies of atomisation.
(ii) Most of the compounds formed by transition metals are coloured.
(iii) Transition metals and most of their compounds show a paramagnetic behaviour.
(iv) Transition metals form complex compounds.
Answer:
A transition element is defined as an element which has incompletely filled d-orbitals in its ground state or in any one of its common oxidation states.
(i) Transition metals have a large number of unpaired electrons in their (n − 1)d and ns orbitals. These electrons take part in strong interatomic metallic bonding. Hence, their enthalpies of atomisation are high.
(ii) Transition metal ions have partly filled d-orbitals. In the presence of ligands, the d-orbitals split into two sets of different energy. An electron absorbs light of a particular wavelength from the visible region and jumps from a lower to a higher d-orbital (d-d transition). The colour observed is the complementary colour of the light absorbed.
(iii) Transition metal ions and most of their compounds contain unpaired electrons in the d-orbitals. Each unpaired electron has a magnetic moment due to its spin and orbital motion. Hence, they are paramagnetic. μ = \( \sqrt{n(n+2)} \) BM.
(iv) Transition metals form complexes because of (a) small size of the metal ions, (b) their high ionic charges and (c) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons from ligands, e.g., [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺.

Teacher's Note:
1. All four properties are linked to the partly filled d-subshell; in a 5-mark answer, connect each reason back to it.