CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Give reasons for the following: Eu²⁺ is a strong reducing agent.
Answer:
Eu²⁺ has a strong tendency to lose electrons to attain the more stable +3 oxidation state of lanthanoids, hence it is a strong reducing agent.

Teacher's Note:
1. Eu²⁺ (4f⁷) changes to the common +3 state on losing an electron; this is the same idea as Ce⁴⁺ acting as an oxidising agent in the opposite direction.

Question 2. Write the formula of an oxoanion of manganese (Mn) in which it shows the oxidation state equal to its group number.
Answer:
Formula of oxoanion of manganese is MnO₄⁻.
Oxidation state of Mn in this oxoanion = +7
Group number of Mn is 7.

Teacher's Note:
1. Check the oxidation state quickly: x + 4(−2) = −1 gives x = +7.

Question 3. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Answer:
Lanthanoids showing +4 oxidation state are ₅₈Ce, ₅₉Pr, ₆₅Tb.

Teacher's Note:
1. Ce is the best-known example: Ce⁴⁺ has the stable 4f⁰ configuration.

Question 4. Assign reason for the following: Copper (I) ion is not known in aqueous solution.
Answer:
In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu²⁺ ion.
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.

Teacher's Note:
1. Do not argue from configuration alone: Cu⁺ (3d¹⁰) looks stable, but the much larger hydration enthalpy of Cu²⁺ decides the matter in water.

Question 5. Account for the following: Zn is not considered as a transition element.
Answer:
In the electronic configuration of Zn the d-orbitals are completely filled in the ground state as well as in its common oxidation state. So, it is not regarded as transition metal.

Teacher's Note:
1. The definition checks the atom and its common ions: Zn (3d¹⁰ 4s²) and Zn²⁺ (3d¹⁰) both have a full d subshell.

Question 6. How would you account for the following: Among lanthanoids, La(III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
Answer:
Lanthanum and all the lanthanoids predominantly show +3 oxidation state. However, some of the lanthanoids also show +2 and +4 oxidation states in solution or in solid compounds. This irregularity arises mainly due to attainment of stable empty (4f⁰), half-filled (4f⁷) and fully filled (4f¹⁴) sub shell.
e.g., Ce⁴⁺ : 4f⁰, Eu²⁺ : 4f⁷
Tb⁴⁺ : 4f⁷, Yb²⁺ : 4f¹⁴

Teacher's Note:
1. Pair each unusual ion with its configuration in the answer; that is what earns the marks.

Question 7. Account for the following: Zr and Hf have almost similar atomic radii.
Answer:
Due to lanthanoid contraction the elements of 4d and 5d-series have similar atomic radii e.g., Zr = 145 pm and Hf = 144 pm.

Teacher's Note:
1. The 14 lanthanoids come between La and Hf; their contraction cancels the size increase expected from Zr to Hf.

Question 8. Zn²⁺ salts are white while Cu²⁺ salts are coloured. Why?
Answer:
Zn²⁺ ion has completely filled d-subshell and no d-d transition is possible. So zinc salts are white. Configuration of Cu²⁺ is [Ar] 3d⁹. It has partly filled d-subshell and hence it is coloured due to d-d transition.

Teacher's Note:
1. Write both configurations, Zn²⁺ 3d¹⁰ and Cu²⁺ 3d⁹; the contrast between them is the whole answer.

Question 9. Copper atom has completely filled d-orbitals in its ground state but it is a transition element. Why? [CBSE Chennai 2015]
Answer:
Copper exhibits +2 oxidation state wherein it has incompletely filled d orbitals (3d⁹4s⁰) hence, a transition element.

Teacher's Note:
1. The definition allows an incomplete d subshell in the atom or in any common oxidation state; quote Cu²⁺ = 3d⁹.

Question 10. Give reason: Zn is soft whereas Cr is hard. [CBSE South 2016]
Answer:
Cr (3d⁵4s¹) has five unpaired electrons in its d-orbitals whereas Zn (3d¹⁰4s²) has no unpaired electrons in its d-orbitals. As a result of this weak metallic bonds exist in Zn whereas strong metallic bonds exist in Cr. Hence, Zn is soft whereas Cr is hard.

Teacher's Note:
1. Link hardness, melting point and enthalpy of atomisation to the number of unpaired d electrons available for metallic bonding.

Short Answer Questions (2 Marks)

Question 11. Why is europium (II) more stable than cerium (II)?
Answer:
Europium (II) has electronic configuration [Xe]4f⁷5d⁰ while cerium (II) has electronic configuration [Xe] 4f¹ 5d¹. In Eu²⁺, 4f subshell is half filled and 5d-subshell is empty. Since, half filled and completely filled electronic configurations are more stable, so Eu²⁺ ions is more stable than Ce²⁺ in which neither 4f subshell nor 5d subshell is half filled or completely filled.

Teacher's Note:
1. Questions on unusual lanthanoid oxidation states almost always come down to reaching 4f⁰, 4f⁷ or 4f¹⁴.

Question 12. The magnetic moment of a transition metal ion is found to be 3.87 BM. How many number of unpaired electrons are present in it?
Answer:
Magnetic moment, μ_eff = 3.87 BM corresponds to the number of unpaired electrons, n = 3 by applying the formula.
\( \mu_eff = \sqrt{n(n+2)} \) BM
For n = 1, μ = 1.73 B.M, for n = 2, μ = 2.83 B.M
For n = 3, μ = 3.87 B.M and so on.
Hence, number of unpaired electrons = 3.

Teacher's Note:
1. Remember the spin-only values 1.73, 2.83, 3.87, 4.90 and 5.92 BM for 1 to 5 unpaired electrons; it saves time in the exam.

Teacher's Note:
1. A more negative E°(M³⁺/M²⁺) means M²⁺ is more easily oxidised, i.e. a stronger reducing agent; a large positive value means M²⁺ resists oxidation.

Question 14. How would you account for the following:
(i) 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.
(ii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second (4d) series.
Answer:
(i) Electrode potential (E°) value is the sum of three factors:
(a) Enthalpy of atomisation; Δ_aH for Cu_(s) → Cu_(g)
(b) Ionisation enthalpy; Δ_iH for Cu_(g) → Cu²⁺_(g)
(c) Hydration enthalpy; Δ_hydH for Cu²⁺_(g) → Cu²⁺_(aq)
In case of copper the sum of enthalpy of atomisation and ionisation enthalpy is greater than enthalpy of hydration. This is why E°_M²⁺/M for Cu is positive.
(ii) This is due to lanthanoid contraction.

Teacher's Note:
1. The positive E° of Cu is also why copper does not liberate H₂ from dilute acids.

Question 15. Write the electronic configuration of Ce³⁺ ion, and calculate the magnetic moment on the basis of ‘spin-only’ formula. [Atomic no. of Ce = 58]
Answer:
Ce (Z = 58) = [Xe] 4f¹ 5d¹ 6s²
∴ Ce³⁺ = [Xe] 4f¹ 5d⁰ 6s⁰
Therefore, it has only one unpaired electron. i.e., n = 1
∴ \( \mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \) = 1.73 B.M.

Teacher's Note:
1. Remove the 6s electrons first, then the 5d electron, when writing lanthanoid ion configurations.

Question 16. (i) Which metal in the first transition series (3d-series) exhibits +1 oxidation state most frequently and why?
(ii) Which of following cations are coloured in aqueous solutions and why?
Sc³⁺, V³⁺, Ti⁴⁺, Mn²⁺
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer:
(i) Copper exhibits +1 oxidation state in its compounds. Electronic configuration of Cu in the ground state is 3d¹⁰ 4s¹. So, Cu can easily lose 4s¹ electron to attain a stable 3d¹⁰ configuration. Thus, it shows +1 oxidation state.
(ii) Only those ions will be coloured which have partially filled d-orbitals facilitating d-d transition. Ions with d⁰ and d¹⁰ will be colourless.
From electronic configuration of the ions, V³⁺ (3d²) and Mn²⁺ (3d⁵), are all coloured. Ti⁴⁺ (3d⁰) and Sc³⁺ (3d⁰) are colourless.

Teacher's Note:
1. Count the d electrons as (group number − charge): Sc³⁺ 3 − 3 = 0, Ti⁴⁺ 4 − 4 = 0, V³⁺ 5 − 3 = 2, Mn²⁺ 7 − 2 = 5.

Question 17. How would you account for the following:
(i) Cr²⁺ is reducing in nature while with the same d-orbital configuration (d⁴), Mn³⁺ is an oxidising agent.
(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.
Answer:
(i) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺ couples are
Cr³⁺_(aq) + e⁻ → Cr²⁺_(aq); E° = −0.41 V
Mn³⁺_(aq) + e⁻ → Mn²⁺_(aq); E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising agent.
(ii) Middle of the transition series contains greater number of unpaired electrons in (n − 1)d and ns orbitals.

Teacher's Note:
1. Link the E° values to configurations: Cr²⁺ → Cr³⁺ gives the stable half-filled t_2g³, while Mn³⁺ → Mn²⁺ gives the stable 3d⁵.

Question 18. What is lanthanoid contraction and what is it due to? Write two consequences of lanthanoid contraction.
Answer:
Lanthanoid contraction: The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction.
It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction:
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium (Zr) is same as that of hafnium (Hf).

Teacher's Note:
1. Write “imperfect shielding by 4f electrons” explicitly; just saying “size decreases” does not explain the cause.

Question 19. What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution.
Answer:
Disproportionation reaction involves the oxidation and reduction of the same substance. The examples of disproportionation reaction are:
(i) Aqueous NH₃ when treated with Hg₂Cl₂ (solid) forms mercury aminochloride disproportionatively.
Hg₂Cl₂ + 2NH₃ → Hg + Hg(NH₂)Cl + NH₄Cl
(ii) 2Cu⁺ → Cu + Cu²⁺

Teacher's Note:
1. Mark oxidation states to show it: in 2Cu⁺ → Cu + Cu²⁺, Cu goes from +1 to both 0 and +2.

Question 20. (i) Write two characteristic of the transition elements.
(ii) Which of the 3d-block elements may not be regarded as the transition elements and why?
Answer:
(i) Elements which have incompletely filled d-orbitals in their ground state or in any one of their oxidation states are called transition elements.
Characteristics of transition elements:
(a) They show variable oxidation states.
(b) They exhibit catalytic properties.
(ii) Zn, Cd, Hg are considered as d-block elements but not as transition elements because they do not have partly filled d-orbitals in their atomic state or their common oxidation states (i.e., Zn²⁺, Cd²⁺, Hg²⁺).

Teacher's Note:
1. For the 3d series specifically the answer is zinc; Cd and Hg belong to the 4d and 5d series.

Question 21. (i) Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. [CBSE Delhi Set-1, 3 2017]
(ii) Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number. [CBSE Delhi Set-2 2017]
Answer:
(i) MnO₄⁻/KMnO₄
(ii) Cr₂O₇²⁻/CrO₄²⁻/K₂Cr₂O₇/K₂CrO₄

Teacher's Note:
1. Mn is in group 7 and Cr in group 6; check the answer by computing the oxidation state: Mn in MnO₄⁻ is +7 and Cr in Cr₂O₇²⁻ is +6.

Short Answer Questions (3 Marks)

Question 22. (i) Explain the cause of paramagnetism in lanthanoid ions.
(ii) Nb and Ta exhibit similar properties. Give reason.
(iii) Among the ionic species, Sc³⁺, Ce⁴⁺ and Eu²⁺, which one is a good oxidising agent.
Answer:
(i) Due to presence of unpaired electrons in f-orbital lanthanoid ions are paramagnetic in nature.
(ii) Due to lanthanoid contraction, their sizes are same. Hence, their properties are similar.
(iii) Ce⁴⁺. The stable oxidation state of lanthanoids is +3. Ce⁴⁺ tends to accept an electron to change to +3 state. Hence, it acts as a good oxidising agent.

Teacher's Note:
1. Nb (4d) and Ta (5d) are a pair like Zr/Hf and Mo/W; all such pairs are explained by lanthanoid contraction.

Question 23. Following are the transition metal ions of 3d series:
Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in aqueous solution and why?
(ii) Which ion is strong oxidising agent and why?
(iii) Which ion is colourless and why?
Answer:
(i) Ti⁴⁺ has highest oxidation state among the given ions. Ti⁴⁺ has stable inert gas configuration and hence, most stable in aqueous solution.
On the other hand, V²⁺, Mn³⁺, Cr³⁺ have unstable electronic configuration and hence, are less stable.
(ii) Mn³⁺ is the strongest oxidising agent among the given ions. Mn³⁺ (3d⁴) readily accepts an electron to change to Mn²⁺, which has the stable half-filled 3d⁵ configuration (E°_{Mn³⁺/Mn²⁺} = +1.57 V).
(iii) Due to absence of unpaired electron in Ti⁴⁺, it is a colourless ion.
E.C. of Ti⁴⁺ : [Ar]3d⁰4s⁰

Teacher's Note:
1. A high oxidation state does not by itself make an ion an oxidising agent; Ti⁴⁺ is already at a stable noble-gas configuration, whereas Mn³⁺ gains stability by being reduced.

Question 24. Compare qualitatively the first and second ionisation potentials of copper and zinc. Explain the observation.
Answer:
The electronic configuration of Zn and Cu are:
Zn: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
From the above configuration it is clear that first ionisation energy of Zn is greater than that of Cu (because of 4s² and 4s¹ configuration of Zn and Cu respectively). More energy is needed to remove an electron from 4s² than from 4s¹.
The second I.E. of Cu is higher than that of Zn because for Cu⁺ the configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ and for Zn⁺ the configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹, it is easier to remove 4s¹ electron of Zn⁺ than a 3d-electron from 3d¹⁰ (stable configuration).

Teacher's Note:
1. State the result as a pair: IE₁: Zn > Cu, but IE₂: Cu > Zn, then give the reason for each.

Question 25. (i) Ce (IV) is a good analytical reagent. Why?
(ii) Account for the following: Copper(I) compounds are white whereas copper(II) compounds are coloured.
Answer:
(i) The E° value for Ce⁴⁺/Ce³⁺ is 1.74 V which suggests that it can oxidise water. However the reaction rate is very slow and hence Ce (IV) is a good analytical reagent.
(ii) Cu(I) compounds have completely filled d-orbitals and there are no vacant d-orbitals for promotion of electrons whereas in Cu(II) compounds have one unpaired electron which is responsible for colour formation.

Teacher's Note:
1. Ce(IV) is used as an oxidising titrant because it is a strong oxidant that, in practice, does not react with the water it is dissolved in.

Question 26. How do the oxides of transition elements in lower oxidation states differ from those in higher oxidation states and why?
Answer:
In the lower oxidation state, the transition metal oxides are basic and they are acidic if the metal is in higher oxidation state. The oxides are amphoteric when the metal is in intermediate oxidation state.
For example, Mn₂O₃ (+3, basic), MnO₂ (+4, amphoteric), Mn₂O₇ (+7, acidic)
In case of lower oxide of a transition metal, the metal atom has a low oxidation state. This means some of the valence electrons of the metal atom are not involved in bonding, hence, these can be used for donation. Thus, these are act as bases.

Teacher's Note:
1. Higher oxidation state means a smaller, more highly charged metal centre and more covalent bonding, so the oxide behaves as an acid anhydride (Mn₂O₇ gives HMnO₄).

Question 27. How would you account for the following:
(i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
(ii) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺ couple.
(iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
Answer:
(i) This is due to lanthanoid contraction.
(ii) Much larger third ionisation energy of Mn (where change is d⁵ to d⁴) is mainly responsible for this. This also explains that +3 state of Mn is of little importance.
From the relation, ΔG° = −nFE°
More positive is the value of E°, reaction will be feasible.
Mn³⁺ (3d⁴) + e⁻ → Mn²⁺ (3d⁵, more stable, half filled);   Fe³⁺ (3d⁵, more stable, half filled) + e⁻ → Fe²⁺ (3d⁶)
Hence, E°_value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(iii) Manganese can form pπ-dπ bond with oxygen by utilising 2p-orbital of oxygen and 3d-orbital of manganese due to which it can show highest oxidation state of +7. While with fluorine it cannot form such pπ-dπ bond thus, it can show a maximum oxidation state of +4.

Teacher's Note:
1. For part (iii), also remember the general point: O and F are the smallest, most electronegative atoms, so they can stabilise the highest oxidation states.

Question 28. (a) Assign reasons for the following:
(i) Cu(I) ion is not known to exist in aqueous solutions.
(ii) Transition metals are much harder than the alkali metals.
(b) Name the lanthanoids which show abnormally low value of third ionisation enthalpy.
Answer:
(a) (i) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu²⁺ ion.
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.
(ii) This is attributed to the involvement of greater number of electrons from (n − 1)d in addition to the ns electrons in the interatomic metallic bonding.
(b) Third ionization enthalpy of lanthanoid is low if it leads to stable empty, half filled or completely filled configuration, as indicated by the abnormally low third ionization enthalpies of La, Gd, and Lu.

Teacher's Note:
1. La³⁺ is 4f⁰, Gd³⁺ is 4f⁷ and Lu³⁺ is 4f¹⁴; writing these configurations justifies part (b).

Question 29. Account for the following:
(i) The transition metals and their compounds act as good catalysts.
(ii) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(iii) The magnetic moment (B.M.) of Fe²⁺ ion is \( \sqrt{24} \).
Answer:
(i) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples.
(ii) Lowest oxidation compounds of transition metals are basic due to their ability to get oxidised to higher oxidation states. Whereas, the higher oxidation state of metal and compounds gets reduced to lower ones and hence acts as acidic in nature.
e.g., MnO is basic whereas Mn₂O₇ is acidic.
(iii) Fe²⁺ ⇒ 3d⁶, ∴ n = 4
Magnetic moment = \( \sqrt{n(n+2)} \) B.M. = \( \sqrt{4(4+2)} \) = \( \sqrt{24} \) B.M.

Teacher's Note:
1. In 3d⁶ five electrons go singly into the five d orbitals and the sixth pairs up, leaving 4 unpaired; \( \sqrt{24} \) ≈ 4.90 B.M.

Question 30. (a) Account the following:
(i) Transition metals form large number of complex compounds.
(ii) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(iii) Which of following cations are coloured in aqueous solutions and why?
Sc³⁺, V³⁺, Ti⁴⁺, Mn²⁺
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer:
(a) (i) Transition metals form a large number of complex compounds due to following reasons:
(1) Comparatively smaller size of metal ions.
(2) High ionic charges.
(3) Availability of d-orbitals for bond formation.
(ii) Because Mn²⁺ is stable due to half filled configuration. Thus Mn³⁺ has high tendency to form Mn²⁺ while Cr³⁺ is more stable than Cr²⁺.
(iii) Only those ions will be coloured which have partially filled d-orbitals facilitating d-d transition. Ions with d⁰ and d¹⁰ will be colourless.
From electronic configuration of the ions, V³⁺ (3d²) and Mn²⁺ (3d⁵), are all coloured. Ti⁴⁺ (3d⁰) and Sc³⁺ (3d⁰) are colourless.

Teacher's Note:
1. Cr³⁺ (t_2g³) is stable, so Cr²⁺ tends to be oxidised and its E°(Cr³⁺/Cr²⁺) is negative; Mn behaves the opposite way.

Question 31. (i) Out of the ions Ag⁺, Co²⁺ and Ti⁴⁺ which will be coloured in aqueous solution?
(ii) If each one of the above ionic species is placed in a magnetic field, how will they respond and why?
Answer:
(i) The ionic species which possesses unpaired electron or electrons in (n − 1)d-subshell will show colour. Out of the ions Ag⁺ (4d¹⁰), Co²⁺ (3d⁷) and Ti⁴⁺ (3d⁰), Co²⁺ will be coloured as it contains three unpaired electrons, Ag⁺ and Ti⁴⁺ will be colourless.
(ii) When placed in a magnetic field, Co²⁺ will be attracted because it is paramagnetic due to unpaired electrons. Ag⁺ and Ti⁴⁺ ions will be repelled by the magnetic field as they are diamagnetic.

Teacher's Note:
1. Colour and paramagnetism go together for simple 3d ions: both need unpaired d electrons.

Question 32. (a) Explain the following:
The enthalpies of atomization of transition metals are quite high.
(b) Explain the following observations.
(i) With the same d-orbital configuration (d⁴), Cr²⁺ is a reducing agent while Mn³⁺ is an oxidising agent.
(ii) There is hardly any increase in atomic size with increasing atomic numbers in a series of transition metals.
Answer:
(a) As transition metals have a large number of unpaired electrons in the d-orbitals of their atoms they have strong interatomic attraction or metallic bonds. Hence, they have high enthalpy of atomization.
(b) (i) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺ couples are
Cr³⁺_(aq) + e⁻ → Cr²⁺_(aq); E° = −0.41 V
Mn³⁺_(aq) + e⁻ → Mn²⁺_(aq); E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising agent.
(ii) As one proceeds 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. As a result, the atomic radii remain practically same after chromium.

Teacher's Note:
1. In part (ii), the key phrase is that the new electron enters the inner (n − 1)d subshell, so it shields the outer ns electrons.

Long Answer Questions (5 Marks)

Question 33. (a) Give reason:
(i) Sc (21) is a transition element but Ca (20) is not.
(ii) The Fe²⁺ is much more easily oxidised to Fe³⁺ than Mn²⁺ to Mn³⁺.
(b) How would you account for the following:
(i) Metal-metal bonding is more extensive in the 4d and 5d series of transition elements than the 3d series.
(ii) Mn (III) undergoes disproportionation reaction easily.
(iii) Co (II) is easily oxidised in the presence of strong ligands.
Answer:
(a) (i) Sc (21) is a transition element but Ca (20) is not because of incompletely filled 3d orbitals.
(ii) Electronic configuration of Mn²⁺ is 3d⁵ which is half filled and hence stable. Therefore, third ionization enthalpy is very high, i.e., 3^rd electron cannot be lost easily. In case of Fe²⁺, electronic configuration is 3d⁶. Hence, it can lose one electron easily to give the stable configuration 3d⁵.
(b) (i) The metals of 4d and 5d-series have more frequent metal bonding in their compounds than the 3d-metals because 4d and 5d-orbitals are more exposed in space than the 3d-orbitals. So the valence electrons are less tightly held and form metal-metal bonding more frequently.
(ii) Mn³⁺ is less stable and changes to Mn²⁺ which is more stable due to half-filled d-orbital configuration. That is why, Mn³⁺ undergoes disproportionation reaction.
(iii) The tendency to form complexes is high for Co(III) as compared to Co(II). Co²⁺ ions are very stable and are difficult to oxidise. Co³⁺ ions are less stable and are reduced by water. In contrast many Co(II) complexes are readily oxidised to Co(III) complexes and Co(III) complexes are very stable, e.g.,
[Co(NH₃)₆]²⁺ \( \xrightarrow{\text{Air}} \) [Co(NH₃)₆]³⁺
This happens because the crystal field stabilisation energy of Co(III) with a d⁶ (t_2g⁶) configuration is higher than for Co(II) with a d⁷ (t_2g⁶ e_g¹) arrangement.

Teacher's Note:
1. For Sc, write its configuration [Ar]3d¹4s²; Ca is [Ar]4s² with no d electron at all.
2. In (b)(ii), show the disproportionation: 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺.

Question 34. 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 configuration
(ii) oxidation states
(iii) ionisation enthalpies and
(iv) atomic sizes.
Answer:
(i) Electronic configuration: Electronic configuration of group 3 elements (Sc, Y, La) is [Noble gas] (n − 1)d¹ ns².
Elements of group 6 (Cr, Mo and W) show exception in electronic configuration. For Cr and Mo [Noble gas] (n − 1)d⁵ ns¹ and for W it is [Noble gas] 4f¹⁴ 5d⁴ 6s².
Group 11 elements (Cu, Ag and Au) also show exceptional electronic configuration:
Cu : [Ar] 3d¹⁰ 4s¹, Ag : [Kr]4d¹⁰ 5s¹, Au : [Xe]4f¹⁴ 5d¹⁰ 6s¹.
Group 10 (Ni, Pd and Pt) also show anomalous electronic configuration:
Ni : [Ar] 3d⁸ 4s²
Pd : [Kr] 4d¹⁰ 5s⁰
Pt : [Xe] 4f¹⁴ 5d⁹ 6s¹
(ii) Oxidation states: Elements within the same group show similar oxidation states. Highest number of oxidation states are shown by the elements lying in the middle of the transition series. Minimum oxidation states are shown by the elements lying near to left and far right side of the series. Stability of higher oxidation states increases from first to third series.
(iii) Ionisation enthalpies: Ionisation enthalpies generally decrease down a group. This trend is followed from 3d to 4d-elements but the ionisation enthalpies either remain same or increase in going from 4d to 5d-series with the same group. This reverse trend is due to the poor shielding of the nuclear charge by the inner 4f-electrons. This increases the Z_eff and in turn increases the ionisation enthalpy.
(iv) Atomic size: Due to poor shielding of nuclear charge by 4f-electrons, increase in Z_eff decreases the size. So, the atomic size increase from 3d to 4d but decrease or remain almost the same from 4d to 5d.

Teacher's Note:
1. A good example for (ii): Fe is common as +2/+3, but Ru and Os in the same group reach +8 (RuO₄, OsO₄).

Question 35. (a) The elements of 3d transition series are given as
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(i) Write the element which shows maximum number of oxidation states. Given reason.
(ii) Which element has the highest melting point?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidising agent in +3 oxidation state and why?
(v) Why Mn₂O₃ is basic whereas Mn₂O₇ is acidic?
Answer:
(i) Mn shows maximum no. of oxidation states from +2 to +7 because Mn has maximum number of unpaired electrons in 3d sub-shell.
(ii) Cr has maximum melting point, because it has 6 unpaired electrons in the valence shell, hence it has strong interatomic interaction.
(iii) Sc shows only +3 oxidation state because after losing 3 electrons, it has noble gas electronic configuration.
(iv) Mn is strong oxidising agent in +3 oxidation state because change of Mn³⁺ to Mn²⁺ give stable half filled (d⁵) electronic configuration,
E°_{(Mn³⁺/Mn²⁺)} = 1.5 V.
(v) Basic nature of oxides decreases and acidic nature increases with increase in oxidation state of the metal. Oxidation state of Mn in Mn₂O₃ is +3 while in Mn₂O₇ is +7.

Teacher's Note:
1. Mn reaches +7 because it can use all seven of its 3d⁵ 4s² electrons in bonding.

Question 36. (a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd and Hg are soft metals.
(b) Give reason:
Iron has higher enthalpy of atomization than that of copper.
(c) What are interstitial compounds? Write their properties.
Answer:
(a) (i) Transition elements can use their ns and (n − 1)d orbital electrons for bond formation therefore, they show variable oxidation states.
For example, Sc has ns² (n − 1)d¹ electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.
(ii) In Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, the metallic bonds are weak. That is why they are soft metals with low melting and boiling points.
(b) Greater the number of unpaired electrons, stronger is the metallic bond and therefore, higher is the enthalpy of atomisation. Since, iron has greater number of unpaired electrons than copper hence has higher enthalpy of atomisation.
(c) When small atoms of non metals like H, C, B, N etc. occupy vacant interstitial spaces in the crystal lattices of metals, they give rise to interstitial compounds like hydrides, carbides. Few properties are as follow
(i) They have high melting points than pure metals.
(ii) They are conductive.
(iii) They are chemically inert.

Teacher's Note:
1. Interstitial compounds are usually non-stoichiometric (e.g. TiC, Fe₃H, VH_0.56) and are harder than the parent metal.