Intext Questions
1. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Answer
A transition element is one with an incompletely filled d subshell in the atom or in any of its common oxidation states. Silver (Z = 47, [Kr]4d¹⁰5s¹) shows the +2 oxidation state (e.g. AgF₂, AgO), in which its configuration is 4d⁹ – an incomplete d subshell. Hence silver is a transition element.
2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol⁻¹. Why?
Answer
Enthalpy of atomisation depends on the strength of metallic bonding, which in transition metals involves both ns and unpaired (n−1)d electrons. Zinc (3d¹⁰4s²) has no unpaired d electrons, so only the 4s electrons take part in metallic bonding. The metallic bonds are therefore weak and the enthalpy of atomisation is lowest.
3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Answer
Manganese (3d⁵4s²). It has the largest number of unpaired electrons (five 3d + two 4s electrons can take part in bonding), so it shows all oxidation states from +2 to +7.
4. The E°(M²⁺/M) value for copper is positive (+0.34 V). What is possible reason for this? (Hint: consider its high ΔaH° and low ΔhydH°)
Answer
E°(M²⁺/M) depends on the energy needed to change M(s) into M²⁺(aq): enthalpy of atomisation + ionisation enthalpies − hydration enthalpy. Copper has a high enthalpy of atomisation and high second ionisation enthalpy, and the hydration enthalpy of Cu²⁺ is not large enough to compensate. The overall change is energetically unfavourable, so E° is positive and copper does not liberate H₂ from acids.
5. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Answer
Ionisation enthalpy generally increases along the series as nuclear charge increases, but irregularities arise because removal of electrons changes the relative energies of 4s and 3d orbitals and the stability of the resulting configuration. Extra stable configurations (d⁰, d⁵, d¹⁰) and exchange energy matter: e.g. Cr and Cu have unusually high second ionisation enthalpies because removing the second electron breaks the stable d⁵ and d¹⁰ configurations, while Zn has a low second IE because removal gives the stable 3d¹⁰ ion. Mn²⁺ (d⁵) gives a high third IE.
6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer
Oxygen and fluorine are very small and highly electronegative, so they can oxidise the metal to its highest oxidation state. Fluorine forms very strong bonds (high lattice/bond energy) and oxygen can also form multiple bonds with the metal (e.g. in Mn₂O₇, MnO₄⁻), which stabilises the high oxidation state.
7. Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Answer
Cr²⁺ is the stronger reducing agent. E°(Cr³⁺/Cr²⁺) = −0.41 V is much lower than E°(Fe³⁺/Fe²⁺) = +0.77 V. Cr²⁺ (d⁴)
⇒ Cr³⁺ (d³) gives the very stable half-filled t₂g³ arrangement in water, whereas Fe²⁺ (d⁶)
⇒ Fe³⁺ (d⁵) gives less extra stability in aqueous medium.
8. Calculate the 'spin only' magnetic moment of M²⁺(aq) ion (Z = 27).
Answer
Z = 27 is cobalt: [Ar]3d⁷4s²
⇒ Co²⁺ = [Ar]3d⁷, which has 3 unpaired electrons (n = 3).
μ = √(n(n + 2)) = √(3 × 5) = √15 = 3.87 BM
9. Explain why Cu⁺ ion is not stable in aqueous solutions?
Answer
In water, Cu⁺ disproportionates: 2Cu⁺(aq)
⇒ Cu²⁺(aq) + Cu(s), and E° for this change is favourable. Although the second ionisation enthalpy of Cu is high, the hydration enthalpy of the small, doubly charged Cu²⁺ is much more negative than that of Cu⁺, which more than compensates. So Cu²⁺(aq) is more stable than Cu⁺(aq).
10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer
The 5f orbitals are more diffuse than 4f orbitals, so 5f electrons shield each other from the nuclear charge even more poorly. Hence the effective nuclear charge increases more sharply from one actinoid to the next, and the contraction in size is greater.
Exercises
1. Write down the electronic configuration of: (i) Cr³⁺ (ii) Pm³⁺ (iii) Cu⁺ (iv) Ce⁴⁺ (v) Co²⁺ (vi) Lu²⁺ (vii) Mn²⁺ (viii) Th⁴⁺
Answer
(i) Cr³⁺: [Ar]3d³ (ii) Pm³⁺: [Xe]4f⁴ (iii) Cu⁺: [Ar]3d¹⁰ (iv) Ce⁴⁺: [Xe] (v) Co²⁺: [Ar]3d⁷ (vi) Lu²⁺: [Xe]4f¹⁴5d¹ (vii) Mn²⁺: [Ar]3d⁵ (viii) Th⁴⁺: [Rn]
2. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Answer
Mn²⁺ has the half-filled 3d⁵ configuration, which is extra stable; removing an electron to form Mn³⁺ (3d⁴) needs a very high third ionisation enthalpy. Fe²⁺ (3d⁶) becomes Fe³⁺ (3d⁵) – a half-filled, extra stable configuration – on losing one electron, so Fe²⁺ is easily oxidised.
3. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer
The +2 state arises by loss of the two 4s electrons. Moving from Sc to Mn, the nuclear charge increases and the number of 3d electrons in M²⁺ increases from d¹ to d⁵. The 3d electrons are held more firmly, and the third ionisation enthalpy rises, so oxidation beyond +2 becomes harder. Thus Sc²⁺ is virtually unknown, Ti²⁺, V²⁺ and Cr²⁺ are strong reducing agents, while Mn²⁺ (stable d⁵) is the most stable +2 ion; E°(M³⁺/M²⁺) becomes more positive in this direction.
4. 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.
Answer
Configurations matter significantly: ions with d⁰, d⁵ and d¹⁰ configurations (or the half-filled t₂g³ set in water) are especially stable. Examples: Sc³⁺ (d⁰) and Ti⁴⁺ (d⁰) are the stable states of Sc and Ti; Mn²⁺ (d⁵) and Fe³⁺ (d⁵) are very stable; Zn²⁺ (d¹⁰) is the only ion of zinc; Cr³⁺ (t₂g³) is the stable state of Cr. However, configuration is not the only factor – hydration enthalpy, lattice energy and ionisation enthalpy also decide stability. For example, Cu²⁺ (d⁹) is more stable in water than Cu⁺ (d¹⁰) because of its much larger hydration enthalpy.
5. 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?
Answer
3d³4s² (vanadium): +5 (also +2, +3, +4).
3d⁵4s² (manganese): +2 and +7; 3d⁵4s¹ (chromium): +3 and +6.
3d⁸4s² (nickel): +2.
3d⁴: no element has a 3d⁴4s² ground state (chromium is 3d⁵4s¹), so no such case arises; if considered as chromium, +3 and +6.
6. 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.
Answer
Vanadate VO₄³⁻ / VO₃⁻ (V = +5, group 5); chromate CrO₄²⁻ and dichromate Cr₂O₇²⁻ (Cr = +6, group 6); permanganate MnO₄⁻ (Mn = +7, group 7).
7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer
Lanthanoid contraction is the steady decrease in atomic and ionic (M³⁺) radii from lanthanum to lutetium. It occurs because each added electron enters the 4f subshell, whose shielding is poor, so the effective nuclear charge increases steadily.
Consequences: (i) The second (4d) and third (5d) transition series have almost identical radii (Zr 160 pm, Hf 159 pm), so Zr and Hf have very similar properties and occur together, making separation difficult. (ii) Lanthanoids themselves have very similar properties and are hard to separate. (iii) Basic character of the hydroxides decreases from La(OH)₃ to Lu(OH)₃. (iv) The 5d metals have high densities and their ionisation enthalpies are close to or higher than those of the 4d metals.
8. 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?
Answer
Characteristics: they are hard metals with high melting and boiling points and high enthalpies of atomisation; show variable oxidation states; form coloured ions and compounds; are often paramagnetic; act as good catalysts; form complex compounds; form interstitial compounds and alloys.
They are called transition elements because they lie between the s-block and p-block and their properties are transitional between the highly electropositive s-block and the more electronegative p-block elements.
Zn, Cd and Hg have completely filled (n−1)d¹⁰ orbitals in the ground state and in their common oxidation states, so they are not regarded as transition elements.
9. In what way is the electronic configuration of the transition elements different from that of the non transition elements?
Answer
Transition elements have the general configuration (n−1)d¹⁻¹⁰ ns¹⁻² with an incompletely filled d subshell (in atom or ion); the last electron enters the penultimate (n−1)d subshell. Non-transition (main group) elements have ns¹⁻² or ns²np¹⁻⁶ valence configurations, with the last electron entering the outermost s or p subshell, and any inner d subshell is either absent or completely filled.
10. What are the different oxidation states exhibited by the lanthanoids?
Answer
The principal and most stable oxidation state is +3. Some show +2 (Eu²⁺, Yb²⁺, Sm²⁺) and some +4 (Ce⁴⁺, and also Pr, Nd, Tb, Dy), usually where this gives an empty, half-filled or completely filled 4f subshell (f⁰, f⁷, f¹⁴). Eu²⁺ is a strong reducing agent and Ce⁴⁺ a good oxidising agent, as both tend to change to +3.
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.
Answer
(i) They contain unpaired electrons in their d orbitals; the spin and orbital motion of these electrons produce magnetic moments, so they are attracted by a magnetic field.
(ii) They have many unpaired d electrons which, together with ns electrons, take part in strong metallic bonding (strong interatomic interactions), giving high enthalpies of atomisation.
(iii) Their ions have partly filled d orbitals. In the presence of ligands the d orbitals split, and an electron absorbs visible light to jump to a higher d level (d–d transition). The colour seen is complementary to the light absorbed.
(iv) They can show variable oxidation states and form unstable intermediate compounds, and they provide a large surface for adsorption of reactants (e.g. Fe in Haber process, V₂O₅ in Contact process, Ni in hydrogenation). This lowers the activation energy.
12. What are interstitial compounds? Why are such compounds well known for transition metals?
Answer
Interstitial compounds are formed when small atoms such as H, C, N or B are trapped in the voids (interstices) of a metal crystal lattice, e.g. TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇. They are usually non-stoichiometric, very hard, have high melting points (higher than the pure metals), retain metallic conductivity and are chemically inert.
Transition metals have close-packed crystal lattices with many interstitial voids of suitable size, so these small atoms fit easily into them.
13. How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
Answer
In transition metals, the (n−1)d and ns energies are close, so different numbers of electrons can be lost and the oxidation states typically differ by one unit, e.g. Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺, Mn from +2 to +7, V from +2 to +5. In non-transition (p-block) elements, oxidation states usually differ by two units because of the inert pair effect, e.g. Sn²⁺/Sn⁴⁺, Pb²⁺/Pb⁴⁺, Tl⁺/Tl³⁺. Also, in transition elements higher states become more stable down a group, whereas in p-block lower states become more stable down a group.
14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer
(i) Fusion of chromite ore with sodium carbonate in free air gives sodium chromate:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂
⇒ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
(ii) The yellow chromate solution is filtered and acidified with H₂SO₄ to give orange sodium dichromate: 2Na₂CrO₄ + 2H⁺
⇒ Na₂Cr₂O₇ + 2Na⁺ + H₂O
(iii) Sodium dichromate is treated with KCl; the less soluble potassium dichromate crystallises out: Na₂Cr₂O₇ + 2KCl
⇒ K₂Cr₂O₇ + 2NaCl
Effect of pH: dichromate and chromate are interconvertible. On increasing pH (adding alkali), orange dichromate changes to yellow chromate: Cr₂O₇²⁻ + 2OH⁻
⇒ 2CrO₄²⁻ + H₂O.
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
Answer
In acidic medium dichromate is a strong oxidising agent: Cr₂O₇²⁻ + 14H⁺ + 6e⁻
⇒ 2Cr³⁺ + 7H₂O (E° = 1.33 V); orange colour changes to green.
(i) Cr₂O₇²⁻ + 14H⁺ + 6I⁻
⇒ 2Cr³⁺ + 3I₂ + 7H₂O
(ii) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺
⇒ 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(iii) Cr₂O₇²⁻ + 8H⁺ + 3H₂S
⇒ 2Cr³⁺ + 3S + 7H₂O
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.
Answer
Preparation: pyrolusite (MnO₂) is fused with KOH in the presence of air or an oxidising agent like KNO₃ to give dark green potassium manganate: 2MnO₂ + 4KOH + O₂
⇒ 2K₂MnO₄ + 2H₂O. The manganate disproportionates in neutral or acidic solution: 3MnO₄²⁻ + 4H⁺
⇒ 2MnO₄⁻ + MnO₂ + 2H₂O. Commercially, manganate is oxidised electrolytically in alkaline solution: MnO₄²⁻
⇒ MnO₄⁻ + e⁻.
In acid, MnO₄⁻ + 8H⁺ + 5e⁻
⇒ Mn²⁺ + 4H₂O.
(i) MnO₄⁻ + 8H⁺ + 5Fe²⁺
⇒ Mn²⁺ + 5Fe³⁺ + 4H₂O
(ii) 2MnO₄⁻ + 5SO₂ + 2H₂O
⇒ 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
(iii) 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻
⇒ 2Mn²⁺ + 10CO₂ + 8H₂O
17. For M²⁺/M and M³⁺/M²⁺ systems the E° values for some metals are as follows: Cr²⁺/Cr −0.9 V, Cr³⁺/Cr²⁺ −0.4 V; Mn²⁺/Mn −1.2 V, Mn³⁺/Mn²⁺ +1.5 V; Fe²⁺/Fe −0.4 V, Fe³⁺/Fe²⁺ +0.8 V. 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 or manganese metal.
Answer
(i) E°(M³⁺/M²⁺): Cr −0.4 V < Fe +0.8 V < Mn +1.5 V. The more positive this value, the more easily M³⁺ is reduced to M²⁺. Mn³⁺ is very easily reduced (least stable); Cr³⁺ is hardly reduced (most stable, since Cr²⁺ is readily oxidised). Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺. Order of stability: Cr³⁺ > Fe³⁺ > Mn³⁺.
(ii) Oxidation potentials of M
⇒ M²⁺: Mn +1.2 V, Cr +0.9 V, Fe +0.4 V. So the ease of oxidation of the metal is Mn > Cr > Fe; iron is the least easily oxidised of the three.
18. Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.
Answer
Ions with partly filled d orbitals show d–d transitions and are coloured; d⁰ and d¹⁰ ions are colourless.
Ti³⁺ (3d¹) – coloured (purple); V³⁺ (3d²) – coloured (green); Mn²⁺ (3d⁵) – coloured (pale pink); Fe³⁺ (3d⁵) – coloured (yellow); Co²⁺ (3d⁷) – coloured (pink).
Cu⁺ (3d¹⁰) and Sc³⁺ (3d⁰) – colourless, since no d–d transition is possible.
19. Compare the stability of +2 oxidation state for the elements of the first transition series.
Answer
Except Sc (which forms Sc³⁺), all first-series metals show +2. In the first half, Ti²⁺, V²⁺ and Cr²⁺ are unstable and act as strong reducing agents, readily going to +3. Mn²⁺ (d⁵) is particularly stable. Fe²⁺ is fairly stable but is oxidised by air to Fe³⁺ (d⁵). Co²⁺, Ni²⁺ and Cu²⁺ are stable in water, and Zn²⁺ (d¹⁰) is the only state of zinc. In general E°(M²⁺/M) becomes less negative across the series (except Mn, Ni, Zn due to stable configurations), and stability of +2 relative to +3 increases from left to right (Mn²⁺, Ni²⁺, Zn²⁺ being exceptionally stable).
20. Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes (iii) oxidation state and (iv) chemical reactivity.
Answer
(i) Lanthanoids: [Xe]4f¹⁻¹⁴5d⁰⁻¹6s²; actinoids: [Rn]5f¹⁻¹⁴6d⁰⁻¹7s². 4f and 5d (and 5f and 6d) are close in energy, but 5f orbitals are more diffuse and take part in bonding more than 4f.
(ii) Both show a steady decrease in atomic and M³⁺ ionic radii (lanthanoid and actinoid contraction). The actinoid contraction is larger from element to element because of poorer shielding by 5f electrons.
(iii) Lanthanoids mainly show +3, with occasional +2 and +4. Actinoids show a much wider range (+3 to +7, e.g. U up to +6, Np and Pu up to +7), because 5f, 6d and 7s levels are close in energy.
(iv) Both are highly reactive, electropositive metals. Actinoids are more reactive, especially when finely divided (react with hot water, form mixtures of oxide and hydride, combine with most non-metals at moderate temperatures); many actinoids are radioactive, which makes their study difficult. Lanthanoids resemble calcium in reactivity, form M₂O₃ and liberate H₂ from dilute acids.
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.
Answer
(i) Cr²⁺ (d⁴) loses an electron to become Cr³⁺ (d³, stable half-filled t₂g³ level), so it acts as a reducing agent. Mn³⁺ (d⁴) gains an electron to become Mn²⁺ (d⁵, stable half-filled), so it acts as an oxidising agent.
(ii) In the presence of strong field ligands, the crystal field stabilisation energy of Co³⁺ (d⁶, low spin t₂g⁶) is much greater than that of Co²⁺, which more than compensates for the third ionisation enthalpy. So Co(II) complexes are easily oxidised to Co(III) complexes.
(iii) An ion with d¹ configuration can easily lose that single electron to attain the stable d⁰ (noble gas) configuration, or it disproportionates; hence d¹ ions (e.g. Ti³⁺, V⁴⁺) tend to be unstable and reducing.
22. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.
Answer
Disproportionation is a redox reaction in which the same species is simultaneously oxidised and reduced, forming products in a higher and a lower oxidation state.
(i) 2Cu⁺(aq)
⇒ Cu²⁺(aq) + Cu(s) (Cu: +1
⇒ +2 and 0)
(ii) 3MnO₄²⁻ + 4H⁺
⇒ 2MnO₄⁻ + MnO₂ + 2H₂O (Mn: +6
⇒ +7 and +4)
(Also 3CrO₄³⁻ + 8H⁺
⇒ 2CrO₄²⁻ + Cr³⁺ + 4H₂O, Cr: +5
⇒ +6 and +3.)
23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Answer
Copper. Its configuration is 3d¹⁰4s¹, so losing the single 4s electron gives Cu⁺ with a stable, completely filled 3d¹⁰ configuration. (In water, however, Cu⁺ disproportionates to Cu²⁺ and Cu.)
24. 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?
Answer
Mn³⁺: 3d⁴
⇒ 4 unpaired; Cr³⁺: 3d³
⇒ 3 unpaired; V³⁺: 3d²
⇒ 2 unpaired; Ti³⁺: 3d¹
⇒ 1 unpaired.
Cr³⁺ is the most stable in aqueous solution because its three electrons occupy the half-filled t₂g³ level in the octahedral aqua complex (high CFSE).
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.
Answer
(i) In the lowest oxide the metal has a low oxidation state, with some valence electrons free, and the bonding is ionic, so it gives OH⁻ ions (basic). In the highest oxide the metal is in a high oxidation state, is small and highly charged, and the bonding is covalent, so the oxide is acidic. Example: MnO (+2) basic, Mn₂O₃/MnO₂ amphoteric, Mn₂O₇ (+7) acidic; CrO basic, Cr₂O₃ amphoteric, CrO₃ acidic; V₂O₅ amphoteric.
(ii) O and F are small and highly electronegative and can oxidise the metal to its highest state; F gives high lattice/bond energies and O forms multiple bonds. Examples: VF₅, CrF₆, Mn₂O₇ (Mn +7), OsO₄ (+8).
(iii) In oxoanions the metal is bonded to several oxygen atoms with multiple (π) bonding, which stabilises the high oxidation state. Examples: MnO₄⁻ (Mn +7), CrO₄²⁻ and Cr₂O₇²⁻ (Cr +6), VO₄³⁻ (V +5).
26. Indicate the steps in the preparation of: (i) K₂Cr₂O₇ from chromite ore. (ii) KMnO₄ from pyrolusite ore.
Answer
(i) Step 1 – fuse chromite with Na₂CO₃ in air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂
⇒ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂. Step 2 – acidify the filtered chromate solution: 2Na₂CrO₄ + 2H⁺
⇒ Na₂Cr₂O₇ + 2Na⁺ + H₂O. Step 3 – add KCl: Na₂Cr₂O₇ + 2KCl
⇒ K₂Cr₂O₇ + 2NaCl; orange K₂Cr₂O₇ crystallises.
(ii) Step 1 – fuse MnO₂ with KOH in air (or with KNO₃): 2MnO₂ + 4KOH + O₂
⇒ 2K₂MnO₄ + 2H₂O (green manganate). Step 2 – oxidise manganate to permanganate, either by disproportionation in neutral/acid solution: 3MnO₄²⁻ + 4H⁺
⇒ 2MnO₄⁻ + MnO₂ + 2H₂O, or commercially by electrolytic oxidation in alkaline solution: MnO₄²⁻
⇒ MnO₄⁻ + e⁻. Purple KMnO₄ crystallises out.
27. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer
An alloy is a homogeneous blend (solid solution) of two or more metals, or of a metal with a non-metal, made by mixing the components in the molten state, e.g. brass, bronze, steel.
Misch metal is a well-known lanthanoid alloy: about 95% lanthanoid metals, about 5% iron, with traces of S, C, Ca and Al. A magnesium-based alloy containing about 3% misch metal is used to make bullets, shells and lighter flints.
28. 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.
Answer
Inner transition elements are the f-block elements (lanthanoids, Z = 58–71, and actinoids, Z = 90–103) in which the last electron enters the (n−2)f subshell.
Inner transition elements from the list: 59 (Pr), 95 (Am) and 102 (No). (29 = Cu, 74 = W and 104 = Rf are d-block elements.)
29. 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.
Answer
Lanthanoids show mainly +3, with only a few +2 and +4 states, so their chemistry changes smoothly. Actinoids show a wide and irregular range of oxidation states because 5f, 6d and 7s levels have comparable energies. For example, Th shows +4, Pa +5, U up to +6, Np and Pu up to +7, Am +3 to +6, while the later actinoids mainly show +3 (No also +2). The maximum oxidation state first rises up to the middle of the series and then falls, so the chemistry is far less regular. Radioactivity of these elements adds to the difficulty.
30. 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.
Answer
Lawrencium, Lr (Z = 103). Configuration: [Rn]5f¹⁴6d¹7s². Its most likely oxidation state is +3, obtained by losing the 6d¹ and 7s² electrons to give the stable 5f¹⁴ configuration.
31. Use Hund's rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
Answer
Ce (Z = 58): [Xe]4f¹5d¹6s². Ce³⁺ is formed by losing the 5d¹ and 6s² electrons
⇒ [Xe]4f¹, with one unpaired electron (n = 1).
μ = √(n(n + 2)) = √(1 × 3) = √3 = 1.73 BM
32. 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.
Answer
+4: Ce, Pr, Nd, Tb, Dy. Ce⁴⁺ ([Xe]4f⁰) and Tb⁴⁺ ([Xe]4f⁷) attain empty and half-filled 4f subshells, which explains their +4 state.
+2: Sm, Eu, Tm, Yb. Eu²⁺ ([Xe]4f⁷) and Yb²⁺ ([Xe]4f¹⁴) attain half-filled and fully filled 4f subshells, so they form +2 ions most readily.
So the extra stability of f⁰, f⁷ and f¹⁴ configurations favours these unusual oxidation states, although thermodynamic factors also contribute (e.g. Pr, Nd, Dy, Sm, Tm).
33. Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.
Answer
(i) Lanthanoids: [Xe]4f¹⁻¹⁴5d⁰⁻¹6s²; actinoids: [Rn]5f¹⁻¹⁴6d⁰⁻¹7s². The 5f orbitals are less deeply buried than 4f, so 5f electrons can take part in bonding more readily.
(ii) Lanthanoids mostly show +3 (some +2, +4). Actinoids show +3 too but a wider range up to +7 (U +6, Np/Pu +7), because of the small energy gap between 5f, 6d and 7s levels.
(iii) Both are reactive metals. Actinoids are more reactive, especially finely divided; they react with boiling water, are attacked by HCl, and combine readily with most non-metals. Lanthanoids react with water and dilute acids to give H₂ and burn to M₂O₃. Actinoids are radioactive, lanthanoids (except Pm) are not.
34. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
Answer
Z = 61 (Pm): [Xe]4f⁵6s²
Z = 91 (Pa): [Rn]5f²6d¹7s²
Z = 101 (Md): [Rn]5f¹³7s²
Z = 109 (Mt): [Rn]5f¹⁴6d⁷7s²
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.
Answer
(i) Electronic configurations: first series (n−1)d¹⁻¹⁰4s¹⁻², second 4d¹⁻¹⁰5s⁰⁻², third 4f¹⁴5d¹⁻¹⁰6s⁰⁻². Elements in the same group usually have similar outer configurations, but there are exceptions (e.g. Cr 3d⁵4s¹ but W 5d⁴6s²; Ni 3d⁸4s² but Pd 4d¹⁰5s⁰).
(ii) Oxidation states: in the first series lower states (+2, +3) are common and stable; in the heavier series higher oxidation states become more stable (e.g. Mo(VI), W(VI) are more stable than Cr(VI); Fe gives +2/+3 while Ru and Os reach +8 in RuO₄, OsO₄).
(iii) Ionisation enthalpies: first IE rises gradually along each series. The 5d series has higher IEs than the 3d and 4d series because the poorly shielding 4f electrons increase the effective nuclear charge; 4d and 3d values are comparable.
(iv) Atomic sizes: radii increase from the first to the second series (e.g. Ti 147 pm
⇒ Zr 160 pm), but the second and third series have almost the same radii (Zr 160, Hf 159 pm) because of lanthanoid contraction.
36. 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).
Answer
In the octahedral aqua complexes, water is a weak field ligand, so high-spin arrangements are expected (d orbitals split into lower t₂g and higher e_g sets):
Ti²⁺: 3d² → t₂g² e_g⁰
V²⁺: 3d³ → t₂g³ e_g⁰
Cr³⁺: 3d³ → t₂g³ e_g⁰
Mn²⁺: 3d⁵ → t₂g³ e_g²
Fe²⁺: 3d⁶ → t₂g⁴ e_g²
Fe³⁺: 3d⁵ → t₂g³ e_g²
Co²⁺: 3d⁷ → t₂g⁵ e_g²
Ni²⁺: 3d⁸ → t₂g⁶ e_g²
Cu²⁺: 3d⁹ → t₂g⁶ e_g³
37. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer
The statement is correct. (i) Sizes: first-series atoms are smaller; 4d and 5d elements of a group have nearly equal radii due to lanthanoid contraction. (ii) Oxidation states: +2 and +3 are common in the first series, while higher oxidation states are more stable for heavier elements (e.g. Cr(VI) is strongly oxidising but Mo(VI), W(VI) are stable). (iii) Ionisation enthalpy: 5d elements have higher IEs than 3d and 4d. (iv) Enthalpies of atomisation and melting points are higher for 4d and 5d metals, which show much more metal–metal bonding. (v) First-series complexes are often high spin with weak field ligands, while 4d/5d complexes are generally low spin. (vi) Heavier metals rarely form simple aqua ions and their halides are often covalent or polymeric.
38. What can be inferred from the magnetic moment values of the following complex species? K₄[Mn(CN)₆]: 2.2 BM; [Fe(H₂O)₆]²⁺: 5.3 BM; K₂[MnCl₄]: 5.9 BM
Answer
Using μ = √(n(n + 2)): n = 1
⇒ 1.73, n = 4
⇒ 4.90, n = 5
⇒ 5.92 BM.
K₄[Mn(CN)₆]: Mn²⁺ is 3d⁵; 2.2 BM indicates 1 unpaired electron. CN⁻ is a strong field ligand that pairs electrons
⇒ low spin, inner-orbital d²sp³ octahedral complex (t₂g⁵).
[Fe(H₂O)₆]²⁺: Fe²⁺ is 3d⁶; 5.3 BM indicates 4 unpaired electrons (with some orbital contribution). H₂O is a weak field ligand
⇒ high spin, outer-orbital sp³d² octahedral complex (t₂g⁴e_g²).
K₂[MnCl₄]: Mn²⁺ is 3d⁵; 5.9 BM indicates 5 unpaired electrons
⇒ high spin, sp³ hybridised tetrahedral complex.
Important Extra Questions
Why do transition elements show high melting points (except Zn, Cd, Hg)?
Answer
Strong metallic bonding involving both ns and (n−1)d electrons; the more unpaired d electrons, the stronger the bonding. Melting points rise to a maximum around d⁵ and then fall; Zn, Cd, Hg have no unpaired d electrons and so have low melting points.
Why is Zr similar to Hf in properties?
Answer
Because of lanthanoid contraction, the atomic radius of Hf (159 pm) becomes almost equal to that of Zr (160 pm). Having the same outer configuration and almost identical size, they show very similar chemistry and are difficult to separate.
Why do transition metal ions form a large number of complex compounds?
Answer
They have small size, high nuclear charge and vacant d orbitals of suitable energy to accept lone pairs from ligands, and they can exist in several oxidation states.
What happens when KMnO₄ is heated?
Answer
On heating to about 513 K it decomposes: 2KMnO₄
⇒ K₂MnO₄ + MnO₂ + O₂.
Why is the E°(Mn³⁺/Mn²⁺) value very high (+1.57 V) compared to that of Fe³⁺/Fe²⁺?
Answer
Mn²⁺ has the extra stable half-filled d⁵ configuration, so Mn³⁺ (d⁴) readily gains an electron. For iron, Fe³⁺ is itself d⁵ and stable, so its reduction to Fe²⁺ is less favourable, giving a lower E° (+0.77 V).