CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain. [NCERT Exemplar]
Answer:
It is due to regular increase in ionisation enthalpy.

Teacher's Note:
1. Higher ionisation enthalpy means the metal loses electrons less easily, so it is less reactive.

Question 2. Arrange the following in increasing order of acidic character: CrO₃, CrO, Cr₂O₃ [HOTS]
Answer:
CrO < Cr₂O₃ < CrO₃. Higher the oxidation state, more will be acidic character.

Teacher's Note:
1. Oxidation states of Cr here are +2, +3 and +6: CrO is basic, Cr₂O₃ amphoteric and CrO₃ acidic.

Question 3. Why does copper not replace hydrogen from acids? [NCERT Exemplar]
Answer:
Cu shows E° positive value.

Teacher's Note:
1. For full marks state the value: E°(Cu²⁺/Cu) = +0.34 V, so Cu cannot reduce H⁺ to H₂.

Question 4. Which divalent metal ion has maximum paramagnetic character among the first transition metals? Why?
Answer:
Mn²⁺ has the maximum paramagnetic character because of the maximum number of unpaired electrons, viz., 5.

Teacher's Note:
1. Fe³⁺ is also 3d⁵, but the question asks for a divalent ion.

Question 5. Although Cr³⁺ and Co²⁺ ions have same number of unpaired electrons but the magnetic moment of Cr³⁺ is 3.87 BM and that of Co²⁺ is 4.87 BM. Why? [NCERT Exemplar] [HOTS]
Answer:
Due to symmetrical electronic configuration there is no orbital contribution in Cr³⁺ ion. However, appreciable orbital contribution takes place in Co²⁺ ion.

Teacher's Note:
1. The spin only formula gives 3.87 BM for both; a higher observed value shows orbital contribution.

Question 6. Out of Cu₂Cl₂ and CuCl₂, which is more stable and why? [NCERT Exemplar] [HOTS]
Answer:
CuCl₂ is more stable than Cu₂Cl₂. The stability of Cu²⁺(aq) is more than Cu⁺(aq) due to the much more negative Δ_hydH° of Cu²⁺(aq) than Cu⁺(aq).

Teacher's Note:
1. The smaller, more highly charged Cu²⁺ ion is hydrated much more strongly than Cu⁺.

Question 7. Write any one use of pyrophoric alloys.
Answer:
Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters.

Teacher's Note:
1. Mischmetall (about 95% lanthanoid metals with iron) is the common pyrophoric alloy; it is also used in bullets and shells.

Teacher's Note:
1. Option (f) is a distractor: decomposition of KClO₃ is catalysed by MnO₂, which is not in Column I.

Question 9. Explain the following :
(i) The enthalpies of atomization of transition metals are quite high.
(ii) The transition metals and many of their compounds act as good catalysts. [CBSE Comptt. Delhi 2012]
Answer:
(i) The transition elements exhibit high enthalpies of atomization because they have large number of unpaired electrons in their atoms. Due to which they have stronger interatomic interaction and hence stronger bonding between atoms.
(ii) Transition elements and their compounds shows good catalytic properties because :
(a) They have variable valencies, show multiple oxidation states, forms unstable intermediate compounds and provides a new path with lower activation energy for the reaction.
(b) In some cases, transition elements provide a suitable surface for the reaction to take place.

Teacher's Note:
1. Link catalysis to two features: variable oxidation states (intermediate formation) and a large surface area for adsorption.

Question 10. Explain the following observation :
Most of the transition metal ions exhibit characteristic colour in aqueous solution. [CBSE Delhi 2012]
Answer:
Most of complexes of transition elements are coloured due to d-d transition. This is because of the absorption of radiation from visible light region to promote an electron from one of the d-orbitals to another. The ions of transition elements absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

Teacher's Note:
1. The colour seen is the complementary colour of the light absorbed; ions with d⁰ or d¹⁰ configuration (Sc³⁺, Ti⁴⁺, Zn²⁺) are colourless.

Short Answer Questions (2 Marks)

Question 11. (i) Why Lanthanoids cannot be easily separated?
(ii) Why does Zr (Z = 40) and Hf (Z = 72) exhibit almost identical atomic radii?
Answer:
(i) Due to lanthanoid contraction, the change in the atomic or ionic radii of these elements is very small. So, their chemical properties are similar. Hence, Lanthanoids cannot be easily separated.
(ii) Lanthanoid contraction is responsible for almost same atomic radii of 4d and 5d transition series elements.

Teacher's Note:
1. For (ii), say why: the 14 poorly shielding 4f electrons filled before Hf cancel the expected increase in size from Zr to Hf.

Question 12. What are the transition elements? Write two characteristics of the transition elements. [CBSE Delhi 2015]
Answer:
These atoms whose d-orbitals are incomplete in ground state or in one of the most common oxidation state are called transition elements or d-block elements.
The valence shell electronic configuration of transition elements is (n−1)d¹⁻¹⁰ns¹⁻².
Two characteristics of transition elements:
(i) Transition metals show variable oxidation states.
(ii) All transition metals act as catalyst.

Teacher's Note:
1. Write the definition precisely ("incomplete d subshell in the atom or in its common ions") and give characteristics, not general descriptions or occurrence.

Question 13. What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution.
OR
Suggest reasons for the following features of transition metal chemistry:
(i) The transition metals and their compounds are usually paramagnetic.
(ii) The transition metals exhibit variable oxidation states. [CBSE Comptt. Delhi 2015]
Answer:
Disproportionation is the reaction in which an element undergoes self-oxidation and self-reduction simultaneously. For example –
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
(Or any other correct equation)
OR
(i) Due to presence of unpaired electrons in d-orbitals.
(ii) Due to incomplete filling of d-orbitals. Due to very small energy difference between (n − 1)d and n s-orbitals.

Teacher's Note:
1. In a disproportionation equation, show the same element going both up and down in oxidation state: Cu(+1) → Cu(+2) and Cu(0).

Question 14. In the following ions:
Mn³⁺, V³⁺, Cr³⁺, Ti⁴⁺
(Atomic no: Mn = 25, V = 23, Cr = 24, Ti = 22)
(a) Which ion is most stable in an aqueous solution?
(b) Which ion is the strongest oxidizing agent?
(c) Which ion is colourless?
(d) Which ion has the highest number of unpaired electrons? [CBSE Foreign Set-1, 2, 3 2017]
Answer:
(a) Cr³⁺
(b) Mn³⁺
(c) Ti⁴⁺
(d) Mn³⁺

Teacher's Note:
1. Write the d-count first: Mn³⁺ d⁴, V³⁺ d², Cr³⁺ d³, Ti⁴⁺ d⁰; every part then follows from these numbers.

Question 15. Explain the following observations:
(i) Copper atom has completely filled d orbitals (3d¹⁰) in its ground state, yet it is regarded as a transition element.
(ii) Cr²⁺ is a stronger reducing agent than Fe²⁺ in aqueous solution. [CBSE Comptt. OD Set-1 2017]
Answer:
(i) Because it has incompletely filled d orbitals in one of its oxidation state (Cu²⁺).
(ii) Cr²⁺(d⁴) changes to Cr³⁺(d³) while Fe²⁺(d⁶) changes to Fe³⁺ (d⁵). In aqueous medium d³ is more stable than d⁵.

Teacher's Note:
1. The key phrase for (ii) is "half-filled t_2g³": Cr³⁺ in water is stabilised by crystal field, which makes Cr²⁺ a strong reducing agent.

Question 16. Name the following:
(i) A transition metal which does not exhibit variation in oxidation state in its compounds.
(ii) A compound where the transition metal is in the +7 oxidation state.
(iii) A member of the lanthanoid series which is well known to exhibit +4 oxidation state. [CBSE SQP 2016]
Answer:
(i) Scandium (Sc).
(ii) KMnO₄ or any other suitable example.
(iii) Cerium (Ce) or any other example.

Teacher's Note:
1. Sc shows only +3 (it loses all three outer electrons to reach the argon core), which is why it shows no variation in oxidation state.

Question 17. Explain the following observation:
(i) Zn²⁺ salts are colourless.
(ii) Copper has exceptionally positive \( E^{\circ}_{M^{2+}/M} \) value. [CBSE Comptt. OD Set-3 2017]
Answer:
(i) Due to absence of unpaired electrons.
(ii) Due to high Δ_aH° and low Δ_hydH°.
Detailed Answer:
(i) Zinc has no unpaired electrons in its d orbital and has a stable fully filled d orbital. Thus, due to absence of unpaired electrons, there will not be the excitation of electrons from lower energy level to higher level to exhibit complementary colour and hence Zn²⁺ salts are colourless.
(ii) As copper has high energy of atomisation Δ_aH° and low hydration energy Δ_hydH°, due to which E° value is positive.

Teacher's Note:
1. Copper is the only 3d metal with a positive E°(M²⁺/M) (+0.34 V): the energy needed for atomisation and ionisation is not compensated by its hydration enthalpy.

Question 18. Give reasons:
(i) Zn is not regarded as a transition element.
(ii) Cr²⁺ is a strong reducing agent. [CBSE Comptt. Delhi 2016]
Answer:
(i) In both, Zn and Zn²⁺ ions absence of incompletely filled d-orbital therefore, Zn is not regarded as a transition element.
(ii) Cr²⁺ has d⁴ configuration while Cr³⁺ has more stable d³ (t_2g³) configuration. Thus, Cr has a tendency to acquire Cr³⁺ due to greater stability of +3 oxidation state. Therefore, Cr²⁺ acts as a strong reducing agent.

Teacher's Note:
1. For (i), mention both the atom (3d¹⁰4s²) and its only ion Zn²⁺ (3d¹⁰); Cu differs because Cu²⁺ is 3d⁹.

Question 19. Explain the following observation:
(i) Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state, yet it is regarded as a transition element.
(ii) E° value for Mn³⁺/Mn²⁺ couple is much more positive than Cr³⁺/Cr²⁺. [CBSE Comptt. OD Set-2 2017]
Answer:
(i) Silver can exhibit +2 oxidation state, wherein it will have incompletely filled d-orbital.
(ii) Much higher third ionisation energy of Mn where the required change is from d⁵ to d⁴.

Teacher's Note:
1. A large positive E°(M³⁺/M²⁺) means M²⁺ is very stable; Mn²⁺ (d⁵) resists oxidation while Cr²⁺ (d⁴) is easily oxidised to d³.

Question 20. Write one similarity and one difference between the chemistry of lanthanoids and actinoids. [CBSE OD 2015]
Answer:
Similarity: (i) Both show contraction in size. (ii) Both show irregularity in their electronic configuration. (iii) Both are stable in +3 oxidation state.
Difference: (i) Actinoids are mainly radioactive but lanthanoids are not. (ii) Actinoids show wide range of oxidation states but lanthanoids do not. (iii) Actinoid contraction is greater than lanthanoid contraction.
(Write any one of these or any other one similarity and one difference)

Teacher's Note:
1. Compare chemical properties, not occurrence; a good difference is that 5f, 6d and 7s are close in energy, so actinoids show many more oxidation states.

Question 21. Identify the following:
(i) Oxo anion of chromium which is stable in acidic medium.
(ii) The lanthanoid element that exhibits +4 oxidation state. [CBSE SQP 2017]
Answer:
(i) Cr₂O₇²⁻
(ii) Cerium

Teacher's Note:
1. Chromate and dichromate interconvert with pH: 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O, so acid favours dichromate.

Short Answer Questions (3 Marks)

Question 22. (i) Transition metals have very high melting and boiling points. Why?
(ii) In d-block element, ionic radii of ions of the same charge decreases progressively with increasing atomic number in a series. Why?
Answer:
(i) The high melting and boiling points of transition metals are attributed to the involvement of greater number of electrons from (n − 1) d-orbital in addition to the ns electrons in the interatomic metallic bonding (d-d overlap).
(ii) As the atomic number increases the new electron enters the d-orbital and expected to increase in atomic size, but due to poor shielding effect of d-orbitals the electrostatic attraction between nucleus and outermost orbital increases and hence, the ionic radii decreases.

Teacher's Note:
1. For ions of the same charge the extra electron is always a d electron, which shields poorly, so the effective nuclear charge rises steadily.

Teacher's Note:
1. For part (a)(i), say the high atomisation and ionisation enthalpies of Cu are not balanced by its hydration enthalpy; this is what makes E° positive.

Question 24. Give reasons for the following:
(i) Mn³⁺ is a good oxidising agent.
(ii) E°_M²⁺/M values are not regular for first row transition metals (3d-series).
(iii) d-block elements exhibit more oxidation states than f-block elements.
Answer:
(i) Mn²⁺ is more stable due to half filled d⁵ configuration and Mn³⁺ easily changes to Mn²⁺ hence, it is a good oxidising agent.
(ii) The E°_M²⁺/M values are not regular which can be explained from the irregular variation of ionisation enthalpies i.e., IE₁ + IE₂ and also the sublimation enthalpies which are relatively much less for manganese and vanadium.
(iii) All transition elements except the first and the last member in each series show a large number of variable oxidation states. This is because difference of energy in the (n − 1)d and ns orbitals is very little.
Hence, electrons from both the energy levels can be used for bond formation.

Teacher's Note:
1. For part (iii), add the contrast: in lanthanoids the 4f electrons are buried deep and rarely take part in bonding, so +3 dominates.

Question 25. How would you account for the following:
(i) The oxidising power of oxoanions are in the order VO₂⁺ < Cr₂O₇²⁻ < MnO₄⁻
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺.
Answer:
(i) Change in Cr₂O₇²⁻ to Cr(III) is 3 and in MnO₄⁻ to Mn (II) is 5.
Change in oxidation state is large and the stability of reduced products in V(III) < Cr(III) < Mn(II). This is why oxidising power of VO₂⁺ < Cr₂O₇²⁻ < MnO₄⁻.
(ii) Third ionization enthalpy of Mn is very high because the third electron has to be removed from the stable half-filled 3d-orbitals [Mn²⁺ (Z = 25) = 3d⁵].
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺. E°_{Cr³⁺/Cr²⁺} is negative (−0.41 V) whereas E°_{Fe³⁺/Fe²⁺} is positive (+0.77 V). Thus Cr²⁺ is easily oxidized to Cr³⁺ but Fe²⁺ cannot be easily oxidized to Fe³⁺. Hence, Cr²⁺ is stronger reducing agent than Fe²⁺.

Teacher's Note:
1. The oxidation states of the central atom in the three oxoanions are +5, +6 and +7; the higher the oxidation state, the greater the oxidising power across this series.

Teacher's Note:
1. Use μ = √[n(n + 2)] BM backwards: 4.90 BM → n = 4, 2.84 BM → n = 2, 1.73 BM → n = 1, 0 BM → n = 0.

Teacher's Note:
1. Read the table as a trend: values become less negative from Ti to Ni; Mn and Zn break the trend because d⁵ and d¹⁰ are extra stable.

Question 28. (i) (a) Transition metals form alloys. Why?
(b) Chromium is typically hard metal but mercury is liquid. Why?
(ii) Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state.
\( E^{\circ}_{Cr^{3+}/Cr^{2+}} = -0.4\ V \)
\( E^{\circ}_{Mn^{3+}/Mn^{2+}} = +1.5\ V \)
\( E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.8\ V \)
Answer:
(i) (a) Due to the almost similar atomic sizes of elements, atoms can easily occupy the position in the crystal lattice of the other atom of another element (metal). Thus, formation of alloys is the property of transition metals.
(b) Due to the presence of large number of unpaired electrons, metal-metal interaction is strong whereas mercury does not have unpaired electrons and forms weak metallic bonds.
(ii) Cr²⁺ < Fe²⁺ < Mn²⁺

Teacher's Note:
1. The more positive E°(M³⁺/M²⁺), the harder it is to oxidise M²⁺, so the more stable the +2 state.

Question 29. Account for the following:
(i) CuCl₂ is more stable than Cu₂Cl₂.
(ii) Atomic radii of 4d and 5d series elements are nearly same.
(iii) Hydrochloric acid is not used in permanganate titration. [CBSE Foreign Set-1, 2 2017]
Answer:
(i) In CuCl₂, Cu is in +2 oxidation state which is more stable due to high hydration enthalpy as compared to Cu₂Cl₂ in which Cu is in +1 oxidation state.
(ii) Due to lanthanoid contraction.
(iii) Because HCl is oxidised to chlorine.

Teacher's Note:
1. Permanganate titrations are done in dilute H₂SO₄, which is not oxidised by KMnO₄; HCl would use up some permanganate and give a wrong titre.

Question 30. (i) Give reasons for the following:
(a) Compounds of transition elements are generally coloured.
(b) MnO is basic while Mn₂O₇ is acidic.
(ii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) (a) Due to d-d transition.
(b) Due to higher oxidation state of Mn₂O₇ / Due to high polarizing power of Mn(VII).
(ii) \( \mu = \sqrt{4(4+2)} \) = 4.90 B.M.

Teacher's Note:
1. For (ii), first write Fe²⁺ = [Ar]3d⁶ to get n = 4 unpaired electrons; only then substitute in μ = √[n(n + 2)].

Question 31. Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations. [CBSE Delhi Set-1, 2016]
Answer:
(i) Mn can form pπ-dπ bond with oxygen by using 2p orbital of oxygen and 3d-orbital of Mn because of which it shows highest oxidation state of +7. With fluorine, Mn cannot form pπ-dπ bond thus shows the highest oxidation state of +4.
(ii) Transition metal show variable oxidation state due to comparable energies of ns and (n − 1)d orbitals and partially filled d orbitals. So, both these orbitals take part in the reactions.
(iii) Due to comparable energies of 5f, 6d and 7s orbitals and the relative stabilities of f⁰, f⁷ and f¹⁴ occupancies of the 5f orbitals.

Teacher's Note:
1. Be specific: name the orbitals (ns and (n − 1)d, or 5f, 6d and 7s) rather than writing a vague "energy difference is small".

Question 32. (i) Account for the following:
(a) Cu⁺ is unstable in an aqueous solution.
(b) Transition metals form complex compounds.
(ii) What is mischmetal? Give one use.
(iii) Complete the following equation:
Cr₂O₇²⁻ + 8H⁺ + 3NO₂⁻ → [CBSE OD 2015]
Answer:
(i) (a) Because Cu⁺ undergoes disproportionation as 2Cu⁺ → Cu + Cu²⁺.
Hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ which compensates the I.E.₂ of Cu involved in the formation of Cu²⁺ ions.
(b) Because of small size of metal, high ionic charge and availability of vacant d-orbital.
(ii) Misch metal is an alloy with composition (95%) lanthanide metal, (5%) iron and traces of S, C, Ca and Al.
It’s alloy of magnesium is used to produce bullets, shells, flints.
(iii) Cr₂O₇²⁻ + 8H⁺ + 3NO₂⁻ → 2Cr³⁺ + 3NO₃⁻ + 4H₂O

Teacher's Note:
1. Check a redox equation by charge: left side −2 + 8 − 3 = +3, right side +6 − 3 = +3, so the equation is balanced.

Long Answer Questions (5 Marks)

Question 33. Give reasons for the following:
(i) Silver bromide is used in photography.
(ii) Most transition metal compounds are coloured.
(iii) Zinc and not copper is used for the recovery of metallic silver from complex [Ag(CN)₂]⁻. Explain.
(iv) The colour of mercurous chloride, Hg₂Cl₂, changes from white to black when treated with ammonia.
(v) The species [CuCl₄]²⁻ exists while [CuI₄]²⁻ does not.
Answer:
(i) Silver bromide is used in photography because of its sensitivity to sunlight. In light, AgBr reduces to metallic silver.
(ii) The colour of transition metal compound is due to the presence of incompletely filled d-orbitals in transition metal ions/atoms, because of this d-d transition can occur in them. The colour is due to d-d transition for which the energy is absorbed from visible region. The visible colour of a compound is the complementary colour of the absorbed light.
(iii) Zinc is a cheaper and stronger reducing agent as compared to copper.
(iv) Mercurous chloride (white) changes to black on treatment with ammonia because of the formation of finely divided mercury (grey).
(v) Cu²⁺ is reduced to Cu⁺ by I⁻ and thus CuI₂ gets converted to Cu₂I₂. This change cannot be brought about by Cl⁻.

Teacher's Note:
1. For (iv), write the equation: Hg₂Cl₂ + 2NH₃ → Hg + Hg(NH₂)Cl + NH₄Cl.
2. For (v), iodide is a much stronger reducing agent than chloride: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂.

Question 34. (i) Complete the following equations:
(a) Cr₂O₇²⁻ + 2OH⁻ →
(b) MnO₄⁻ + 4H⁺ + 3e⁻ →
(ii) Account for the following:
(a) Zn is not considered as a transition element.
(b) Transition metals form a large number of complexes.
(c) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ couple. [CBSE OD 2014]
Answer:
(i) (a) Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O
(b) MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O
(ii) (a) Because Zn/Zn²⁺ has fully filled d-orbitals.
(b) This is due to smaller ionic sizes, higher ionic charge and availability of d-orbitals.
(c) Because Mn²⁺ is more stable (3d⁵) than Mn³⁺ (3d⁴), while Cr³⁺ is more stable due to t_2g³/d³ configuration.

Teacher's Note:
1. In (i)(a) Cr stays +6 on both sides; it is an acid–base change (dichromate to chromate), not a redox reaction.

Question 35. 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 is not regarded as a transition element. Give reason.
(ii) Which element has the highest m.p.?
(iii) Write the element which can show an oxidation state of +1.
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why? [CBSE OD Set-2 2016]
Answer:
(i) Zn, because it does not have partially filled d-orbital in its ground state or ionic state.
(ii) Cr has the highest melting point. As the number of unpaired electrons increases upto d⁵ configuration, it results in the increase in the strength of metallic bonds. To break the metallic bond, significant energy is required thus Cr with highest number of unpaired electrons i.e., 6 has the highest melting point.
(iii) Cu can show +1 oxidation state as it can loose one electron present in 4s orbital.
(iv) Mn is a strong oxidising agent in +3 oxidation state because change of Mn³⁺ to Mn²⁺ give stable half filled (d⁵) configuration.

Teacher's Note:
1. Cr (3d⁵4s¹) has six unpaired electrons, the most in the 3d series, which is why it is the expected answer for the highest melting point.

Question 36. (i) (a) How is the variability in oxidation states of transition metals different from that of the p-block elements?
(b) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?
(c) Orange colour of Cr₂O₇²⁻ ion changes to yellow when treated with an alkali. Why?
(ii) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons. [CBSE Delhi Set-1, 2, 3 2017]
Answer:
(i) (a) In p-block elements the difference in oxidation state is 2 and in transition metals the difference is 1.
(b) Cu⁺, due to disproportionation reaction and low hydration enthalpy.
(c) Due to formation of chromate ion/CrO₄²⁻ ion, which is yellow in colour.
(ii) Actinoids are radioactive, actinoids show wide range of oxidation states.

Teacher's Note:
1. For (i)(a) give an example of each: Fe shows +2 and +3 (steps of one), while Sn shows +2 and +4 and P shows +3 and +5 (steps of two).

Question 37. (i) Account for the following:
(a) Transition metals show variable oxidation states.
(b) Zn, Cd and Hg are soft metals.
(c) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [CBSE OD Set-1, 2, 3 2017]
Answer:
(i) (a) The valence electrons of transition metals are in (n − 1)d and ns orbitals. As there is almost little energy difference between these orbitals, both the energy levels can be used for bond formation. Thus, they exhibit variable oxidation states.
(b) Because they contain fully filled d-orbitals, no unpaired d electrons are present resulting in weak metallic bonding.
(c) On gaining one electron, Mn³⁺ (3d⁴) changes to Mn²⁺ which has the very stable, exactly half-filled 3d⁵ configuration, so E°(Mn³⁺/Mn²⁺) is highly positive. For Cr³⁺ + e⁻ → Cr²⁺, the stable 3d³ (half-filled t_2g) configuration of Cr³⁺ is converted to 3d⁴, so E°(Cr³⁺/Cr²⁺) is much less (negative).
(ii) Similarity: Both lanthanoid and actinoid elements are reactive metals and show +3 as the most common oxidation state in their respective series.
Difference: Actinoids have more tendency for complex formation and form oxocations like UO₂²⁺, PuO₂²⁺, whereas lanthanoids have less tendency for complex formation and do not form oxocations. (Or: magnetic properties of lanthanoids can be easily interpreted, while those of actinoids are very difficult to explain.)

Teacher's Note:
1. In (i)(c), quote both E° values when you can: +1.57 V for Mn³⁺/Mn²⁺ against −0.41 V for Cr³⁺/Cr²⁺.