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

Case Study 1

In transition elements, generally, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d orbital each time the nuclear charge increases by unity. But the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction.
The following questions refer to these transition metal ions of 3d series: Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺ (Atomic number: Ti = 22, V = 23, Mn = 25, Cr = 24)

Case Study 2

The d-block elements are those elements in which the last electron enters the d-subshell of the penultimate shell. The general electronic configuration of these elements is (n − 1)d^1–10ns^1–2, where n is outermost shell. The d-block consisting of groups 3–12 occupies the large middle section of the periodic table. The elements of d-block are also known as transition elements as they possess properties that are transitional between the s and p block elements. A transition element is defined as an element which has incompletely filled d-orbitals in its ground state or any one of its oxidation states. There are four series of transition elements spread between group 3 and 12. First transition series or 3d-series: Scandium (₂₁Sc) to Zinc (₃₀Zn), Second transition series or 4d-series: Yttrium (₃₉Y) to Cadmium (₄₈Cd), Third transition series or 5d-series: Lanthanum (₅₇La) and Hafnium (₇₂Hf) to Mercury (₈₀Hg) (Omitting ₅₈Ce to ₇₁Lu), Fourth transition series or 6d-series: Begins with Actinium (₈₉Ac) and is still incomplete.

Question 5. In what way is the electronic configuration of transition elements different from that of the non-transition elements?
Answer:
Transition elements contain incompletely filled d-subshell, i.e., their electronic configuration is (n − 1)d^1–10ns^1–2 whereas non-transition elements have no d-subshell or their subshell is completely filled and have ns^1–2 or ns²np^1–6 in their outermost shell.

Teacher's Note:
1. In transition elements the differentiating electron enters the penultimate shell, not the outermost shell.

Question 6. Why is zinc not regarded as a transition element? [CBSE Guwahati 2015]
Answer:
As zinc atom has completely filled d-orbitals (3d¹⁰) in its ground state as well as in its oxidised state, therefore, it is not regarded as a transition element.

Teacher's Note:
1. Mention both the ground state and the Zn²⁺ state; copper is different because Cu²⁺ is 3d⁹.

Question 7. Why does a transition series contain 10 elements?
Answer:
There are five d-orbitals in an energy level and each orbital can contain two electrons. As we move from one element to the next, an electron is added and for complete filling of the five d-orbitals, 10 electrons are required.

Teacher's Note:
1. Similarly, seven f orbitals hold 14 electrons, which is why each inner transition series has 14 elements.

Question 8. Name the element which shows outer electronic configuration 3d³4s².
Answer:
Vanadium

Teacher's Note:
1. Count: [Ar] + 5 electrons = 23 electrons, which is vanadium (Z = 23).

Question 9. What are interstitial compounds?
Answer:
Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice.

Teacher's Note:
1. Give examples such as TiC, Mn₄N, Fe₃H; these are hard, have high melting points and retain metallic conductivity.

Case Study 3

Potassium permanganate, (KMnO₄) is prepared by fusion of pyrolusite, MnO₂ with KOH in the presence of an oxidising agent like KNO₃. This produces the dark green potassium manganate, K₂MnO₄ which disproportionates in a neutral or acidic solution to give purple permanganate ion. Potassium permanganate is an important oxidising agent in acidic, alkaline as well as neutral medium.

Question 10. What is the state of hybridisation of Mn in MnO₄⁻?
Answer:
sp³

Teacher's Note:
1. MnO₄⁻ is tetrahedral; strictly the Mn–O π bonds use d orbitals (d³s type), but CBSE accepts sp³ for the tetrahedral shape.

Question 11. Write an application of potassium permanganate.
Answer:
It is used for the estimation of hydrogen peroxide.

Teacher's Note:
1. Other uses: estimation of Fe²⁺ and oxalates in volumetric analysis, and as a disinfectant.

Question 12. What are the products formed after heating potassium permanganate?
Answer:
K₂MnO₄, O₂ and MnO₂ will be formed after heating of potassium permanganate.

Teacher's Note:
1. Write the equation if marks allow: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂ (at 513 K).

Question 13. How many electrons are involved in oxidation by KMnO₄ in an acidic medium?
Answer:
Five electrons are involved in oxidation by KMnO₄ in an acidic medium.

Teacher's Note:
1. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O: Mn goes from +7 to +2.

Question 14. Draw the structure of permanganate ion. Is it paramagnetic or diamagnetic?
Answer:

It is diamagnetic.

Teacher's Note:
1. Mn is in the +7 state (3d⁰) with no unpaired electrons, hence diamagnetic; its colour is due to charge transfer.

Case Study 4

From the study of properties of transition elements, it is observed that they show some similarities and some trends along the horizontal row (3d, 4d, 5d and 6d series) and in the vertical groups (3 – 12). The properties of transition elements which have been extensively studied include their atomic radii, metallic character, melting points and boiling points, their heats of atomisation, standard electrode potentials, oxidation states, formation of coloured ions and catalytic and magnetic properties. The reason for similarity in properties is that in all of them (n − 1) d-orbitals are progressively filled but have similar ns² configuration, i.e., they contain incompletely filled d-orbitals. The trend in properties arises due to difference in the number of electrons present in (n − 1) d-orbitals. Exceptional behaviour at some points is exhibited because of exceptional electronic configuration shown by some elements.

Question 15. Au lies below Ag in the same group but still they have same atomic size. Why ?
Answer:
Ag lies in the 4d series while gold lies in the 5d series. In the 5d series, after lanthanum, there is lanthanoid contraction which affects the radii of the remaining elements of the series. Hence, Au has nearly the same size as Ag.

Teacher's Note:
1. Normally size increases down a group; the 14 lanthanoids placed before the 5d metals cancel this expected increase.

Question 16. Why do Mn and Zn have low melting points than the other elements of the first transition series ?
Answer:
In the 1st transition series, Mn has exactly half filled 3d-orbitals and Zn has fully filled d-orbitals. Hence, their electronic configurations are stable, i.e., electrons are held tightly by the nucleus. As a result, delocalization of electrons is less and metallic bonds are much weaker.

Teacher's Note:
1. Melting point follows the strength of metallic bonding; stable d⁵ and d¹⁰ configurations keep the d-electrons out of bonding.

Question 17. Why Cu²⁺ and Co³⁺ salts have the same colour viz blue ?
Answer:
Both of them absorb red wavelength of the white light falling on them. The complementary colour, blue, is reflected back in both cases.

Teacher's Note:
1. The observed colour is the complementary colour of the light absorbed: red absorbed means blue (blue-green) seen.

Question 18. Why do transition elements form complexes ?
Answer:
When ligands donate electron pairs to a transition metal ion, a complex is formed. Transition metals form complexes because of the following reasons :
(i) availability of vacant d-orbitals.
(ii) comparatively smaller size of their metal ion
(iii) their high ionic charges or large size/charge ratio.

Teacher's Note:
1. Remember the three factors as a set: vacant d-orbitals, small size and high charge of the metal ion.

Case Study 5

Potassium permanganate is an important salt containing transition metal, Mn as one of its constituents. This salt is widely used as an oxidizing agent in the laboratory as well as industry. It is often used in volumetric analysis for estimation of ferrous salts, oxalates, iodides etc. It is a well known disinfectant and germicide. Alkaline KMnO₄ solution is used for testing unsaturation in organic compounds. It is prepared from the mineral pyrolusite. A variety of reagents are available to oxidize the ore to potassium manganate which in the neutral or acidic solution undergoes disproportionation to produce permanganate. But permanganate is mainly obtained from manganate produced in the first step by chemical oxidation or electrolytic oxidation. It reacts differently with well cooled conc. H₂SO₄ and warm conc. H₂SO₄. However, the most important use of KMnO₄ is as an oxidizing agent in the presence of dilute H₂SO₄, though it also acts as oxidizing agent in neutral and alkaline medium. In acidic medium, it can oxidize substances like H₂S, SO₂, KNO₂, SO₃²⁻, S₂O₃²⁻, oxalates, ferrous salts, iodides etc.

Question 19. Write the reaction involved when powdered pyrolusite mineral is fused with potassium carbonate in the presence of air.
Answer:
2MnO₂ (Pyrolusite) + 2K₂CO₃ + O₂ → 2K₂MnO₄ (Potassium manganate) + 2CO₂

Teacher's Note:
1. K₂CO₃ here plays the same role as KOH in the usual equation; CO₂ is released instead of H₂O.

Question 20. What are the equivalent weights of potassium permanganate in neutral, alkaline and acidic medium, giving reactions involved.
Answer:
Molecular mass of KMnO₄ = 39 + 55 + 4 × 16 = 158
In neutral solution :
MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻,
Eq. wt. = \( \frac{158}{3} \) = 52.67
In alkaline medium : Net reaction is same, Hence, Eq. wt. = 52.67
In acidic medium :
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O,
Eq. wt. = \( \frac{158}{5} \) = 31.6

Teacher's Note:
1. Equivalent weight = molar mass ÷ number of electrons gained per formula unit; in strongly alkaline medium (MnO₄⁻ → MnO₄²⁻, 1 electron) it would be 158.

Question 21. What is the action of conc. H₂SO₄ on KMnO₄ ?
Answer:
With cold conc. H₂SO₄, Mn₂O₇ is formed which on warming explodes to form MnO₂
2 KMnO₄ + 2 H₂SO₄ (conc.) → Mn₂O₇ + 2 KHSO₄ + H₂O
2 Mn₂O₇ \( \xrightarrow{\Delta} \) 4 MnO₂ + 3 O₂

Teacher's Note:
1. Mn₂O₇ is an explosive oily liquid, which is why KMnO₄ should never be mixed with concentrated H₂SO₄.

Question 22. How many moles of KMnO₄ will be used to react completely with one mole of ferrous oxalate, Fe(C₂O₄) in acidic solution ? Write the balanced equation for the reaction.
Answer:
[MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O] × 3
[Fe²⁺ + C₂O₄²⁻ → Fe³⁺ + 2 CO₂ + 3 e⁻] × 5
3 MnO₄⁻ + 5 FeC₂O₄ + 24 H⁺ → 3 Mn²⁺ + 5 Fe³⁺ + 10 CO₂ + 12 H₂O
∴ KMnO₄ needed to react with 1 mole of FeC₂O₄ = \( \frac{3}{5} \) mole

Teacher's Note:
1. Both Fe²⁺ and oxalate are oxidised, so one FeC₂O₄ gives 1 + 2 = 3 electrons; this is the step most students miss.

Case Study 6

There are some elements in which the last electron enters into (n − 2) f-orbital. These are called 'f-block elements' or 'inner-transition elements' because last electron enters into ante-penultimate i.e. shell inner to the penultimate shell and also because they form a transition series within the transition series (d-block). These are placed separately at the bottom of the periodic table in two rows. The first row is called first inner transition series or 4f series or lanthanoids as they come after lanthanum. The second row is called second inner-transition series or 5f-series or actinoids as they come after actinium. Their general electronic configuration is (n − 2) f¹⁻¹⁴ (n − 1) d⁰⁻¹ ns². They show variable oxidation states. As we move along the series, their atomic and ionic radii decrease regularly. This is called lanthanoid/actinoid contraction. Lanthanoid contraction affects the size of the second and third transition series and also affects the basic strength of their hydroxides. In the pure state, they do not find any use. However, their oxides and alloys are useful. Actinoids show greater number of oxidation states than lanthanoids. Actinoids differ from lanthanoids in a number of properties such as colour, tendency to form complexes, basic nature etc.

Question 23. Why are Sm²⁺, Eu²⁺ and Yb²⁺ in solution are good reducing agents while aqueous solution of Ce⁴⁺ and Tb⁴⁺ are good oxidizing agent ?
Answer:
+3 oxidation state of lanthanoids is most stable. Hence, Sm²⁺, Eu²⁺ and Yb²⁺ tend to change to +3 oxidation state by loss of electron and hence are good reducing agents while Ce⁴⁺ and Tb⁴⁺ tend to change to +3 oxidation state by gain of electron and hence are good oxidizing agent.

Teacher's Note:
1. Any lanthanoid ion not in the +3 state tends to return to +3: +2 ions lose an electron (reducing), +4 ions gain one (oxidising).

Question 24. Name two actinoids which show oxidation state of +7 ?
Answer:
Neptunium (₉₃Np) and Plutonium (₉₄Pu).

Teacher's Note:
1. +7 is the highest oxidation state shown by any actinoid, and only Np and Pu reach it.

Question 25. Why are complexes of Gd (III) colourless inspite of the presence of unpaired electrons ?
Answer:
₆₄Gd = [Xe] 4f⁷ 5d¹ 6s²
Gd³⁺ = [Xe] 4f⁷
This is extra stable due to extra stability of half-filled 4f orbitals. Hence, it does not absorb light in the visible region and is colourless.

Teacher's Note:
1. Unpaired electrons make Gd³⁺ strongly paramagnetic, but colour needs an f–f transition in the visible region, which the stable f⁷ ion does not have.

Question 26. What is the cause of lanthanoid contraction ?
Answer:
As we move along the lanthanoid series, the nuclear charge increases by one unit at each step and one electron is added to the 4f subshell. Due to the diffused shape of f-orbitals, the shielding of one 4f electron by another is very imperfect. Hence, the effective nuclear charge increases and the atomic and ionic radii decrease steadily. This is the cause of lanthanoid contraction.

Teacher's Note:
1. The key words examiners look for are "imperfect (poor) shielding by 4f electrons" and "increase in effective nuclear charge".

Case Study 7

Transition elements are called transition metals because they are metallic in their nature. Both (n − 1) d as well as ns electrons participate in the formation of metallic bonds (M–M). Greater the number of unpaired d-electrons, stronger is the metallic bond formed. Strong metallic bonds are responsible for their high melting and boiling points. Further, stronger is the metallic bonding, higher is the enthalpy of atomisation. The first ionization enthalpies of d-block elements lie between those of s-block and p-block elements. The ionization enthalpy gradually increases along the series but the increase is not regular. The standard electrode potentials, E°_M²⁺,M and E°_{M³⁺,M²⁺} show an increase along the series but the increase is not regular. Transition metals and their oxides show catalytic properties in a number of reactions. Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid state and in aqueous solution in contrast to compounds of s and p-block elements. Further, as transition metals contain unpaired electrons in the (n − 1) d orbitals, most of the transition metals and their ions are paramagnetic. When no unpaired electron is present, it is diamagnetic. Lastly, transition metals form interstitial compounds and alloys.

Case Study 8

Potassium dichromate is an important oxidizing agent used in the laboratory as well as industry. It is a salt containing oxoanion of the transition metal, Cr (Cr₂O₇²⁻). It is prepared from chromite ore. The ore is first converted into sodium chromate which is then converted into sodium dichromate and finally into potassium dichromate by suitable treatments. On adding alkali to orange coloured potassium dichromate solution, it changes its colour to yellow which on acidifying again changes to orange. It behaves differently on its reaction with conc. H₂SO₄ in cold and in hot. Its most important use is as a strong oxidizing agent in dilute H₂SO₄. It oxidizes KI, FeSO₄, H₂S, nitrites, halogen acids, SO₂, ethyl alcohol etc. It is used in testing a chloride salt by heating alongwith conc. H₂SO₄ when red vapours are evolved. With hydrogen peroxide, acidified K₂Cr₂O₇ gives a deep blue colour. Its structure consists of two tetrahedra linked together sharing one oxygen atom at one corner.