CBSE Class 12 Chemistry D and F Block Exam Notes

CBSE Class 12 Chemistry D and F Block Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Class 12 Chemistry exam notes on d and f-Block elements, covering the definition and electronic configuration of d-block elements, the 3d, 4d and 5d transition series, properties of transition metals (variable oxidation states, paramagnetism, coloured ions, catalytic behaviour, alloy and interstitial compound formation), lanthanoid and actinoid contraction, the chemistry of potassium permanganate and potassium dichromate, and the comparison of lanthanoids and actinoids.

D-BLOCK ELEMENTS AND TRANSITION SERIES

Q1. What are d-Block elements?
Ans. These are those elements in which the last electron enters the d-subshell.

Q2. Why are d-block elements called transition elements?
Ans. Because in this block the elements lie between s and p block, i.e., between group 12 and group 13.

Q3. Give the general electronic configuration of d-block elements.
Ans. (n - 1)d^1-10 ns^1-2

Q4. Name the 3 series of transition elements.
Ans. (i) 3d series: First transition series. Has 10 elements from scandium (Sc) to zinc (Zn).
(ii) 4d series: Second transition series. Has 10 elements from yttrium (Y) to cadmium (Cd).
(iii) 5d series: Third transition series. Has 10 elements from lanthanum (La) to mercury (Hg).

Q5. Which is the most commonly known feature/characteristic of d-block elements?
Ans. It is the presence of unpaired electrons in the d-orbitals, i.e., an incompletely filled set of d-orbitals.

Q6. Why are Zn, Hg and Cd often not regarded as transition elements?
Ans. Being the end members of their respective transition series, their configuration is generally given by (n - 1)d¹⁰ ns², i.e., these elements have completely filled d-orbitals and hence do not fulfil the general characteristic of d-block elements, which is the presence of incompletely filled d-orbitals. Hence they are not regarded as transition elements.

Q7. On what grounds can you say that Scandium (Z = 21) is a transition element but Zn is not?
Ans. Electronic configuration of Sc (Z = 21): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
Electronic configuration of Zn (Z = 30): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
We can say this on the basis of filling of d-orbitals. Sc fulfils the general condition of d-block elements due to the presence of incompletely filled d-orbitals (presence of unpaired electrons), whereas zinc does not, as its d-orbitals are completely filled.

PROPERTIES OF TRANSITION ELEMENTS

Q8. Why does ionic size or atomic size decrease in any transition series?
Ans. As we move from left to right in the transition series, the nuclear charge increases by one unit and the additional electron goes into the inner d-orbital. Since d-electrons are poor at shielding, the effective nuclear charge increases and the electrons are attracted towards the nucleus, due to which a decrease in atomic and ionic radius is observed.

Q9. Why do transition metals show variable oxidation states?
Ans. This is because the outer ns electrons also participate in bonding along with the inner (n - 1)d electrons, as their energies are nearly the same. Also, due to the presence of incompletely filled d-orbitals, different numbers of electrons can be lost by the atoms.

Q10. Why are most of the transition elements paramagnetic? How is their magnetic moment calculated?
Ans. This is due to the presence of unpaired electrons, and each electron has a magnetic moment associated with its spin and its rotation around the axis.
Magnetic moment (μ) = \( \sqrt{n(n+2)} \) B.M., where n = number of unpaired electrons in the d-orbital.

Q11. Calculate the magnetic moment of the compound with Z = 25.
Ans. Z = 25: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
n = 5
μ = \( \sqrt{n(n+2)} \) = \( \sqrt{5(5+2)} \) = \( \sqrt{35} \) = 5.92 B.M.

Q12. Why do transition metals form complex compounds?
Ans. (i) Due to high charge/size ratio (due to small size).
(ii) Availability of vacant d-orbitals for bond formation.

Q13. What is Lanthanoid Contraction? Give its cause and consequence. Show with the help of graph.
Ans. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic number is called lanthanoid contraction.

Causes:
As one moves from one element to another in the lanthanoid series, the nuclear charge increases by one unit and the additional electron goes into the inner 4f orbital. Since 4f electrons are poor at shielding, the effective nuclear charge increases and the electrons are attracted towards the nucleus, due to which a decrease in atomic and ionic size is observed.

[Figure: Graph of ionic radii (pm) on the y-axis (marked 90 and 100) against atomic number on the x-axis (57 to 71, labelled Atomic number). The line of M3+ ions falls steadily from La3+ (above 100 pm, at atomic number 57) to Lu3+ (below 90 pm, at atomic number 71). Separate labelled points are Ce3+ and Pr4+ near atomic numbers 58 to 59 (near 90 pm), Tb4+ near atomic number 65 (below the line), and Tm3+ and Yb3+ near atomic numbers 69 to 70 (above the line).]

Consequences:

(i) Due to lanthanoid contraction the increase in radius from the second to the third transition series almost vanishes. For example, Zr and Hf, and Nb and Ta, have similar radii.
(ii) Due to the very small change in radii of the lanthanoids, their chemical properties are quite similar. Hence, separation from a mixture is very difficult.
(iii) Due to the decrease in ionic radius (due to lanthanoid contraction), the size of the ion decreases and hence the covalent character of the bond between Ln³⁺ (lanthanoid ion) and OH^- ion increases from La³⁺ to Lu³⁺. Therefore, the basic strength of hydroxides decreases with increasing atomic number.

La(OH)₃ - most basic
Lu(OH)₃ - least basic
[La = Lanthanum, Lu = Lutetium]

Q14. Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have the same d⁴ configuration?
Ans. Cr²⁺ is reducing as its configuration changes from d⁴ to d³, i.e., it loses an electron to form Cr³⁺, which has a half-filled t_2g set. On the other hand, when Mn³⁺ changes to Mn²⁺, it results in an extra stable half-filled d⁵ configuration, due to which Mn³⁺ is oxidising.

Q15. Why does the melting point increase from 3d to 4d to 5d series? Show the graph for m.pt vs atomic no. for the 3 series.
Ans. This is because metal-metal bonding becomes more frequent from 3d to 4d to 5d series due to the availability of more valence electrons.

[Figure: Graph of melting point (m.pt, in 10³ K, scale marked 1 to 4) against atomic number for three series. The 3d series (Ti, V, Cr, Mn, Fe, Co, Ni, Cu) has a peak at Cr and a dip at Mn, and ends lowest at Cu and Ag. The 4d series (Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag) peaks at Mo. The 5d series (Hf, Ta, W, Re, Os, Ir, Pt, Au) has the highest peak at W, and each series lies above the one before it. Labels: Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Hf, Ta, W, Re, Os, Ir, Pt, Au.]

Q16. What happens to density from Ti to Cu? Why?
Ans. The density increases from titanium to copper because when we move from Ti to Cu, the atomic mass increases but the metallic radius decreases, because the additional electron enters the d-orbital and hence there is a poor shielding effect.
Density = Atomic Mass / Atomic Volume

Q17. Why do transition metals have high enthalpies of atomisation?
Ans. Because of strong interatomic interactions, which are a result of the large number of valence electrons involved in metallic bonding.

Q18. Name a transition element that is liquid at room temperature.
Ans. Mercury (Hg).

Q19. Why does ionisation enthalpy increase from left to right?
Ans. Because of the increase in nuclear charge, due to which the inner d-orbitals are filled up (poor shielding effect).

Q20. Why do transition metals form coloured ions?
Ans. Due to the presence of unpaired electrons in the d-orbital.
Example: Ti³⁺ forms a purple coloured ion, and Cu²⁺ forms blue coloured ions.
Ti⁴⁺, Sc³⁺ and Zn²⁺ are colourless due to the absence of unpaired electrons.

Q21. Why are transition metals and their compounds used as catalysts?
Ans. (i) Ability to form complexes.
(ii) Ability to form multiple oxidation states.
Example: Iron(III) catalyses the reaction between iodide and persulphate ions.
2I^- + S₂O₈^2- → I₂ + 2SO₄^2- (in presence of Fe³⁺)

Q22. What are interstitial compounds? Why do transition metals form interstitial compounds?
Ans. Compounds which are formed when small non-metal atoms like H, N, C and B are trapped inside the voids (interstitial sites) of the crystal lattice of the metals are called interstitial compounds.
Transition metals form interstitial compounds because the crystal lattices of these metals have empty spaces called voids, where small non-metal atoms like H, N, C and B can get trapped.

Characteristics of interstitial compounds / transition metals:
(i) These metals are non-stoichiometric.
(ii) They have high melting points and are pure metals.
(iii) They are hard, conducting and chemically inert.
(iv) They are neither typically covalent nor ionic.

Q23. Why do transition metals form alloys?
Ans. Because transition metals have almost similar radii, hence atoms of one element can easily get distributed among the atoms of other elements and form alloys. For example, ferrous alloys.

Q24. Explain why E° value for Mn³⁺/Mn²⁺ couple is larger than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺.
Ans. This is because when we move from Mn²⁺ to Mn³⁺, the electronic configuration changes from 3d⁵ to 3d⁴. Mn²⁺ with the stable half-filled 3d⁵ configuration resists removal of the third electron, so the third ionisation enthalpy is very high and hence the E° value is also high (positive).

Q25. Explain the irregular behaviour of the E° values shown in the table below.

Ans. The irregular behaviour can be explained from the irregular variation in ionisation energies (IE₁ + IE₂) and also the sublimation energy, which are very low for manganese and vanadium.

Q26. Is actinoid contraction larger than lanthanoid contraction? Why?
Ans. Yes. This is because the shielding effect of 5f electrons is poorer than that of 4f electrons.

POTASSIUM PERMANGANATE (KMnO₄)

Q27. Give the methods of preparation of potassium permanganate (KMnO₄).
Ans. (i) By fusion of MnO₂ (pyrolusite ore) with an alkali metal hydroxide in the presence of O₂:
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
The dark green K₂MnO₄ (potassium manganate) produced undergoes disproportionation reaction in acid medium:
3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂ + 2H₂O
(from K₂MnO₄) (permanganate)
(ii) Commercial preparation: KMnO₄ is prepared by alkaline oxidative fusion of MnO₂, followed by the electrolytic oxidation of manganate ion (MnO₄^2-) in alkaline solution.
MnO₂ → (KOH, O₂, fusion, oxidation) → MnO₄^2- → (electrolytic oxidation in alkaline solution) → MnO₄^- (permanganate)

Q28. Write the equations to show the oxidising action of KMnO₄ in acidic and basic medium.
Ans. In acidic medium:
MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O
In basic medium:
MnO₄^- + 2H₂O + 3e^- → MnO₂ + 4OH^-

Q29. Give two reactions to show the oxidising action of KMnO₄ in acid medium.
Ans. (i) MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(ii) 2MnO₄^- + 16H⁺ + 10I^- → 2Mn²⁺ + 5I₂ + 8H₂O

Q30. Give two reactions to show the oxidising action of KMnO₄ in basic medium.
Ans. (i) MnO₄^- + 2H₂O + 3Fe²⁺ → MnO₂ + 3Fe³⁺ + 4OH^-
(ii) 2MnO₄^- + 4H₂O + 6I^- → 2MnO₂ + 3I₂ + 8OH^-

Q31. Give the properties and uses of KMnO₄.
Ans. Properties:
(i) Strong oxidising agent in acidic and basic medium.
(ii) It is a purple solid.
Uses:
(i) Used for decolourisation of oils.
(ii) Used as an oxidant in the laboratory.

Q32. Draw the structure of MnO₄^-.
Ans. (Tetrahedral)

[Figure: Structure of the permanganate ion MnO₄^-. Mn is at the centre, with four O atoms around it in a tetrahedral arrangement. One O is at the top, one at the left, one at the right and one at the bottom. Mn is joined to one O by a double bond, and the other O atoms are shown with single bonds, with one labelled O^-.]

POTASSIUM DICHROMATE (K₂Cr₂O₇)

Q33. Why do transition metals have high melting and boiling points?
Ans. Due to strong interatomic bonding, which involves the outer ns electrons together with the inner (n - 1)d electrons in bonding. Greater the number of valence electrons, greater is the bonding.

Q34. How is potassium dichromate prepared?
Ans. K₂Cr₂O₇ is prepared from chromite ore (FeCr₂O₄).
(i) Fusion of chromite ore with Na₂CO₃ in air:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
(ii) Acidification of sodium chromate formed:
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
(iii) Treatment of sodium dichromate with KCl:
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl (potassium dichromate, orange crystals)

Q35. Give the chemical equation to show the oxidising action of K₂Cr₂O₇ in acidic medium.
Ans. Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O (chromic ion)

Q36. Give three reactions to show the oxidising action of K₂Cr₂O₇ in acidic medium.
Ans. (i) Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) Cr₂O₇^2- + 14H⁺ + 6I^- → 2Cr³⁺ + 3I₂ + 7H₂O
(iii) Cr₂O₇^2- + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Q37. Explain how chromate and dichromate ions are interconvertible. Or explain the effect of pH on chromate and dichromate ions.
Ans. 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
Cr₂O₇^2- + 2OH^- → 2CrO₄^2- + H₂O

Q39. Draw the structures of chromate and dichromate ions.
Ans. Chromate ion (CrO₄^2-): tetrahedral.
Dichromate ion (Cr₂O₇^2-): two tetrahedral units sharing one O.

[Figure: Chromate ion CrO₄^2- drawn in square brackets with charge 2-. Cr is at the centre with four O atoms around it in a tetrahedral arrangement (one O at top, one at left, one at right and one at bottom). Dichromate ion Cr₂O₇^2- drawn in square brackets with charge 2-. Two Cr atoms are each surrounded by three terminal O atoms, and the two tetrahedra share one bridging O atom in the middle.]

Q40. What is the highest oxidation state of Cr?
Ans. +6 in CrO₄^2-.

Q41. What is the highest oxidation state of Mn?
Ans. +7 in KMnO₄.

INNER TRANSITION ELEMENTS: LANTHANOIDS AND ACTINOIDS

Q42. What are inner transition elements?
Ans. These are those elements which have partly filled f-orbitals in their neutral atom or ion.

Q43. What are lanthanoids?
Ans. Elements from La (lanthanum) to Lu (lutetium) are called lanthanoids. These are characterised by the filling of 4f-orbitals in their atoms.

Q44. Give the general electronic configuration of lanthanoids.
Ans. [Xe] 4f^0-14 5d^0-1 6s²

Q45. What is the most common oxidation state of lanthanoids?
Ans. +3, however some elements exhibit +2 and +4 also.

Q46. What are actinoids?
Ans. The elements from Ac (actinium) to Lr (lawrencium) are called actinoids. These are characterised by the filling of 5f-orbitals.

Q47. What is the general electronic configuration of actinoids?
Ans. [Rn] 5f^0-14 6d^0-2 7s²

Q48. Why among lanthanoids, Ce(III) can be easily oxidised to Ce(IV)?
Ans. Electronic configuration:
Ce: [Xe] 4f¹ 5d¹ 6s²
Ce³⁺: [Xe] 4f¹
Ce⁴⁺: [Xe]
In Ce³⁺, one electron is present in the 4f orbital. This electron can be lost to form Ce⁴⁺, which has the stable configuration of a noble gas. Therefore, Ce(III) can be easily oxidised to Ce(IV).

Q49. Compare the chemistry of lanthanoids and actinoids on the basis of electronic configuration, oxidation state, atomic and ionic sizes, and chemical reactivity.
Ans.