CBSE Class 12 Chemistry Notes Chapter 8 The d and f Block Elements

Chapter 7 The d and f Block Elements Class 12 Chemistry Notes for CBSE Students

8.1 Silver atom has completely filled d orbitals (4d 10) in its ground state. How can you say that it is a transition element?

Ans: Ag has a completely filled 4d orbital (4d 10 5s 1 ) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s- orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d 9).
Hence, it is a transition element

8.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol–1. Why?

Ans: The enthalpy of atomization is directly linked with the stability of the crystal lattice and also the strength of the metallic bond. In case of zinc (3d104s2 configuration), no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. However, in all other elements belonging to 3d series one or more d-electrons are involved in the metallic bonds. This means that the metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomization in the 3d series.

8.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Ans: Mn (Z = 25) = 3d 5 4s 2

Manganese (Z = 25) shows a maximum number of O.S. This is because its outer EC is 3d54s2. As 3d and 4s are close in energy, it has a maximum number of e-1 s to lose or share. Hence, it shows O.S. from +2 to +7 which is the maximum number.

8.4 The E V (M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high ∆aH V and low ∆hydH V)


The d and f Block Elements

8.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Ans: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d 0 , d 5 , d

Since these states are exceptionally stable, their ionization enthalpies are very high. In case of first ionization energy, Cr has low ionization energy.

This is because after losing one electron, it attains the stable configuration (3d 5 ). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d 10 4s 2 ).

Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out.

(Cr+ : 3d 5 and Cu+ : 3d 10).

Hence, taking out one electron more from this stable configuration will require a lot of energy.

8.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Ans: Both fluorine and oxygen have very high electronegativity values. They can oxidize the metals to the highest oxidation state. As a result, the highest oxidation states are shown by the fluorides and oxides of the metals; transition metals in particular.

8.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Ans: Cr2+ is stronger reducing agent than Fe2+ Reason: d4 → d3 occurs in case of Cr2+ to Cr2+. But d6 → d5 occurs in case of Fe2+ to Fe2+.

In a medium (like water) d’1 is more stable as compared to d5.

8.8 Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).

Ans: Z = 27 = [Ar] 3d7 4s2
M2+ = [Ar] 3d7

The d and f Block Elements

8.9 Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable.

Cu+ in aqueous solution undergoes disproportionation, i.e.,

2Cu+(aq) → Cu2+(aq) + Cu(s)

The stability of Cu2+(aq) rather than Cu+(aq) is due to the much more negative ∆hyd H° of Cu2+(aq) than Cu+, which more than compensates for the second ionization enthalpy of Cu.

8.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans: The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the series.

In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids.

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