Chapter 4 The d-block and f-block Elements Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 91
4.1. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
Solution
Ag has a completely filled 4d orbital (4d10 5s1) 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-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element.
Page No. 95
4.2. 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?
Solution
In 3d series from Sc to Zn, all elements have one or more unpaired e-1 s except Zn which has no unpaired electron as its outer EC is 3d104s2. Hence, the intermetallic bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.
Page No. 97
4.3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Solution
Mn (Z = 25) = 3d5 4s2
Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
Page No. 98
4.4. The Eθ(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHθ and low ΔhydHθ)
Solution
The Eθ(M2+/M) value of a metal depends on the energy changes involved in the following:
1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.M(s) → M(g) (△sH: Sublimation energy)
M(g) → M2+(aq) (△iH: Ionization energy)
M2+(g) → M2+(aq) (△hydH: Hydration energy)
Now, copper has a high energy of atomization and low hydration energy. Hence, the Eθ(M2+/M) value for copper is positive.
Page No. 100
4.5. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?
Solution
There is a irregularity in the IE’s of 3d-series due to alternation of energies of 4s and 3d orbitals when an e-1 is removed. Thus, there is a reorganisation energy accompanying ionization. This results into release of exchange energy which increases as the number of e-1s increases in the dn configuration. Cr has low 1st IE because loss of 1 e- gives stable EC (3d6). Zn has very high IE because e~ has to be removed from 4s orbital of the stable configuration (3d10 4s2). After the loss of one e–, removal of 2nd e–, becomes difficult. Hence, 2nd IE’s are higher and in general, increase from left to right. However, Cr and Cu show much higher values because 2nd e– has to be removed from stable configuration of Cr+(3d5) and Cu+(3d10).
Page No. 101
4.6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Solution
Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
4.7. Which is a stronger reducing agent Cr2+ or Fe2+ and why?
Solution
The following reactions are involved when Cr2+ and Fe2+ act as reducing agents.
Cr2+ → Cr3+Fe2+ → Fe3+
The 
value is -0.14 V and 
is + 0.77 V. This means that Cr2+ can be easily oxidized to Cr3+, but Fe2+ does not get oxidized to Fe3+ easily. Therefore, Cr2+ is a better reducing agent that Fe3+.
Page No. 103
4.8. Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).
Solution
Z = 27
⇒ [Ar] 3d7 4s2
∴ M2+ = [Ar] 3d7
i.e., 3 unpaired electrons
∴ n = 3
μ ≈ 4 BM
Page No. 105
4.9. Explain why Cu+ ion is not stable in aqueous solutions?
Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why?
Solution
In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ion in an aqueous solution is unstable. It disproportionates to give Cu2+ and Cu.
2Cu2+(aq) → Cu2+(aq) + Cu(s)
Page No. 114
How would you account for the following: Actinoid contraction is greater than lanthanoid contraction.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Solution
In actinoids, 5 f-orbitals are filled. These 5 f-orbitals have a poorer shielding effect than 4 f-orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more than that experienced by electrons in valence shells in case of lanthanoids. Hence, the size contraction in actinoids is greater than that in lanthanoids.