February 22, 2024

Explore the NCERT Solutions for Class 12 Science Chemistry Chapter 8: “The D And F Block Elements,” featuring clear and concise step-by-step explanations. Widely embraced by class 12 Science students, these solutions prove invaluable for completing homework efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of class 12 Science Chemistry Chapter 8 is available here, making it a convenient resource for students. Page No 212: Question 8.1: Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? ANSWER: 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 215: Question 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? ANSWER: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization. Page No 217: Question 8.3: Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? ANSWER: 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 217: Question 8.4: 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) ANSWER: 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. 2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state. 3. Hydration: The energy released when one mole of ions are hydrated. Now, copper has a high energy of atomization and low hydration energy. Hence, the Eθ(M2+/M) value for copper is positive. Page No 219: Question 8.5: How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? ANSWER: 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 d0, d5, d10. 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 (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2). Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+: 3d5 and Cu+: 3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy. Page No 220: Question 8.6: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? ANSWER: 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. Page No 220: Question 8.7: Which is a stronger reducing agent Cr2+ or Fe2+ and why? ANSWER: The following reactions are involved when Cr2+ and Fe2+ act as reducing agents. Cr2+ Cr3+ Fe2+ Fe3+ The value is −0.41 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 222: Question 8.8: Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27). ANSWER: Z = 27  [Ar] 3d7 4s2  M2+ = [Ar] 3d7 3d7 =  i.e., 3 unpaired electrons n = 3 μ ≈ 4 BM Page No 224: Question 8.9: Explain why Cu+ ion is not stable in aqueous solutions? ANSWER: 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. Page No 232: Question 8.10: Actinoid contraction is greater from element to element than lanthanoid contraction. Why? ANSWER: 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. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids. Page No 234: Question 8.1: Write down the electronic configuration of: (i) Cr3++ (iii) Cu+(v) Co2+ (vii) Mn2+ (ii) Pm3+(iv) Ce4+ (vi) Lu2+(viii) Th4+ ANSWER: (i) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or, [Ar]18 3d3 (ii) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4 Or, [Xe]54 3d3 (iii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 (iv) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or, [Xe]54 (v) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar]18 3d7 (vi) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1 Or, [Xe]54 2f14 3d3 (vii) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5 (viii) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6s6 Or, [Rn]86 Page No 234: Question 8.2: Why are Mn2+compounds more stable than Fe2+towards oxidation to their +3 state? ANSWER: Electronic configuration of Mn2+ is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one …

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