Questions & explanations
1. Why does copper not liberate H₂ from dilute HCl, unlike zinc?
- Because Cu has a high sublimation enthalpy and high ionization enthalpies, making E°(Cu²⁺/Cu) positive
- Because Cu has a low hydration enthalpy, making E°(Cu²⁺/Cu) negative
- Because Cu is less reactive than H₂ in the reactivity series
- Because Cu forms a protective oxide layer that prevents reaction
Answer: Because Cu has a high sublimation enthalpy and high ionization enthalpies, making E°(Cu²⁺/Cu) positive
The standard electrode potential E°(Cu²⁺/Cu) is +0.34 V, which is positive. This arises because the sum of sublimation enthalpy and ionization enthalpies is high, and hydration enthalpy is insufficient to make the overall process spontaneous. Since E°(Cu²⁺/Cu) > E°(H⁺/H₂) = 0 V, Cu cannot reduce H⁺ to H₂. The thermodynamic cycle (Born-Haber) explains this.
2. For which of the following Ln³⁺ ions is the magnetic moment close to the spin-only value?
- Gd³⁺
- Nd³⁺
- Pr³⁺
- Er³⁺
Answer: Gd³⁺
The spin-only formula μ = √(n(n+2)) BM works only when orbital angular momentum is zero (L=0). Gd³⁺ has a half-filled 4f⁷ configuration, giving L=0 and S=7/2, so its magnetic moment is close to the spin-only value of √(7×9) ≈ 7.94 BM. Other Ln³⁺ ions have non-zero L due to incomplete f-subshells, leading to orbital contribution.
3. Which of the following statements about actinoids is correct?
- All actinoids are radioactive and show only +3 oxidation state.
- Actinoids exhibit a greater variety of oxidation states than lanthanoids because 5f, 6d, and 7s orbitals have comparable energies.
- The actinoid contraction per element is less than the lanthanoid contraction.
- Naturally occurring actinoids include Th, U, and Pu.
Answer: Actinoids exhibit a greater variety of oxidation states than lanthanoids because 5f, 6d, and 7s orbitals have comparable energies.
Actinoids show a wide range of oxidation states (e.g., U: +3 to +6, Np: +3 to +7) because the 5f, 6d, and 7s orbitals are close in energy, allowing all to participate in bonding. This contrasts with lanthanoids, where 4f, 5d, and 6s energies are more separated, limiting oxidation states mostly to +3.
4. Which of the following pairs of catalyst and industrial process is correctly matched?
- V₂O₅ – Haber process
- Fe – Contact process
- Ni – Hydrogenation of vegetable oils
- MnO₂ – Contact process
Answer: Ni – Hydrogenation of vegetable oils
Nickel is used as a catalyst for the hydrogenation of vegetable oils to produce vanaspati ghee. This is a standard application of transition metals as catalysts. The other pairs are mismatched: V₂O₅ is used in the contact process, Fe in the Haber process, and MnO₂ in the decomposition of KClO₃.
5. Which of the following is NOT a transition element according to the IUPAC definition?
- Zinc (Zn)
- Copper (Cu)
- Scandium (Sc)
- Iron (Fe)
Answer: Zinc (Zn)
A transition element has a partially filled d-subshell in its ground state or in any common oxidation state. Zinc (Zn) has d¹⁰ in both ground state (3d¹⁰4s²) and Zn²⁺ (3d¹⁰), so it is not a transition element. Sc, Cu, and Fe all have partially filled d-subshells in at least one common state.
6. Which property of transition metals makes them effective catalysts in industrial processes?
- Variable oxidation states and ability to adsorb reactants
- High melting points and densities
- Formation of coloured ions and complexes
- High electrical and thermal conductivity
Answer: Variable oxidation states and ability to adsorb reactants
Transition metals exhibit catalytic properties due to their ability to adopt multiple oxidation states, allowing them to act as electron relays, and their capacity to adsorb reactant molecules on their surface through partially filled d-orbitals, weakening bonds and facilitating reactions.
7. Down a group in the d-block, the stability of higher oxidation states generally:
- increases
- decreases
- remains the same
- first increases then decreases
Answer: decreases
Down a group, the increasing atomic size and poor shielding of d-electrons make higher oxidation states less stable. Relativistic effects also stabilize lower oxidation states. For example, Cr(VI) is strongly oxidizing, but Mo(VI) and W(VI) are less stable and more reducing.
8. Which pair of elements has nearly identical atomic radii due to lanthanoid contraction?
- V and Nb
- Ti and Zr
- Zr and Hf
- Cr and Mo
Answer: Zr and Hf
Lanthanoid contraction causes the atomic radii of 4d and 5d transition elements in the same group to become nearly equal. Zr (4d) and Hf (5d) have radii of 145 pm and 144 pm respectively, making them almost identical. This is a direct consequence of the contraction.
9. The intense purple colour of KMnO₄ arises from which type of electronic transition?
- d-d transition in Mn(VII)
- f-f transition in Mn
- Metal-to-ligand charge transfer
- Ligand-to-metal charge transfer
Answer: Ligand-to-metal charge transfer
In KMnO₄, Mn is in +7 oxidation state with d⁰ configuration, so d-d transitions are impossible. The intense purple colour is due to ligand-to-metal charge transfer (LMCT): an electron from O 2p orbital jumps to an empty Mn 3d orbital, absorbing green-yellow light.
10. Which of the following statements about manganese is correct?
- Mn shows only +2 and +7 oxidation states because 3d electrons are too tightly bound.
- Mn shows all oxidation states from +2 to +7 because all seven outer electrons can participate in bonding.
- Mn shows oxidation states from +2 to +7, but +7 is the most stable in aqueous solution.
- Mn shows oxidation states from +2 to +7, but +3 is the most stable due to half-filled d-subshell.
Answer: Mn shows all oxidation states from +2 to +7 because all seven outer electrons can participate in bonding.
Manganese has the ground-state configuration [Ar]3d^5 4s^2. All seven outer electrons (5 from 3d and 2 from 4s) can be ionized, allowing oxidation states from +2 to +7. The +2 state (d^5) is the most stable in aqueous solution due to half-filled stability.
11. The intense purple colour of aqueous KMnO4 is due to which type of electronic transition?
- d-d transition
- Intra-ligand π→π* transition
- Metal-to-ligand charge transfer
- Ligand-to-metal charge transfer
Answer: Ligand-to-metal charge transfer
In KMnO4, Mn is in +7 oxidation state with d0 configuration, so d-d transitions are impossible. The intense purple colour arises from ligand-to-metal charge transfer (LMCT) where an electron from oxygen (ligand) is excited to the empty d orbitals of Mn.
12. Which of the following oxides of chromium is amphoteric?
- CrO
- Cr2O3
- CrO3
- CrO2
Answer: Cr2O3
Cr2O3 has chromium in +3 oxidation state, which is intermediate. For the same element, lower oxides are basic, intermediate oxides are amphoteric, and higher oxides are acidic. Cr2O3 dissolves in both acids and bases, confirming its amphoteric nature.