Questions & explanations
1. Which of the following is a correct statement regarding the use of KMnO₄ and K₂Cr₂O₇ as oxidants in titrations?
- Both KMnO₄ and K₂Cr₂O₇ act as self-indicators in titrations.
- KMnO₄ is a primary standard while K₂Cr₂O₇ is a secondary standard.
- K₂Cr₂O₇ can be used in titrations involving hydrochloric acid, whereas KMnO₄ cannot.
- Both require an external indicator like diphenylamine.
Answer: K₂Cr₂O₇ can be used in titrations involving hydrochloric acid, whereas KMnO₄ cannot.
KMnO₄ is a strong oxidant that oxidises Cl⁻ ions present in HCl to Cl₂, leading to side reactions and inaccurate results; hence it cannot be used in HCl medium. K₂Cr₂O₇, on the other hand, does not oxidise Cl⁻ under usual titration conditions, so it can be used in the presence of HCl. KMnO₄ acts as a self-indicator (purple to colourless), so option A is wrong. K₂Cr₂O₇ is a primary standard, while KMnO₄ is a secondary standard, making option B incorrect. KMnO₄ does not need an external indicator, and K₂Cr₂O₇ requires an external indicator (e.g., diphenylamine), so option D is also false.
2. The colour exhibited by transition metal ions in aqueous solution is primarily due to:
- absorption of light caused by s-d transitions
- absorption of light caused by d-d transitions
- emission of light caused by d-d transitions
- reflection of light due to metallic luster
Answer: absorption of light caused by d-d transitions
Transition metal ions have partially filled d-orbitals. When visible light falls on them, electrons absorb energy and undergo excitation from lower to higher d-orbitals (d-d transitions). The complementary colour of the absorbed light is transmitted, giving the ion its characteristic colour. The other options are incorrect: s-d transitions do not occur in the visible range; the colour is due to absorption, not emission; metallic luster is a property of bulk metals, not aqueous ions.
3. Which property of transition metals makes them particularly suitable as heterogeneous catalysts?
- Variable oxidation states
- High electrical conductivity
- High melting points
- Formation of coloured compounds
Answer: Variable oxidation states
Transition metals can exhibit multiple oxidation states, which allows them to form reactive intermediates with reactants and provide alternative pathways with lower activation energy. This versatility is the key to their catalytic activity. High electrical conductivity, high melting points, and formation of coloured compounds are not the primary reasons for catalytic behaviour.
4. Which of the following is a reason for the tendency of transition metals to form stable complex compounds?
- Small size and high charge on the cations
- High melting points
- High densities
- High ionization energies
Answer: Small size and high charge on the cations
Transition metal cations are small and have high nuclear charge, which enables them to attract electron-rich ligands strongly. This, along with the availability of vacant d-orbitals for accepting lone pairs from ligands, leads to the formation of stable coordination complexes. High melting points, densities, and ionization energies are not direct reasons for complex formation.
5. The spin-only magnetic moment (in Bohr magneton) of Fe³⁺ ion (high spin) is:
- 5.92
- 4.90
- 3.87
- 2.83
Answer: 5.92
Fe³⁺ has the electronic configuration [Ar]3d⁵. In high spin condition, all five d-electrons are unpaired (n = 5). The spin-only magnetic moment is given by μ = √(n(n+2)) = √(5×7) = √35 ≈ 5.92 BM. The other options correspond to different numbers of unpaired electrons: 4.90 BM (n = 4), 3.87 BM (n = 3), 2.83 BM (n = 2).
6. Which of the following is an example of an interstitial compound formed by a transition metal?
- Fe₃C
- NaCl
- CaO
- MgCl₂
Answer: Fe₃C
In interstitial compounds, small non-metal atoms (like C, H, B, N) occupy the interstitial voids in the crystal lattice of a transition metal. Fe₃C (cementite) is a classical example where carbon atoms fit into the interstices of the iron lattice. NaCl, CaO, and MgCl₂ are ionic compounds, not interstitial compounds.
7. Lanthanoid contraction is responsible for which of the following?
- Similar atomic radii of Zr and Hf
- Increasing atomic size from La to Lu
- High melting points of lanthanoids
- All lanthanoids exhibit +3 oxidation state
Answer: Similar atomic radii of Zr and Hf
Lanthanoid contraction causes the 4d and 5d series elements to have nearly identical atomic radii, as the size increase due to an extra shell is offset by the contraction from poor shielding of 4f electrons. Thus Zr (4d) and Hf (5d) have similar sizes.
8. Which of the following correctly describes the trend of atomic radii across the 3d series from Sc to Zn?
- Increases gradually
- Decreases gradually
- Remains constant
- Increases then decreases
Answer: Decreases gradually
Atomic radii decrease slightly across a transition series because the increase in nuclear charge is not fully shielded by the electrons added to the inner d-orbitals, leading to a stronger pull on the outer electrons.
9. Which of the following elements is not considered a transition element according to IUPAC?
- Fe
- Cu
- Zn
- Mn
Answer: Zn
Zn has completely filled d10 configuration in its atomic state (3d10 4s2) and in its common +2 oxidation state, so it does not have partially filled d-orbitals, which is a requirement for transition elements.
10. In the preparation of potassium dichromate from chromite ore (FeCr₂O₄), the ore is first fused with:
- NaOH and air
- Na₂CO₃ and air
- KOH and KNO₃
- H₂SO₄ and Na₂SO₄
Answer: Na₂CO₃ and air
Chromite ore is fused with sodium carbonate (Na₂CO₃) in the presence of air (oxygen) to form sodium chromate (Na₂CrO₄). This is followed by leaching, acidification, and treatment with KCl to obtain K₂Cr₂O₇.
11. Why are zinc, cadmium, and mercury not classified as typical transition metals?
- They have low boiling points.
- They are liquids at room temperature.
- They have completely filled d-orbitals in ground state and common oxidation states.
- They do not form colored compounds.
Answer: They have completely filled d-orbitals in ground state and common oxidation states.
Transition metals require partially filled d-orbitals. Zn, Cd, and Hg have d10 configurations in both the ground state and all common oxidation states, so they are not typical transition metals.
12. In the preparation of potassium permanganate from pyrolusite, the first step involves:
- Fusion of pyrolusite with KOH and KNO₃ to form K₂MnO₄
- Reduction of MnO₂ to Mn metal
- Direct oxidation of MnO₂ with H₂SO₄
- Electrolysis of MnSO₄ solution
Answer: Fusion of pyrolusite with KOH and KNO₃ to form K₂MnO₄
Pyrolusite (MnO₂) is fused with KOH and an oxidizing agent like KNO₃ to convert it into potassium manganate (K₂MnO₄). This is the first step; later, K₂MnO₄ is electrolytically oxidized to KMnO₄.