i. Which one of the following species has d¹⁰ sub-shell?
Answer: b) Zn²⁺
Explanation: Zinc (Zn) has the electronic configuration [Ar] 3d¹⁰ 4s². When it forms Zn²⁺, it loses the two 4s electrons, leaving a stable 3d¹⁰ configuration.
Part A: Multiple Choice Questions (MCQs)
ii. Which one is correct statement about zinc element?
1. It has complete 3d sub-shell
2. It is transition metal
3. It forms colourless complexes
4. It can exist in one stable oxidation state
Answer: b) 1, 3 and 4 only
Explanation: Zinc has a complete 3d subshell (1), forms colourless complexes because there are no d-d transitions (3), and exists only in the +2 oxidation state (4). It is not a typical transition metal (2) because it lacks an incomplete d subshell.
iii. The electronic configuration of iron cation in the complex, [Fe(CN)₆]³⁻, is
Answer: d) [Ar]3d⁵ 4s⁰
Explanation: Fe in [Fe(CN)₆]³⁻ has an oxidation state of +3. The electronic configuration of Fe³⁺ is [Ar] 3d⁵ (the 4s electrons are lost first).
iv. Why is the hexaaquacopper(II) ion blue in colour?
Answer: d) The d-d electronic transition absorbs red wavelengths and transmit the remaining visible light
Explanation: The hexaaquacopper(II) ion absorbs light in the red region of the spectrum. The transmitted light is blue, which is why the solution appears blue.
v. The cyanide ligand (CN⁻) can form two complexes with iron with formulae, [Fe(CN)₆]⁴⁻ and [Fe(CN)₆]³⁻. What is the oxidation state of iron in them?
Answer: a) +2 and +3 respectively
Explanation: In [Fe(CN)₆]⁴⁻, let x be the oxidation state of Fe: x + 6(-1) = -4 [source symbol unavailable] x = +2. In [Fe(CN)₆]³⁻: x + 6(-1) = -3 [source symbol unavailable] x = +3.
vi. Transition metals are different from alkali metals and alkaline earth metals in many ways. Which one is incorrect statement about transition metal when compared with group 1 and group 2 metals?
Answer: d) they are more reactive
Explanation: Transition metals are generally less reactive than alkali and alkaline earth metals (Group 1 and 2). They have higher ionization energies and are harder, denser, and less reactive.
vii. The species that can act as ligand in transition metal complex is
Answer: d) C₆H₅OH₂ (Note: Phenol can act as a weak ligand through its oxygen atom. The other options are not typical ligands.)
viii. Transition metals can show different oxidation states. Which one is the most common oxidation state in first row transition elements?
Answer: b) +2
Explanation: The +2 oxidation state is the most common for first-row transition metals because the 4s electrons are lost relatively easily.
ix. Which one of the following properties of transition metals is held responsible for their catalytic behaviour?
Answer: b) they show variable oxidation states
Explanation: The ability to easily change oxidation states allows transition metals to act as catalysts by providing alternative reaction pathways with lower activation energies.
x. The coordination number of nickel in the complex [Ni(en)₂(OH₂)₂]²⁺ is
Answer: c) 6
Explanation: The complex has two bidentate ethylenediamine (en) ligands (2 × 2 = 4 coordination sites) and two water ligands (2
coordination sites). Total coordination number = 4 + 2 = 6.
xi. Which one of the following complexes cannot show stereoisomerism?
1. [Cr(C₂O₄)₃]³⁻
2. [CuCl₄]²⁻
3. [Cu(H₂O)₆]²⁺
4. [Ni(en)₃]²⁺
Answer: a) 2 and 3 only
Explanation: [CuCl₄]²⁻ is tetrahedral and [Cu(H₂O)₆]²⁺ is octahedral but often distorted; neither shows stereoisomerism. [Cr(C₂O₄)₃]³⁻ and [Ni(en)₃]²⁺ are octahedral with bidentate ligands and show optical isomerism.
xii. The weakest field ligand among the following is
Answer: a) H₂O
Explanation: According to the spectrochemical series, H₂O is a weaker field ligand than NH₃, CN⁻, and OH⁻.
xiii. The highest oxidation state shown by first row transition metals is shown by
Answer: b) manganese
Explanation: Manganese (Mn) can show oxidation states up to +7 (e.g., in KMnO₄), which is the highest for first-row transition metals.
Part B: Short Answer Questions
i. The melting point of titanium is higher than calcium in the same period. Justify this statement.
Answer: Titanium has a higher melting point than calcium because titanium has more unpaired d-electrons available for metallic bonding. These d-electrons participate in strong covalent-like metallic bonds. Calcium, with a filled 4s orbital and no d-electrons in its metallic state, has weaker metallic bonding, resulting in a lower melting point.
ii. The first-row elements of d-block exist in more than one oxidation states. However, zinc shows only +2 oxidation state in its complexes. Why?
Answer: Zinc has the electronic configuration [Ar] 3d¹⁰ 4s². After losing the two 4s electrons to form Zn²⁺, it attains a stable, completely filled 3d¹⁰ configuration. Removing electrons from the stable d¹⁰ configuration requires a very high amount of energy. Therefore, zinc shows only the +2 oxidation state.
iii. Define the terms: a. ligand, b. coordination number, c. complex ion
Answer:
Ligand: An ion or molecule that donates a pair of electrons to a central metal atom or ion to form a coordinate bond in a
complex. Coordination Number: The total number of ligand donor atoms bonded to the central metal ion in a complex. Complex Ion: A charged species consisting of a central metal ion surrounded by ligands bonded through coordinate covalent bonds.
iv. Explain why hydrated Ti³⁺ complexes are colourless and hydrated Ti³⁺ complexes are coloured?
Answer: (Note: The question likely meant Ti⁴⁺ and Ti³⁺)
Ti⁴⁺ (d⁰): Titanium in the +4 oxidation state has a d⁰ configuration. With no d-electrons, no d-d transitions are possible, so its complexes are colourless. Ti³⁺ (d¹): Titanium in the +3 oxidation state has a d¹ configuration. This single d-electron can undergo d-d transitions by absorbing visible light, making its complexes coloured (usually purple).
v. Why do the melting points of first-row transition metals increase up to the middle and then decrease?
Answer: The melting points of first-row transition metals increase up to the middle of the series (e.g., Cr, Mn) because the number of unpaired d-electrons increases, leading to stronger metallic bonding. After the middle, the number of unpaired d-electrons decreases, and the metallic bonds weaken, causing a decrease in melting points.
vi. Why are transition metal used as catalysts in industries for performing different reactions? Give one example.
Answer: Transition metals are used as catalysts because they can easily change their oxidation states and provide a surface for reactions to occur. This allows them to lower the activation energy of reactions. Example: Iron is used as a catalyst in the Haber process for the synthesis of ammonia (N₂ + 3H₂ [source symbol unavailable] 2NH₃).
vii. The blue cobalt chloride paper is used to test the presence of water. If water is present, the paper turns pink. This is because six water ligands exchange for four chloride ligands present in cobalt chloride, [CoCl₄]²⁻. Write the equation to show the ligand substitution reaction when the test is positive.
Answer: [CoCl₄]²⁻(aq) + 6H₂O(l) [source symbol unavailable] [Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) (Blue) (Pink)
viii. Octahedral complexes of copper(II) ion have different colours. Explain why?
Answer: Octahedral complexes of copper(II) (d⁹) have different colours because the crystal field splitting energy (Δ) varies depending on the ligands. Different ligands cause different degrees of splitting of the d-orbitals. As a result, the energy of light absorbed during d-d transitions changes, leading to different colours of transmitted light.
Part C: Long Answer Questions
i. The reduction potential for Cr³⁺ ions being reduced to chromium metal is -0.91V and that for Cr²⁺ ions being reduced to Cr ions in -0.41V as shown:
Cr³⁺ + 3e⁻ [source symbol unavailable] Cr E° = -0.91V
Cr²⁺ + 2e⁻ [source symbol unavailable] Cr E° = -0.41V Predict whether the chromium(II) ion will disproportionate to chromium(III) ion and chromium metal. If yes, then write the possible ionic equation.
Answer: To determine if disproportionation occurs, we calculate the standard cell potential (E°cell) for the reaction: 3Cr²⁺ [source symbol unavailable] 2Cr³⁺ + Cr
This can be broken into two half-reactions:
Oxidation (Anode): Cr²⁺ [source symbol unavailable] Cr³⁺ + e⁻ (Reverse of Cr³⁺ + 3e⁻ [source symbol unavailable] Cr, but for 1 electron)
E°oxidation = -E°reduction = -(-0.91 V) = +0.91 V (but this is for 3 electrons)
For 1 electron: E° = +0.91 V.
Reduction (Cathode): Cr²⁺ + 2e⁻ [source symbol unavailable] Cr E°reduction = -0.41 V
To balance electrons, multiply the oxidation reaction by 2: 2Cr²⁺ [source symbol unavailable] 2Cr³⁺ + 2e⁻ Cr²⁺ + 2e⁻ [source symbol unavailable] Cr
Overall: 3Cr²⁺ [source symbol unavailable] 2Cr³⁺ + Cr
E°cell = E°cathode - E°anode = -0.41 V - (-0.91 V) = +0.50 V
Since E°cell is positive, the disproportionation reaction is spontaneous (feasible).
Ionic Equation: 3Cr²⁺(aq) [source symbol unavailable] 2Cr³⁺(aq) + Cr(s)
ii. Explain the difference in the splitting pattern of the 3d atomic orbitals in the two copper(II) complexes, [Cu(H₂O)₆]²⁺ and [CuCl₄]²⁻.
Answer: The splitting pattern of d-orbitals depends on the geometry of the complex and the nature of the ligands.
[Cu(H₂O)₆]²⁺: This is an octahedral complex. The six water ligands approach along the axes. The d-orbitals pointing directly at the ligands (dz² and dx²-y², eg) experience greater repulsion and are raised in energy. The d-orbitals pointing between the ligands (dxy, dyz, dxz, t2g) experience less repulsion and are lower in energy. This creates a gap Δo (octahedral splitting).
[CuCl₄]²⁻: This is a tetrahedral complex. The splitting pattern is inverted compared to octahedral. The d-orbitals pointing between the ligands (e set) are higher in energy, and those pointing at the ligands (t2 set) are lower in energy. The splitting energy Δt is smaller than Δ_o.
In summary, the geometry (octahedral vs. tetrahedral) and the ligand field strength determine the splitting pattern and the magnitude of Δ.
iii. The following scheme shows some reactions of copper(II) ion. Cu²⁺(aq) [source symbol unavailable] (concentrated HCl) [source symbol unavailable] solution C [source symbol unavailable] (heat with Cu(s)) [source symbol unavailable] [CuCl₂]⁻(aq) Cu²⁺(aq) [source symbol unavailable] (excess of NH₃) [source symbol unavailable] pale blue precipitate A [source symbol unavailable] (excess of NH₃) [source symbol unavailable] solution B
Answer:
A: Cu(OH)₂ (pale blue precipitate) B: [Cu(NH₃)₄]²⁺ (deep blue solution) C: [CuCl₄]²⁻ (yellow-green solution)
Answer: Solution B is deep blue. Solution C is yellow-green.
Answer: The reaction is a reduction (or disproportionation). Copper acts as a reducing agent, reducing Cu²⁺ to Cu⁺.
Answer: A white precipitate of CuCl (copper(I) chloride) is formed. The colour is white because Cu⁺ has a d¹⁰ configuration (no d-d transitions).
iv. Transition metal complexes such as [Cu(H₂O)₆(NH₃)₂]²⁺ and Pt(NH₃)₂Cl₂ show same type of stereoisomerism.
Answer: Geometrical (cis-trans) isomerism.
Answer: (Diagram Description)
[Cu(H₂O)₄(NH₃)₂]²⁺ (Octahedral):
cis-isomer: The two NH₃ ligands are adjacent (90° apart). trans-isomer: The two NH₃ ligands are opposite (180° apart).
[Pt(NH₃)₂Cl₂] (Square Planar):
cis-isomer: The two Cl ligands are adjacent (90° apart). This is the anticancer drug cisplatin. trans-isomer: The two Cl ligands are opposite (180° apart).
v. Aqueous copper(II) ion can form complexes with the ligands, ammonia and 1,2-diaminoethane as shown.
[Cu(H₂O)₆]²⁺(aq) + en(aq) [source symbol unavailable] [Cu(H₂O)₄(en)]²⁺(aq) + 2H₂O(l) Kstab = 3.98 × 10¹⁰
[Cu(H₂O)₆]²⁺(aq) + 2NH₃(aq) [source symbol unavailable] [Cu(H₂O)₄(NH₃)₂]²⁺(aq) + 2H₂O(l) Kstab = 5.01 × 10⁷
Answer:
Expression 1: Kstab = [[Cu(H₂O)₄(en)]²⁺] / [[Cu(H₂O)₆]²⁺][en]
Expression 2: Kstab = [[Cu(H₂O)₄(NH₃)₂]²⁺] / [[Cu(H₂O)₆]²⁺][NH₃]² Units: Both are in mol⁻¹ dm³.
Answer: The most stable complex is [Cu(H₂O)₄(en)]²⁺. This is because it has the highest stability constant (Kstab = 3.98 × 10¹⁰). The bidentate ligand (en) forms a more stable complex than the monodentate ligand (NH₃) due to the chelate
effect.