MCQ practice for d & f Block Elements and Coordination Compounds — click an option to check your answer.
🔬 d & f Block Elements 7 marks
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Q1
Lanthanoid contraction is primarily caused by:
Correct: A. 4f orbitals have a diffuse shape with very poor shielding ability. As protons are added across La→Lu, the effective nuclear charge (Z_eff) on outer electrons increases significantly with each element → electrons pulled inward → radius decreases steadily across the lanthanoid series.
Q2
The magnetic moment of Cr³⁺ (electronic configuration [Ar]3d³) is approximately:
Correct: B — 3.87 BM. Cr³⁺ is [Ar]3d³ — three unpaired electrons. μ = √(n(n+2)) = √(3×5) = √15 ≈ 3.87 BM. Remember: n=1→1.73, n=2→2.83, n=3→3.87, n=4→4.90, n=5→5.92 BM.
Q3
KMnO₄ acts as an oxidising agent in acidic medium. Mn is reduced to:
Correct: A — Mn²⁺. In acidic medium (H₂SO₄): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Mn goes from +7 to +2 (gain of 5e⁻). In neutral: MnO₄⁻ → MnO₂ (brown, Mn = +4). In alkaline: MnO₄⁻ → MnO₄²⁻ (green, Mn = +6).
Q4
Which pair of elements has nearly identical atomic radii due to lanthanoid contraction?
Correct: B — Zr (160 pm) and Hf (159 pm). Lanthanoid contraction causes 5d elements to be smaller than expected. Hf (5d²) should be larger than Zr (4d²), but 14 lanthanoids intervene between them, whose contraction exactly compensates → nearly identical radii → very difficult to separate.
Q5
Which of the following is a characteristic property of transition metals?
Correct: D — All of the above. Transition metals show: (i) Variable OS — due to comparable energies of (n-1)d and ns electrons; (ii) Coloured ions — due to d-d electronic transitions absorbing visible light; (iii) Catalytic activity — due to variable OS and ability to adsorb reactants.
Q6
The number of unpaired electrons in Fe³⁺ (in high-spin state) is:
Correct: C — 5 unpaired electrons. Fe is [Ar]3d⁶4s². Fe³⁺ loses 3 electrons → [Ar]3d⁵. In high-spin 3d⁵: ↑ ↑ ↑ ↑ ↑ (each of 5 d-orbitals has one electron, by Hund's rule) → 5 unpaired electrons. μ = √(5×7) = √35 ≈ 5.92 BM.
Q7
The highest oxidation state shown by Mn is +7 (in KMnO₄). This is due to:
Correct: C. Mn has [Ar]3d⁵4s² = 7 electrons available for bonding. In +7 state, all 5 d-electrons and 2 s-electrons participate in bond formation. The highest OS equals the total number of (n-1)d + ns electrons for early d-block elements.
Q8
Which of the following electronic configurations does NOT belong to a transition metal in its ground state?
Correct: B. [Ar]3d⁹4s² is incorrect for any element. Cu is [Ar]3d¹⁰4s¹ (not 3d⁹4s²) due to extra stability of completely filled 3d subshell. Zn is [Ar]3d¹⁰4s², not a transition metal (fully filled d in all oxidation states). Cr is [Ar]3d⁵4s¹ (half-filled 3d, extra stable).
🔗 Coordination Compounds 7 marks
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Q1
The IUPAC name of [Co(NH₃)₆]³⁺ is:
Correct: A. IUPAC rule: ligands are named alphabetically before the metal. NH₃ = ammine (double m). Six = hexa. Co³⁺ = cobalt(III). Hence: hexaamminecobalt(III) ion. Note: if it were an anion (like [Co(CN)₆]³⁻), the metal would be named cobaltate(III).
Q2
The coordination number of Pt in [Pt(en)₂Cl₂]²⁺ (en = ethylenediamine) is:
Correct: C — 6. en (ethylenediamine) is a bidentate ligand (donates 2 lone pairs). Two en ligands donate 2×2 = 4 bonds. Two Cl⁻ donate 2×1 = 2 bonds. Total coordination number = 4 + 2 = 6. Geometry: octahedral, which can show cis-trans and optical isomerism.
Q3
Which of the following complexes can exhibit geometrical isomerism?
Correct: B — [PtCl₂(NH₃)₂]. This square planar complex has cis (Cl atoms on same side) and trans (Cl atoms on opposite sides) isomers. [Co(NH₃)₆]³⁺ has all identical ligands — no isomers. Tetrahedral complexes with 2 types of ligands [MA₂B₂] do NOT show geometrical isomerism. [Co(en)₃]³⁺ shows optical isomerism only.
Q4
Crystal Field Stabilisation Energy (CFSE) is maximum for which d-configuration?
Correct: C — d⁶ low spin. In low spin d⁶: all 6 electrons are in t₂g (↑↓ ↑↓ ↑↓). CFSE = 6×(−0.4Δo) = −2.4Δo (most negative = maximum stabilisation). d³: CFSE = −1.2Δo. d⁵ high spin: CFSE = 0 (no net stabilisation). d⁰: CFSE = 0.
Q5
Which of the following is a chelating ligand?
Correct: C — en (ethylenediamine). A chelating ligand has two or more donor atoms that can simultaneously bond to the same metal ion, forming a ring structure. en (H₂N–CH₂–CH₂–NH₂) has two NH₂ groups → bidentate chelate. EDTA is hexadentate. Cl⁻, NH₃, H₂O are monodentate (unidentate).
Q6
The hybridisation of Ni in [Ni(CO)₄] is:
Correct: A — sp³, tetrahedral. Ni in [Ni(CO)₄] is Ni(0): [Ar]3d¹⁰4s⁰. CO is a strong field ligand. The 3d¹⁰ configuration is fully filled; no d-orbitals available for hybridisation. So hybridisation uses 4s and three 4p orbitals → sp³, tetrahedral, diamagnetic.
Q7
Linkage isomers differ in:
Correct: B. Linkage isomers arise when an ambidentate ligand (one that can bind through two different atoms) uses different donor atoms. Example: NO₂⁻ can bind through N (nitro, –NO₂) or through O (nitrito, –ONO). [Co(NO₂)(NH₃)₅]²⁺ vs [Co(ONO)(NH₃)₅]²⁺.
Q8
The oxidation state of Co in [Co(CN)₆]³⁻ is:
Correct: B — +3. Let OS of Co = x. CN⁻ = −1 (charge on each cyanide). Total charge on complex = −3. So: x + 6(−1) = −3 → x − 6 = −3 → x = +3. This is a strong-field complex: d²sp³, octahedral, diamagnetic.
Q9
Which of the following is a strong-field ligand that causes maximum d-orbital splitting?
Correct: D — CO. Spectrochemical series (increasing field strength): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. CO and CN⁻ are strongest field ligands (π-backbonding). F⁻ is a weak field ligand despite being highly electronegative.
Q10
The magnetic moment of [Ni(CN)₄]²⁻ is 0 BM. This is because:
Correct: C. Ni²⁺ is [Ar]3d⁸. Free ion has 2 unpaired electrons. But CN⁻ is a strong-field ligand → large crystal field splitting → electrons pair up in dsp² hybridisation (square planar). All 8 d-electrons are paired in four orbitals → 0 unpaired → diamagnetic, μ = 0 BM.