55+ direct NCERT questions from d & f-Block Elements and Coordination Compounds — click an option to reveal the answer.
⚗️ d & f-Block Elements 9 marks
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Q1
The electronic configuration of Chromium (Cr, Z=24) is an exception. It is:
Correct: B — [Ar] 3d⁵ 4s¹. Expected configuration is [Ar] 3d⁴ 4s², but one electron moves from 4s to 3d to achieve a half-filled d⁵ configuration. Half-filled (d⁵) subshells have extra stability due to: (1) symmetrical distribution of electrons, (2) maximum exchange energy (each electron can exchange with 4 others → 10 exchanges for d⁵). Similarly, Cu ([Ar] 3d¹⁰ 4s¹) has fully-filled d¹⁰ stability.
Q2
The electronic configuration of Copper (Cu, Z=29) is:
Correct: C — [Ar] 3d¹⁰ 4s¹. Expected: [Ar] 3d⁹ 4s². Actual: [Ar] 3d¹⁰ 4s¹, because a completely filled d¹⁰ subshell is extra stable. Note: Cu forms Cu⁺ ([Ar] 3d¹⁰) and Cu²⁺ ([Ar] 3d⁹) ions. Cu⁺ is diamagnetic (no unpaired electrons), Cu²⁺ is paramagnetic (one unpaired electron). Both Cr and Cu are classic exceptions to the Aufbau principle in the first transition series.
Q3
Which of the following is NOT considered a transition element according to NCERT definition?
Correct: D — Zinc (Zn). NCERT defines transition elements as those with incompletely filled d-orbitals in their ground state OR in commonly occurring oxidation states. Zn: [Ar] 3d¹⁰ 4s² (ground state) and Zn²⁺: [Ar] 3d¹⁰. Since d-orbitals are completely filled in both ground state and common ion, Zn is NOT a transition metal. Also, Sc (only Sc³⁺ has d⁰) and Zn are placed at the boundaries but Zn is excluded by the d-orbital criterion.
Q4
The change in oxidation state of Mn in KMnO₄ when it reacts in acidic medium is:
The colour of KMnO₄ (potassium permanganate) is purple/violet even though Mn⁷⁺ has no d-electrons. This is due to:
Correct: B — Charge transfer. Mn in KMnO₄ is Mn⁷⁺ (d⁰) — no d electrons for d-d transitions. The intense purple colour arises from charge transfer: electrons are transferred from O²⁻ ligands to the empty d-orbitals of Mn⁷⁺. These transitions involve higher energy and produce more intense colours than d-d transitions. Similarly, K₂Cr₂O₇ (Cr⁶⁺, d⁰) is orange due to charge transfer from O to Cr. TiO₂ (Ti⁴⁺, d⁰) is white because charge transfer absorbs only UV.
Q6
The change in oxidation state of Cr in K₂Cr₂O₇ in acidic medium is:
Correct: C — Cr₂O₇²⁻ gains 6e⁻ total. Half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Each Cr goes from +6 to +3 (gain of 3e⁻), and with 2 Cr atoms: total 6e⁻ gained. K₂Cr₂O₇ is an orange oxidising agent used for: (1) FeSO₄ titration, (2) oxidising organic compounds, (3) preparing chromium compounds. Dichromate (orange) and chromate (yellow) are interconverted by pH: Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O.
Q7
Transition metals generally show paramagnetism because:
Correct: D — Unpaired d-electrons. Magnetic moment μ = √(n(n+2)) BM, where n = number of unpaired electrons. Mn²⁺ (d⁵, 5 unpaired e⁻): μ = √35 ≈ 5.92 BM. Cu²⁺ (d⁹, 1 unpaired): μ = √3 ≈ 1.73 BM. Zn²⁺ (d¹⁰, 0 unpaired): diamagnetic. Exception: Zn and Sc³⁺ have fully filled/empty d-orbitals → diamagnetic. The element with highest paramagnetism in first transition series is Mn or Cr (5 or 6 unpaired electrons).
Q8
Lanthanide contraction is caused by:
Correct: A — Poor 4f shielding. As we move across the lanthanide series (La to Lu), electrons are added to the inner 4f orbitals. The 4f orbitals are diffuse and provide poor shielding of the outer electrons from the nuclear charge. So effective nuclear charge (Zeff) increases → outer electrons pulled inward → gradual decrease in atomic and ionic radius. This is lanthanide contraction. Consequence: 4d and 5d elements (e.g., Zr and Hf) have nearly identical radii, making them chemically similar and hard to separate.
Q9
Due to lanthanide contraction, the atomic radii of which pair of elements are nearly identical?
Correct: B — Zr and Hf. Zirconium (4d, Period 5) and Hafnium (5d, Period 6) have almost identical radii (~159 pm) due to lanthanide contraction. The 14 lanthanide elements intervene between Period 5 and Period 6, causing contraction that exactly offsets the expected increase in radius going from 4d to 5d. This makes Zr and Hf the most chemically similar pair in all of chemistry — extremely difficult to separate. Similarly, Nb/Ta and Mo/W pairs have similar properties.
Q10
Actinides show more variable oxidation states than lanthanides because:
Correct: C — Close energy 5f/6d/7s orbitals. In actinides, the 5f, 6d, and 7s orbitals have similar energies, so electrons from all three can participate in bond formation → wide range of oxidation states. Uranium (U) shows +3 to +6; Plutonium (Pu) shows +3 to +7. Lanthanides mostly show +3 (since 4f, 5d, 6s energy gap is larger, 4f electrons are less available for bonding). Most stable state for lanthanides is +3; for actinides +3 to +6 are common.
Q11
Catalytic activity of transition metals is attributed to:
Correct: D — Variable oxidation states + adsorption. Transition metals are excellent catalysts because: (1) Variable oxidation states allow them to accept and donate electrons easily (form intermediates and regenerate). (2) Their surfaces can adsorb reactant molecules, holding them in positions that favour reaction (heterogeneous catalysis). Examples: Fe in Haber process (N₂ + H₂ → NH₃), V₂O₅ in Contact process (SO₂ → SO₃), Ni in hydrogenation of oils. Pt and Pd catalyse many reactions.
Q12
Interstitial compounds of transition metals have which of the following properties?
Correct: A — Hard, high MP, conducting, inert. Interstitial compounds are formed when small atoms (H, C, N, B) are trapped in the interstices (voids) of the metallic crystal lattice. Examples: TiC (titanium carbide), Fe₄N (iron nitride), PdH₀.₇ (palladium hydride). Properties: (1) harder than pure metal (distorted lattice); (2) higher melting point; (3) retain metallic conductivity; (4) chemically inert; (5) non-stoichiometric (variable composition). They are NOT ionic (no electron transfer).
Q13
FeSO₄·7H₂O (green vitriol), CuSO₄·5H₂O (blue vitriol), and ZnSO₄·7H₂O (white vitriol) — the vivid colours of Fe and Cu sulphates are due to:
Correct: B — d-d transitions. Fe²⁺ (3d⁶, 4 unpaired e⁻) → green colour. Cu²⁺ (3d⁹, 1 unpaired e⁻) → blue colour. In both, d-orbitals are partially filled, and electrons can absorb visible light energy to jump between d-orbital levels split by the ligand field (here, water molecules). The absorbed colour determines the observed (complementary) colour. Anhydrous CuSO₄ is white because Cu²⁺ without water ligands has different d-splitting; ZnSO₄ is white because Zn²⁺ has d¹⁰ (no d-d transitions possible).
Q14
The highest melting point among all metals belongs to:
Correct: C — Tungsten (W). Tungsten (W, Z=74) has the highest melting point of all metals at 3422°C. This is due to very strong metallic bonding — W has 6 unpaired electrons available for metallic bonding (configuration [Xe] 4f¹⁴ 5d⁴ 6s²). High melting points are a characteristic of transition metals (strong metallic bonds due to participation of d-electrons). Exception: Mercury (Hg) is a liquid at room temperature (d¹⁰ 6s², no d-electron bonding contribution).
Q15
Which transition metal is liquid at room temperature?
Correct: D — Mercury (Hg). Mercury is the only metal that is liquid at room temperature (mp = −38.83°C). Configuration: [Xe] 4f¹⁴ 5d¹⁰ 6s². The d and s orbitals are completely filled → weak metallic bonding. Also, relativistic effects contract the 6s orbital, making it less available for bonding. This is why Hg has an anomalously low melting point. Note: Gallium (Ga) melts at 29.76°C (melts in your palm) but is not a transition metal.
Q16
The number of unpaired electrons in Mn²⁺ (Z=25) ion is:
Correct: A — 5 unpaired electrons. Mn: [Ar] 3d⁵ 4s². Mn²⁺ loses 2 electrons from 4s: [Ar] 3d⁵. A d⁵ configuration has all 5 d-orbitals singly occupied (by Hund's rule, one electron in each d-orbital, all with same spin). So 5 unpaired electrons. Magnetic moment μ = √(5×7) = √35 ≈ 5.92 BM. Mn²⁺ is very lightly coloured (nearly colourless/faint pink) because d⁵ d-d transitions are spin-forbidden (all spins parallel → transitions require spin flip).
Q17
Transition metals form alloys more readily with each other because:
Correct: B — Similar atomic radii. Transition metals across a period have gradually changing but similar atomic sizes. When atoms of similar size are mixed, they can substitute for one another in the metallic lattice forming a substitutional solid solution (alloy) without too much lattice distortion. This is Hume-Rothery's rule. Alloys of transition metals (steel = Fe+C, brass = Cu+Zn, bronze = Cu+Sn) are generally harder and stronger than pure metals.
Q18
In the first transition series, which element has the highest oxidation state?
Correct: C — Mn, +7. Across the first transition series, the maximum oxidation state increases from Sc (+3) to Mn (+7 in KMnO₄), then decreases towards the end. This is because early transition metals can use all their d and s electrons for bonding, but later ones have more d-electrons that are harder to remove. After Mn, the electrons pair up and the high oxidation states become less stable due to increased nuclear charge. Mn shows +2 to +7, maximum is +7 in permanganate.
Q19
When K₂Cr₂O₇ solution is made alkaline, it converts to:
Correct: D — Yellow chromate. The dichromate–chromate equilibrium: Cr₂O₇²⁻ (orange) + 2OH⁻ ⇌ 2CrO₄²⁻ (yellow) + H₂O. In acidic medium: 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O. So: orange in acidic/neutral, yellow in basic. This interconversion is used to test for these ions. Both Cr(VI) species are oxidising agents; dichromate is used more commonly in titrations (more stable in acid).
Q20
The most stable oxidation state for most lanthanides is:
Correct: B — +3. Most lanthanides show +3 as their most stable and common oxidation state. This is because removing 3 electrons (2 from 6s + 1 from 5d or 4f) gives a stable [Xe] core or half-filled/fully-filled 4f configurations. Some exceptions: Ce⁴⁺ (+4, stable — loses 4f¹ to get empty 4f⁰ = Xe configuration), Eu²⁺ (+2, stable — half-filled 4f⁷), Yb²⁺ (+2, stable — fully filled 4f¹⁴). All lanthanides form +3 ions; some also form +2 or +4.
🔗 Coordination Compounds 9 marks
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Q1
The IUPAC name of [Co(NH₃)₆]Cl₃ is:
Correct: C — Hexaamminecobalt(III) chloride. IUPAC rules: (1) Cation before anion (complex cation, Cl⁻ is counter ion). (2) Ligands named before metal: ammine (NH₃, neutral ligand). (3) Number of ligands: hexa (6). (4) Metal with oxidation state: Co³⁺ → cobalt(III). Co³⁺ charge: 3+ = 6×0 (NH₃) + Co, so Co is +3. (5) Chloride is the anion. Note: "ammine" has double 'm' to distinguish from organic "amine" groups.
Q2
The IUPAC name of K₄[Fe(CN)₆] is:
Correct: A — Potassium hexacyanoferrate(II). K₄[Fe(CN)₆]: 4K⁺ are counter ions (potassium). Complex ion [Fe(CN)₆]⁴⁻: 6 CN⁻ (cyano ligands), Fe charge = 4− − 6×(−1) = +2, so Fe(II) → ferrate(II) (complex is anionic → metal suffix is "-ate"). Yellow prussiate of potash, used in blueprinting (Turnbull's blue test). K₃[Fe(CN)₆] = potassium hexacyanoferrate(III) = red prussiate of potash (Fe³⁺ with 6 CN⁻).
Q3
The coordination number of Co in [Co(en)₂Cl₂]⁺ is:
Correct: B — 6. en (ethylenediamine) is a bidentate ligand — each en donates 2 pairs of electrons to the metal. So 2 en ligands donate 2×2 = 4 coordination bonds. Plus 2 Cl⁻ (monodentate) donate 2 bonds. Total coordination number = 4 + 2 = 6. The complex is octahedral. To find Co oxidation state: [x + 0(×2) + (−1)(×2)] = +1 → x = +3. Co is +3.
Q4
EDTA (ethylenediaminetetraacetic acid) is which type of ligand?
Correct: D — Hexadentate. EDTA structure has 2 nitrogen atoms and 4 carboxylate oxygen atoms, all capable of donating electron pairs to the metal. Total 6 donor atoms → hexadentate. EDTA can wrap around a metal completely, forming 5 chelate rings with one metal ion. It is used in medicine (lead poisoning treatment), water softening (binds Ca²⁺, Mg²⁺), and analytical chemistry (complexometric titrations). The chelate effect makes EDTA complexes very stable.
Q5
According to Werner's theory, "primary valence" of a metal in a complex is:
Correct: A — Ionisable valence. Werner proposed two types of valences: Primary valence = oxidation state of the metal, satisfied by ionisable anions (outside coordination sphere). Secondary valence = coordination number, satisfied by ligands inside the coordination sphere (inside square brackets). Example: [Co(NH₃)₆]Cl₃ — primary valence = 3 (three Cl⁻ outside satisfy Co³⁺); secondary valence = 6 (six NH₃ inside satisfy CN = 6). Secondary valence is always fixed and directed in space (explains geometry).
Q6
According to Crystal Field Theory (CFT), in an octahedral complex, d-orbitals split into:
Correct: C — t₂g lower, eg higher. In an octahedral field, 6 ligands approach along x, y, z axes. The dx²-y² and dz² orbitals (eg set) point directly at ligands → greater repulsion → higher energy. The dxy, dxz, dyz orbitals (t₂g set) point between axes → less repulsion → lower energy. Crystal field splitting Δₒ (or 10Dq). Strong field ligands give large Δₒ → low spin; weak field give small Δₒ → high spin.
Q7
The spectrochemical series of ligands is (in order of increasing field strength):
Correct: B — Weak to strong field order. The spectrochemical series ranks ligands by their ability to split d-orbitals (Δ value). Weak field ligands (halides, OH⁻) → small Δ → high spin, less pairing. Strong field ligands (CN⁻, CO) → large Δ → low spin, forced pairing. Memory tip: "I Bring Cold Fluids, Our Hot Nervous English Nation Considers Carbon Monoxide" — I < Br < Cl < F < OH < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO.
Q8
[NiCl₄]²⁻ is tetrahedral and paramagnetic. The hybridisation of Ni²⁺ in this complex is:
Correct: D — sp³. Ni²⁺: [Ar] 3d⁸. Cl⁻ is a weak-field ligand. Does not force pairing. 4 Cl⁻ ligands occupy one 4s + three 4p orbitals → sp³ hybridisation → tetrahedral geometry. 2 unpaired electrons remain in 3d → paramagnetic. Compare: [Ni(CN)₄]²⁻ — CN⁻ is strong field, forces pairing of 3d⁸ electrons, freeing one 3d orbital → dsp² hybridisation → square planar → diamagnetic. Same metal, different geometry depending on ligand field strength.
Q9
Linkage isomerism is exhibited by complexes containing which type of ligand?
Correct: A — Ambidentate ligands. Linkage isomers differ only in which donor atom of an ambidentate ligand is coordinated. NO₂⁻ bonded through N = nitro (−NO₂); through O = nitrito (−ONO). Example: [Co(NH₃)₅(NO₂)]²⁺ vs [Co(NH₃)₅(ONO)]²⁺. SCN⁻ bonded through S = thiocyanato (S-SCN); through N = isothiocyanato (N-NCS). These are not the same compound — they have different physical properties and reactivity.
Q10
Ionisation isomers differ in:
Correct: C — Different ionisation in solution. Ionisation isomers have the same overall composition but different ions inside and outside the coordination sphere. [Co(NH₃)₅Br]SO₄ gives Br⁻ inside (no precipitate with AgNO₃) and SO₄²⁻ outside (precipitate with BaCl₂). [Co(NH₃)₅SO₄]Br gives SO₄²⁻ inside and Br⁻ outside (precipitate with AgNO₃). They can be distinguished by testing their solutions with AgNO₃ and BaCl₂. Same molecular formula, different connectivity.
Q11
Geometrical (cis-trans) isomerism is shown by the square planar complex [Pt(NH₃)₂Cl₂]. Cisplatin (the anticancer drug) is:
Correct: B — cis isomer. In cisplatin, both Cl⁻ ligands are on the same side (cis) of the square planar Pt²⁺ complex. Cisplatin (cis-diamminedichloroplatinum II) is used in cancer chemotherapy. It works by cross-linking with guanine bases in DNA, preventing replication. The trans isomer (transplatin) is pharmacologically inactive. Geometrical isomerism requires: (1) rigid geometry, (2) at least two different kinds of ligands. Square planar [MA₂B₂] and octahedral [MA₄B₂] type complexes show cis-trans isomerism.
Q12
Optical isomerism is exhibited by [Co(en)₃]³⁺. The two optical isomers are designated as:
Correct: D — Δ and Λ. [Co(en)₃]³⁺ is a tris-chelate octahedral complex with a propeller-like (helical) structure. The two non-superimposable mirror image forms are called: Δ (delta, right-handed propeller) and Λ (lambda, left-handed). They rotate plane-polarised light in opposite directions. Optical isomers: same molecular formula, same connectivity, but non-superimposable mirror images (chiral molecules). No plane/centre of symmetry. d-form and l-form are detected by polarimetry.
Q13
The Effective Atomic Number (EAN) of [Ni(CO)₄] equals 36. This corresponds to the configuration of:
Correct: C — Krypton (Kr, Z=36). EAN = atomic number of metal − electrons lost + electrons donated by ligands. Ni (Z=28) in [Ni(CO)₄] is Ni⁰ (CO is neutral, no ionisation). Each CO donates 2 electrons. EAN = 28 − 0 + 4×2 = 28 + 8 = 36 = Kr. This satisfies the EAN rule (complexes tend to achieve noble gas configuration). [Fe(CO)₅]: EAN = 26 + 5×2 = 36 = Kr. [Cr(CO)₆]: EAN = 24 + 6×2 = 36 = Kr.
Q14
Chelate effect: complexes with polydentate ligands are more stable than those with equivalent monodentate ligands. The primary thermodynamic reason is:
Correct: A — Entropy (ΔS > 0). [M(NH₃)₂]²⁺ + en → [M(en)]²⁺ + 2NH₃. One bidentate ligand replaces two monodentate ligands. Number of free particles increases (2 NH₃ released): ΔS > 0. Since ΔG = ΔH − TΔS, a positive ΔS lowers ΔG (makes it more negative) → more stable complex. The enthalpy changes are similar, so entropy is the key driver. This entropy-driven stability is the "chelate effect." Larger chelate rings are less stable (entropy advantage decreases).
Q15
The complex [Fe(CN)₆]³⁻ has 1 unpaired electron (low spin), while [Fe(H₂O)₆]³⁺ has 5 unpaired electrons (high spin). This difference is because:
Correct: B — Large Δₒ for CN⁻ forces pairing. Fe³⁺ is d⁵. CN⁻ (strong field): Δₒ > pairing energy → electrons pair up in t₂g before filling eg. Configuration: t₂g⁵eg⁰ → 1 unpaired electron (low spin). H₂O (weak field): Δₒ < pairing energy → Hund's rule followed → t₂g³eg² → 5 unpaired electrons (high spin). Both are octahedral (coordination number 6). Low spin: inner orbital, d²sp³. High spin: outer orbital, sp³d².