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NEET Chemistry: Inorganic Chemistry · Chapter 4

Coordination Compounds

What to remember

  • A coordination entity is a central metal atom or ion bonded to a fixed number of ligands. The coordination number is the number of ligand donor atoms attached to the metal. Werner's primary valence is the ionisable oxidation state and the secondary valence is the coordination number.
  • Naming: ligands first (alphabetical order), then the metal with its oxidation state in Roman numerals. In an anionic complex the metal name ends in –ate. Neutral and cationic complexes keep the metal name.
  • Valence bond theory links geometry, hybridisation and magnetism: d²sp³ (inner orbital) and sp³d² (outer orbital) octahedral, dsp² square planar, sp³ tetrahedral.
  • Crystal field theory: octahedral splitting gives t₂g (–0.4 Δo each) and eg (+0.6 Δo each); Δt = 4/9 Δo. Strong field ligands (CN⁻, CO) give low spin; weak field ligands (I⁻, Cl⁻, F⁻) give high spin.

Werner's theory

Werner proposed that a metal has two kinds of valence.

  • Primary valence: ionisable, satisfied by negative ions, and equal to the oxidation state of the metal. It is non-directional.
  • Secondary valence: non-ionisable, satisfied by ligands (negative ions or neutral molecules), and equal to the coordination number. It is directional, so it fixes the geometry (six gives octahedral, four gives tetrahedral or square planar).

Evidence from CoCl₃·xNH₃ compounds (excess AgNO₃, and conductivity):

FormulaColourAgCl precipitated per moleIons per formula unitModern formula
CoCl₃·6NH₃Yellow34[Co(NH₃)₆]Cl₃
CoCl₃·5NH₃Purple23[Co(NH₃)₅Cl]Cl₂
CoCl₃·4NH₃Green12[Co(NH₃)₄Cl₂]Cl
CoCl₃·3NH₃—00 (non-conductor)[Co(NH₃)₃Cl₃]

Chloride inside the coordination sphere does not precipitate with Ag⁺. In every case the coordination number of Co(III) is 6.

Definitions

  • Coordination entity: central atom or ion plus ligands, written in square brackets, for example [Fe(CN)₆]⁴⁻.
  • Central atom/ion: the acceptor of electron pairs (Lewis acid).
  • Ligand: an ion or molecule that donates an electron pair to the central atom.
  • Coordination sphere: the central atom and ligands inside the square brackets. Counter ions are outside.
  • Coordination polyhedron: the spatial arrangement of the ligands (octahedral, square planar, tetrahedral).
  • Oxidation number: the charge the central atom would have if all ligands were removed with their electron pairs. In [Co(NH₃)₅Cl]²⁺ it is +3; in [Fe(CN)₆]⁴⁻ it is +2; in [Ni(CO)₄] it is 0.
  • Homoleptic: only one kind of ligand, [Co(NH₃)₆]³⁺. Heteroleptic: more than one kind, [Co(NH₃)₄Cl₂]⁺.
Type of ligandDonor atomsExamples
Unidentate1NH₃, H₂O, Cl⁻, CN⁻, CO
Didentate2en (ethane-1,2-diamine), C₂O₄²⁻ (oxalate), glycinate
PolydentateMore than 2EDTA⁴⁻ (hexadentate: 2 N and 4 O)
Ambidentate1 (either of two atoms)NO₂⁻ (N or O), SCN⁻ (S or N)

A didentate or polydentate ligand that forms a ring with the metal is a chelate ligand. Chelates are more stable than complexes of similar unidentate ligands (the chelate effect). Coordination number is the count of donor atoms: [Fe(C₂O₄)₃]³⁻ has 3 × 2 = 6, [Co(en)₃]³⁺ has 6, [PtCl₄]²⁻ has 4.

Nomenclature and formula writing

  • 1. The cation is named first, then the anion.
  • 2. Inside the entity, ligands are named first in alphabetical order (prefixes di, tri are ignored), then the metal.
  • 3. Anionic ligands end in –o: chlorido, cyanido, oxalato, hydroxido, sulphato, nitrito-N (NO₂⁻ bonded through N is "nitrito-N", through O is "nitrito-O"). Neutral ligands keep their names, except aqua (H₂O), ammine (NH₃), carbonyl (CO) and nitrosyl (NO).
  • 4. Use di, tri, tetra for simple ligands; bis, tris, tetrakis with brackets for complex ligands such as ethane-1,2-diamine.
  • 5. The oxidation state of the metal follows in Roman numerals in brackets, with no space.
  • 6. If the complex ion is an anion, the metal name ends in –ate (ferrate, cuprate, argentate, aurate); for Fe, Cu, Ag, Au, Pb and Sn the Latin stem is used.
  • 7. In a formula, the metal symbol comes first, then anionic ligands and neutral ligands in alphabetical order of the first symbol.
FormulaName
[Co(NH₃)₆]Cl₃Hexaamminecobalt(III) chloride
[Co(NH₃)₅Cl]Cl₂Pentaamminechloridocobalt(III) chloride
K₃[Fe(CN)₆]Potassium hexacyanidoferrate(III)
K₄[Fe(CN)₆]Potassium hexacyanidoferrate(II)
[Cr(en)₃]Cl₃Tris(ethane-1,2-diamine)chromium(III) chloride
[Ni(CO)₄]Tetracarbonylnickel(0)
[Pt(NH₃)₂Cl₂]Diamminedichloridoplatinum(II)
[Cu(NH₃)₄]SO₄Tetraamminecopper(II) sulphate

Isomerism

Compounds with the same formula but different arrangements.

Structural isomerism

  • Linkage: an ambidentate ligand binds through different atoms. [Co(NH₃)₅(NO₂)]Cl₂ (nitrito-N, yellow) and [Co(NH₃)₅(ONO)]Cl₂ (nitrito-O, red).
  • Coordination: exchange of ligands between cation and anion of the same salt. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].
  • Ionisation: exchange of an ion between the inside and outside of the sphere. [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄ give different ions in water (the first gives a precipitate with BaCl₂, the second with AgNO₃).
  • Solvate (hydrate): water is inside the sphere in one isomer and a lattice molecule in another. [Cr(H₂O)₆]Cl₃ (violet), [Cr(H₂O)₅Cl]Cl₂·H₂O (grey-green), [Cr(H₂O)₄Cl₂]Cl·2H₂O (dark green).

Stereoisomerism

  • Geometrical (cis–trans): found in square planar [MA₂B₂] such as [Pt(NH₃)₂Cl₂] and in octahedral [MA₄B₂] such as [Co(NH₃)₄Cl₂]⁺. Tetrahedral complexes show none. In octahedral [MA₃B₃], such as [Co(NH₃)₃Cl₃], the facial (fac) isomer has three identical ligands on one triangular face, and the meridional (mer) isomer has them in a plane through the metal. [MA₆] and [MA₅B] have only one form.
  • Optical: a non-superimposable mirror image pair (d and l, which rotate plane polarised light in opposite directions). [Co(en)₃]³⁺ exists as two optical isomers. In [CoCl₂(en)₂]⁺, the cis form is optically active (two enantiomers) while the trans form is not, so there are three stereoisomers in all.

Worked example 4 (Werner): 0.01 mol of CoCl₃·5NH₃ is [Co(NH₃)₅Cl]Cl₂. Only the two chloride ions outside the sphere react with excess AgNO₃, so the precipitate is 0.02 mol AgCl (about 2.87 g, using 143.5 g mol⁻¹). The solution has 3 ions per formula unit, which agrees with its conductivity.

Worked example 5 (oxidation number): in K₃[Fe(CN)₆] the complex ion has charge –3 and each CN⁻ is –1, so x + 6(–1) = –3 and x = +3. In [Co(NH₃)₄Cl₂]⁺, NH₃ is neutral and Cl⁻ is –1, so x + 0 – 2 = +1 and x = +3.

Bonding: valence bond theory

The metal provides empty orbitals (hybridised) that accept lone pairs from ligands. The geometry follows the hybridisation. Magnetic behaviour follows the unpaired electrons left in the d orbitals.

ComplexMetal ionHybridisationGeometryUnpaired electronsMagnetism
[Co(NH₃)₆]³⁺Co³⁺ d⁶d²sp³ (inner)Octahedral0Diamagnetic
[CoF₆]³⁻Co³⁺ d⁶sp³d² (outer)Octahedral4Paramagnetic, 4.90 BM
[Ni(CN)₄]²⁻Ni²⁺ d⁸dsp²Square planar0Diamagnetic
[NiCl₄]²⁻Ni²⁺ d⁸sp³Tetrahedral2Paramagnetic, 2.83 BM
[Fe(CN)₆]³⁻Fe³⁺ d⁵d²sp³Octahedral1Paramagnetic, 1.73 BM
[Fe(CN)₆]⁴⁻Fe²⁺ d⁶d²sp³Octahedral0Diamagnetic

Inner orbital complexes use (n–1)d orbitals (d²sp³), after strong field ligands force the electrons to pair. Outer orbital complexes use nd orbitals (sp³d²) and keep the electrons unpaired. [Ni(CO)₄] is sp³, tetrahedral and diamagnetic (Ni is 3d¹⁰ in the zero state).

Limitations of VBT: it gives no quantitative explanation of magnetic data, does not explain the colour of complexes, does not give the relative strength of ligands (no spectrochemical series), and does not explain thermodynamic or kinetic stability in a quantitative way.

Crystal field theory

Ligands are treated as point negative charges. The five degenerate d orbitals split in energy.

Octahedral field. The orbitals dxy, dyz and dzx (t₂g) are lower, by 0.4 Δo each. The orbitals dz² and dx²–y² (eg) are higher, by 0.6 Δo each. Δo is the crystal field splitting energy.

Tetrahedral field. The splitting is inverted: e (two orbitals) is lower and t₂ (three orbitals) is higher. Δt = 4/9 Δo, which is always smaller than the pairing energy P, so tetrahedral complexes are high spin.

High and low spin (d⁴ to d⁷, octahedral). If Δo < P (weak field), electrons occupy eg before pairing: high spin. If Δo > P (strong field), electrons pair in t₂g first: low spin.

Spectrochemical series (increasing field strength): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO.

CFSE (octahedral) = (–0.4 × number in t₂g + 0.6 × number in eg) Δo, with pairing energy ignored.

ConfigurationCFSE (units of Δo)
d¹–0.4
d³–1.2
d⁶ low spin (t₂g⁶)–2.4
d⁶ high spin (t₂g⁴ eg²)–0.4
d⁸ (t₂g⁶ eg²)–1.2
d⁹ (t₂g⁶ eg³)–0.6

Worked example 1: Fe²⁺ (d⁶) in [Fe(CN)₆]⁴⁻ is low spin, t₂g⁶ eg⁰, so CFSE = 6 × (–0.4) = –2.4 Δo, with 0 unpaired electrons. Worked example 2: Co²⁺ (d⁷) in a weak field is t₂g⁵ eg²: CFSE = 5(–0.4) + 2(0.6) = –0.8 Δo, with 3 unpaired electrons and μ = √15 = 3.87 BM. Worked example 3: if Δo = 18000 cm⁻¹, then Δt = 4/9 × 18000 = 8000 cm⁻¹.

Colour. A complex absorbs light of one colour for a d–d transition and appears in the complementary colour. A larger Δo means absorption at a shorter wavelength. Changing the ligand changes Δo and so the colour: [Cu(H₂O)₄]²⁺ is light blue and [Cu(NH₃)₄]²⁺ is deep blue. Complexes with d⁰ or d¹⁰ ions (such as Zn²⁺ complexes) are colourless.

Magnetic moment from structure. Count the unpaired electrons from the d configuration and the field strength, then use μ = √(n(n+2)) BM. For [Fe(CN)₆]³⁻ (Fe³⁺ d⁵, strong field, t₂g⁵) n = 1 and μ = √3 = 1.73 BM. For [FeF₆]³⁻ (weak field, t₂g³ eg²) n = 5 and μ = √35 = 5.92 BM. For [CoF₆]³⁻ (Co³⁺ d⁶, weak field, t₂g⁴ eg²) n = 4 and μ = √24 = 4.90 BM. The same ion can be high spin or low spin depending only on the ligand.

Why colour changes with the ligand. Water is a weaker field ligand than ammonia, so Δo of [Cu(NH₃)₄]²⁺ is larger than that of [Cu(H₂O)₄]²⁺. The complex therefore absorbs light of shorter wavelength and appears a different, deeper blue. A complex of a d⁰ ion has no d electron to excite, so it is colourless.

Stability. The more strongly a ligand binds, the larger the stability constant of the complex. Chelating ligands such as en and EDTA give rings and more stable complexes than monodentate ligands.

Importance and applications

  • Qualitative analysis: Cu²⁺ with excess NH₃ gives deep blue [Cu(NH₃)₄]²⁺; Fe³⁺ with SCN⁻ gives blood-red [Fe(SCN)]²⁺; Ni²⁺ with dimethylglyoxime gives a red precipitate.
  • Extraction of metals: silver and gold dissolve as cyanido complexes, [Ag(CN)₂]⁻ and [Au(CN)₂]⁻, and are recovered by zinc. Pure nickel is made by the Mond process through volatile [Ni(CO)₄].
  • Biological systems: chlorophyll is a complex of Mg; haemoglobin is a complex of Fe; vitamin B₁₂ (cyanocobalamin) is a complex of Co.
  • EDTA: used to estimate the hardness of water (Ca²⁺ and Mg²⁺ form EDTA complexes) and to treat lead poisoning.
  • Medicine: cis-platin, cis-[Pt(NH₃)₂Cl₂], is an anticancer drug.
  • Photography: AgBr is dissolved in sodium thiosulphate ("hypo") as [Ag(S₂O₃)₂]³⁻.

NEET traps

  • 1. Cl⁻ inside the coordination sphere does not precipitate AgCl; only the counter ion does.
  • 2. Primary valence is the oxidation state, secondary valence is the coordination number.
  • 3. Ligands are listed alphabetically (ammine before chlorido) and not by charge; prefixes di and tri are ignored.
  • 4. In anionic complexes the metal ends in –ate with a Latin stem for Fe, Cu, Ag, Au (ferrate, cuprate).
  • 5. The coordination number of [Co(en)₃]³⁺ is 6, not 3.
  • 6. Tetrahedral complexes show no geometrical isomerism, and Δt = 4/9 Δo.
  • 7. Weak field ligand does not mean weak bonding: it means small Δo and high spin. The order is I⁻ < Cl⁻ < H₂O < NH₃ < en < CN⁻ < CO.
  • 8. d²sp³ is inner orbital (low spin, uses (n–1)d); sp³d² is outer orbital (uses nd).
  • 9. [Ni(CN)₄]²⁻ is square planar and diamagnetic, but [NiCl₄]²⁻ is tetrahedral and paramagnetic.
  • 10. trans-[CoCl₂(en)₂]⁺ is optically inactive; cis is active.

One-liners

  • 1. Werner studied the series CoCl₃·xNH₃, with x = 6, 5, 4 and 3.
  • 2. [Co(NH₃)₅Cl]Cl₂ gives 2 mol AgCl per mole with excess AgNO₃.
  • 3. EDTA is a hexadentate ligand.
  • 4. NO₂⁻ and SCN⁻ are ambidentate ligands.
  • 5. Linkage isomerism needs an ambidentate ligand.
  • 6. [Co(NH₃)₆]³⁺ is d²sp³, octahedral and diamagnetic.
  • 7. [CoF₆]³⁻ is sp³d², octahedral and has four unpaired electrons.
  • 8. Ni²⁺ with CN⁻ gives square planar dsp²; with Cl⁻ gives tetrahedral sp³.
  • 9. t₂g lies 0.4 Δo below the average energy; eg lies 0.6 Δo above it.
  • 10. Δt = 4/9 Δo.
  • 11. CO and CN⁻ are among the strongest field ligands; I⁻ is among the weakest.
  • 12. High Δo means absorption of shorter wavelengths.
  • 13. Chlorophyll has Mg, haemoglobin has Fe, and vitamin B₁₂ has Co.
  • 14. Cis-platin is used against cancer.
  • 15. EDTA titration measures hardness due to Ca²⁺ and Mg²⁺.

Practice questions

  1. Assertion (A): The colours of [Cu(H₂O)₄]²⁺ and [Cu(NH₃)₄]²⁺ are different. Reason (R): Different ligands cause different crystal field splitting energies.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Δo changes with the ligand, so the wavelength of light absorbed changes.

  2. 20 mL of hard water needs 10 mL of 0.01 M EDTA to react completely with its Ca²⁺ (1:1 complex). What is the molarity of Ca²⁺ in the water?

    1. 0.01 M
    2. 0.05 M
    3. 0.0025 M
    4. 0.005 M
    Answer

    D. 0.005 M

    EDTA = 10 × 0.01 = 0.1 mmol = Ca²⁺ in 20 mL; 0.1 mmol/20 mL = 0.005 M.

  3. Assertion (A): [Co(NH₃)₅Cl]Cl₂ gives 2 mol of AgCl with excess AgNO₃ per mole of complex. Reason (R): Only the two chloride ions outside the coordination sphere are ionisable.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The Cl⁻ bound to Co is part of the coordination sphere and does not precipitate.

  4. Given below are two statements: Statement I: The trans isomer of [Pt(NH₃)₂Cl₂] is the anticancer drug cis-platin. Statement II: cis-Platin is used in the treatment of cancer. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The anticancer drug is the cis isomer.

  5. Given below are two statements: Statement I: Weak field ligands give high-spin octahedral complexes when Δo < P. Statement II: In the spectrochemical series I⁻ is a weaker field ligand than Cl⁻. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Small Δo lets electrons occupy eg before pairing; the series runs I⁻ < Br⁻ < Cl⁻.

  6. Which of the following is the strongest field ligand?

    1. Cl⁻
    2. F⁻
    3. CN⁻
    4. H₂O
    Answer

    C. CN⁻

    In the spectrochemical series CN⁻ lies far above H₂O, F⁻ and Cl⁻.

  7. Assertion (A): [Co(NH₃)₅(NO₂)]Cl₂ shows linkage isomerism. Reason (R): NO₂⁻ is an ambidentate ligand.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    NO₂⁻ can bind through N or O, giving nitrito-N and nitrito-O isomers.

  8. What is the oxidation number of platinum in K₂[PtCl₄]?

    1. +3
    2. +2
    3. 0
    4. +4
    Answer

    B. +2

    2(+1) + x + 4(–1) = 0, so x = +2.

  9. Given below are two statements: Statement I: In a tetrahedral crystal field Δt = 4/9 Δo. Statement II: Tetrahedral complexes are generally low spin. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Δt is smaller than the pairing energy, so tetrahedral complexes are high spin.

  10. When 0.5 mol of CoCl₃·5NH₃ (formula [Co(NH₃)₅Cl]Cl₂) is treated with excess AgNO₃ solution, how many moles of AgCl are precipitated?

    1. 1.5 mol
    2. 1.0 mol
    3. 3.0 mol
    4. 0.5 mol
    Answer

    B. 1.0 mol

    Only the 2 Cl⁻ outside the sphere precipitate: 0.5 × 2 = 1.0 mol.

  11. Match the complex in List I with the number of unpaired electrons in List II. List I: a. [Fe(CN)₆]⁴⁻ b. [Fe(CN)₆]³⁻ c. [FeF₆]³⁻ d. [NiCl₄]²⁻. List II: i. 0 ii. 1 iii. 5 iv. 2.

    1. a-i, b-ii, c-iv, d-iii
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    B. a-i, b-ii, c-iii, d-iv

    Fe²⁺ low spin t₂g⁶ has 0; Fe³⁺ low spin t₂g⁵ has 1; Fe³⁺ high spin has 5; Ni²⁺ tetrahedral has 2.

  12. What is the IUPAC name of K₃[Fe(CN)₆]?

    1. Potassium hexacyanidoferrate(II)
    2. Potassium hexacyanoiron(III)
    3. Potassium hexacyanidoferrate(III)
    4. Tripotassium hexacyanidoferrate(III)
    Answer

    C. Potassium hexacyanidoferrate(III)

    The complex ion is an anion, so iron becomes ferrate; 3 + x – 6 = 0 for the salt gives Fe +3.

  13. Given below are two statements: Statement I: EDTA is a hexadentate ligand. Statement II: Chelate complexes are more stable than complexes of comparable unidentate ligands. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    EDTA binds through 2 N and 4 O atoms; ring formation (chelate effect) adds stability.

  14. Which of the following complexes shows geometrical isomerism?

    1. [Fe(CN)₆]⁴⁻
    2. [Co(NH₃)₅Cl]²⁺
    3. [Ni(CO)₄]
    4. [Pt(NH₃)₂Cl₂] (square planar)
    Answer

    D. [Pt(NH₃)₂Cl₂] (square planar)

    Square planar MA₂B₂ has cis and trans forms; the others have only one arrangement of ligands.

  15. What is the hybridisation of nickel in the diamagnetic complex [Ni(CN)₄]²⁻?

    1. d²sp³
    2. sp³
    3. sp³d²
    4. dsp²
    Answer

    D. dsp²

    Ni²⁺ is 3d⁸; the strong field CN⁻ pairs the electrons, leaving one 3d orbital empty for dsp² (square planar).

  16. [Zn(NH₃)₄]²⁺ is colourless but [Cu(NH₃)₄]²⁺ is deep blue. What is the reason?

    1. Zn²⁺ is a larger ion than Cu²⁺
    2. Zn²⁺ is d¹⁰ and has no d–d transition, but Cu²⁺ is d⁹
    3. NH₃ is a weak field ligand for Zn²⁺ only
    4. Zn²⁺ has five unpaired electrons
    Answer

    B. Zn²⁺ is d¹⁰ and has no d–d transition, but Cu²⁺ is d⁹

    A d–d transition needs a vacant d level; the filled d¹⁰ shell of Zn²⁺ has none.

  17. Assertion (A): [Ni(CN)₄]²⁻ is diamagnetic. Reason (R): The CN⁻ ion carries a negative charge.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    It is diamagnetic because the strong field CN⁻ pairs the d electrons of Ni²⁺ (dsp²); the negative charge is not the reason.

  18. Assertion (A): Tetrahedral complexes are high spin. Reason (R): Δt is larger than the pairing energy in tetrahedral complexes.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Δt = 4/9 Δo is smaller than the pairing energy, which is why they are high spin.

  19. What is the spin-only magnetic moment of the tetrahedral complex [NiCl₄]²⁻?

    1. 1.73 BM
    2. 3.87 BM
    3. 0 BM
    4. 2.83 BM
    Answer

    D. 2.83 BM

    Ni²⁺ (3d⁸) with Cl⁻ is sp³ with 2 unpaired electrons: √(2×4) = 2.83 BM.

  20. 0.01 mol of a chromium chloride hydrate, CrCl₃·6H₂O, gives 0.02 mol AgCl with excess AgNO₃. What is its formula?

    1. [Cr(H₂O)₃Cl₃]·3H₂O
    2. [Cr(H₂O)₄Cl₂]Cl·2H₂O
    3. [Cr(H₂O)₅Cl]Cl₂·H₂O
    4. [Cr(H₂O)₆]Cl₃
    Answer

    C. [Cr(H₂O)₅Cl]Cl₂·H₂O

    Two Cl⁻ per formula unit are ionisable (0.02/0.01), so one Cl is inside the sphere with five H₂O (CN = 6) and one H₂O is outside.

  21. Which metal is present in vitamin B₁₂ (cyanocobalamin)?

    1. Cobalt
    2. Zinc
    3. Iron
    4. Magnesium
    Answer

    A. Cobalt

    Vitamin B₁₂ is a cobalt complex; chlorophyll has Mg and haemoglobin has Fe.

  22. What is the coordination number of platinum in [Pt(en)Cl₂]? (en = ethane-1,2-diamine)

    1. 2
    2. 3
    3. 6
    4. 4
    Answer

    D. 4

    en is didentate (2 donor atoms) and each Cl⁻ is unidentate: 2 + 2 = 4.

  23. Given below are two statements: Statement I: [Ni(CN)₄]²⁻ is tetrahedral. Statement II: [Ni(CN)₄]²⁻ is paramagnetic. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    It is square planar (dsp²) and diamagnetic, whereas [NiCl₄]²⁻ is tetrahedral and paramagnetic.

  24. How many unpaired electrons are there in [CoF₆]³⁻?

    1. 4
    2. 0
    3. 2
    4. 5
    Answer

    A. 4

    F⁻ is a weak field ligand, so Co³⁺ (3d⁶) stays high spin: t₂g⁴ eg² gives 4 unpaired electrons (outer orbital, sp³d²).

  25. Which of the following octahedral complexes has the greatest number of stereoisomers?

    1. [Co(en)₃]³⁺
    2. [Co(NH₃)₅Cl]²⁺
    3. [Co(en)₂Cl₂]⁺
    4. [Co(NH₃)₃Cl₃]
    Answer

    C. [Co(en)₂Cl₂]⁺

    [Co(en)₂Cl₂]⁺ has trans plus the cis pair of enantiomers = 3; [Co(en)₃]³⁺ has 2, [Co(NH₃)₃Cl₃] has 2 (fac, mer), [Co(NH₃)₅Cl]²⁺ has 1.

  26. Given below are two statements: Statement I: Tetrahedral complexes show cis–trans isomerism. Statement II: [CoF₆]³⁻ is a low-spin complex. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Tetrahedral complexes have no geometrical isomers; F⁻ is a weak field ligand so [CoF₆]³⁻ is high spin.

  27. Match the pair of isomers in List I with the type of isomerism in List II. List I: a. [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂ b. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] c. [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br d. [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O. List II: i. Coordination ii. Linkage iii. Solvate (hydrate) iv. Ionisation.

    1. a-i, b-ii, c-iv, d-iii
    2. a-iv, b-i, c-ii, d-iii
    3. a-ii, b-i, c-iii, d-iv
    4. a-ii, b-i, c-iv, d-iii
    Answer

    D. a-ii, b-i, c-iv, d-iii

    NO₂ versus ONO is linkage; exchange of metal-ligand sets is coordination; SO₄ and Br swapping is ionisation; H₂O versus Cl swapping is solvate.

  28. Given below are two statements: Statement I: [Co(NH₃)₆]³⁺ is diamagnetic. Statement II: [Co(NH₃)₆]³⁺ is an inner orbital complex with d²sp³ hybridisation. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Co³⁺ (3d⁶) pairs all electrons with strong field NH₃, using two 3d orbitals.

  29. Match List I with List II. List I: a. [Co(NH₃)₆]³⁺ b. [CoF₆]³⁻ c. [Ni(CN)₄]²⁻ d. [NiCl₄]²⁻. List II: i. sp³ ii. dsp² iii. d²sp³ iv. sp³d².

    1. a-iv, b-iii, c-ii, d-i
    2. a-ii, b-iv, c-iii, d-i
    3. a-iii, b-iv, c-ii, d-i
    4. a-iii, b-iv, c-i, d-ii
    Answer

    C. a-iii, b-iv, c-ii, d-i

    Low-spin Co³⁺ d²sp³; high-spin Co³⁺ sp³d²; Ni²⁺ with CN⁻ dsp²; Ni²⁺ with Cl⁻ sp³.

  30. In Werner's theory, the primary valence of the central metal is equal to its

    1. number of counter ions outside the sphere
    2. number of ligand donor atoms
    3. coordination number
    4. oxidation state
    Answer

    D. oxidation state

    Primary valence is ionisable and equals the oxidation state; secondary valence equals the coordination number.

  31. A high-spin octahedral complex of Co²⁺ (d⁷) has what CFSE (pairing energy ignored)?

    1. –1.2 Δo
    2. –0.8 Δo
    3. –0.4 Δo
    4. –1.8 Δo
    Answer

    B. –0.8 Δo

    t₂g⁵ eg²: 5(–0.4) + 2(0.6) = –2.0 + 1.2 = –0.8 Δo.

  32. Given below are two statements: Statement I: In names of coordination compounds the ligands are cited in alphabetical order. Statement II: The name of an anionic complex ends with –ide. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Anionic complexes end with –ate (for example ferrate); ligands are alphabetical.

  33. For a metal ion with Δo = 22500 cm⁻¹ in an octahedral complex, what is Δt for the same ligand in a tetrahedral complex?

    1. 12500 cm⁻¹
    2. 33750 cm⁻¹
    3. 5000 cm⁻¹
    4. 10000 cm⁻¹
    Answer

    D. 10000 cm⁻¹

    Δt = 4/9 Δo = 4/9 × 22500 = 10000 cm⁻¹.

  34. Match the formula in List I with its IUPAC name in List II. List I: a. [Co(NH₃)₆]Cl₃ b. K₄[Fe(CN)₆] c. [Ni(CO)₄] d. [Pt(NH₃)₂Cl₂]. List II: i. Tetracarbonylnickel(0) ii. Diamminedichloridoplatinum(II) iii. Potassium hexacyanidoferrate(II) iv. Hexaamminecobalt(III) chloride.

    1. a-iii, b-iv, c-i, d-ii
    2. a-iv, b-iii, c-ii, d-i
    3. a-iv, b-iii, c-i, d-ii
    4. a-i, b-iii, c-iv, d-ii
    Answer

    C. a-iv, b-iii, c-i, d-ii

    Match each formula with its oxidation state and ligand names: Co(III), Fe(II) (4 + x – 6 = 0 gives 2), Ni(0), Pt(II).

  35. Which is the correct order of increasing ligand field strength?

    1. I⁻ < H₂O < Cl⁻ < NH₃ < CN⁻
    2. I⁻ < Cl⁻ < H₂O < NH₃ < CN⁻
    3. Cl⁻ < I⁻ < H₂O < NH₃ < CN⁻
    4. I⁻ < Cl⁻ < NH₃ < H₂O < CN⁻
    Answer

    B. I⁻ < Cl⁻ < H₂O < NH₃ < CN⁻

    Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO.

  36. Given below are two statements: Statement I: [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are coordination isomers. Statement II: The two isomers give different precipitates: BaCl₂ precipitates the sulphate of one and AgNO₃ precipitates the bromide of the other. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    They are ionisation isomers because SO₄²⁻ and Br⁻ exchange places between inside and outside the sphere; Statement II is true.

  37. Which of the following statements are correct? 1. [Co(en)₃]³⁺ shows optical isomerism. 2. cis-[CoCl₂(en)₂]⁺ is optically active. 3. trans-[CoCl₂(en)₂]⁺ is optically active.

    1. 1, 2 and 3
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1 and 3 only
    Answer

    B. 1 and 2 only

    [Co(en)₃]³⁺ and the cis form lack a mirror plane (they have enantiomers); the trans form has a mirror plane and is optically inactive.

  38. An octahedral Fe²⁺ complex has a magnetic moment of 4.90 BM. What is its CFSE (pairing energy ignored)?

    1. –0.4 Δo
    2. –2.4 Δo
    3. –1.2 Δo
    4. –0.8 Δo
    Answer

    A. –0.4 Δo

    4.90 BM means 4 unpaired electrons, so d⁶ high spin: t₂g⁴ eg² gives –1.6 + 1.2 = –0.4 Δo.

  39. Match the compound in List I with the metal it contains in List II. List I: a. Chlorophyll b. Haemoglobin c. Vitamin B₁₂ d. cis-Platin. List II: i. Co ii. Pt iii. Mg iv. Fe.

    1. a-iii, b-iv, c-i, d-ii
    2. a-iv, b-iii, c-i, d-ii
    3. a-iii, b-iv, c-ii, d-i
    4. a-i, b-iv, c-iii, d-ii
    Answer

    A. a-iii, b-iv, c-i, d-ii

    Chlorophyll Mg, haemoglobin Fe, vitamin B₁₂ Co, cis-platin Pt.

  40. Match the octahedral configuration in List I with its CFSE in List II. List I: a. d³ b. d⁶ (low spin) c. d⁶ (high spin) d. d⁹. List II: i. –0.4 Δo ii. –0.6 Δo iii. –1.2 Δo iv. –2.4 Δo.

    1. a-iii, b-iv, c-ii, d-i
    2. a-i, b-iv, c-iii, d-ii
    3. a-iii, b-iv, c-i, d-ii
    4. a-iv, b-iii, c-i, d-ii
    Answer

    C. a-iii, b-iv, c-i, d-ii

    d³ t₂g³ = –1.2; d⁶ low t₂g⁶ = –2.4; d⁶ high t₂g⁴eg² = –1.6 + 1.2 = –0.4; d⁹ t₂g⁶eg³ = –2.4 + 1.8 = –0.6 Δo.

  41. A cobalt(III) complex CoCl₃·4NH₃ gives 1 mol AgCl per mole and conducts as a 1:1 electrolyte. What is its formula and the number of ions it gives in solution?

    1. [Co(NH₃)₄Cl₂]Cl, 3 ions
    2. [Co(NH₃)₄Cl]Cl₂, 3 ions
    3. [Co(NH₃)₃Cl₃], 0 ions
    4. [Co(NH₃)₄Cl₂]Cl, 2 ions
    Answer

    D. [Co(NH₃)₄Cl₂]Cl, 2 ions

    One ionisable Cl⁻ and coordination number 6 (4 NH₃ + 2 Cl⁻) gives [Co(NH₃)₄Cl₂]⁺ and Cl⁻: 2 ions.

  42. Which of the following is an ambidentate ligand?

    1. C₂O₄²⁻
    2. en
    3. NO₂⁻
    4. NH₃
    Answer

    C. NO₂⁻

    NO₂⁻ can bind through N (nitrito-N) or O (nitrito-O); en and oxalate are didentate, NH₃ is unidentate.

  43. Assertion (A): [Fe(CN)₆]³⁻ is a high-spin complex. Reason (R): CN⁻ is a strong field ligand.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The strong field CN⁻ forces pairing, so [Fe(CN)₆]³⁻ is low spin with one unpaired electron.

  44. What is the crystal field stabilisation energy of a low-spin d⁶ octahedral complex (pairing energy ignored)?

    1. –0.4 Δo
    2. –2.4 Δo
    3. –1.2 Δo
    4. –1.6 Δo
    Answer

    B. –2.4 Δo

    t₂g⁶ eg⁰: 6 × (–0.4) = –2.4 Δo.

  45. How many donor atoms does the ligand EDTA⁴⁻ have?

    1. 6
    2. 2
    3. 4
    4. 8
    Answer

    A. 6

    EDTA is hexadentate: two nitrogen atoms and four oxygen atoms of carboxylate groups.

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