Coordination Compounds Class 12: Theory, Naming, Isomerism and Board Answers

What Coordination Compounds is about

Weightage: This chapter is Unit V, Coordination Compounds, which carries 7 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit V Coordination Compounds

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Werner's theory

NCERT section 5.1 Werner's Theory of Coordination Compounds

Alfred Werner studied cobalt(III) chloride ammines and found that they gave different numbers of AgCl moles per mole with excess AgNO₃ at room temperature:

• CoCl₃·6NH₃ (yellow): 3 AgCl, so it is $[Co(N{H}_3{)}_6]C{l}_3$
• CoCl₃·5NH₃ (purple): 2 AgCl, so it is $[CoCl(N{H}_3{)}_5]C{l}_2$
• CoCl₃·4NH₃ (green and violet forms): 1 AgCl, so it is $[CoC{l}_2(N{H}_3{)}_4]Cl$

Conductivity measurements in solution supported the same number of ions.

Postulates:
1. Metals show two types of linkages (valences): primary and secondary.
2. Primary valences are ionisable and are satisfied by negative ions; they correspond to the oxidation state.
3. Secondary valences are non-ionisable and are satisfied by neutral molecules or negative ions. The secondary valence equals the coordination number and is fixed for a metal.
4. The ions or groups bound by secondary linkages have characteristic spatial arrangements (coordination polyhedra): octahedral, tetrahedral or square planar.

The species in square brackets is the coordination entity; the ions outside are counter ions.

Double salts versus complexes: a double salt such as carnallite (KCl·MgCl₂·6H₂O) or Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O) dissociates completely into simple ions in water. A complex ion such as $[Fe(CN{)}_6{]}^{4-}$ of K₄[Fe(CN)₆] does not break into Fe²⁺ and CN⁻.

Worked example

0.1 mol of a complex with empirical formula CoCl₃·5NH₃ is treated with excess AgNO₃ solution, and 0.2 mol of AgCl is precipitated. (a) Write the structural formula of the complex. (b) How many ions will it give in solution? (c) What mass of AgCl is formed? (AgCl = 143.5 g mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. 0.2 mol AgCl from 0.1 mol complex means 2 Cl⁻ per formula are ionisable (primary valences); the third Cl is inside the coordination sphere.
2. (a) $[CoCl(N{H}_3{)}_5]C{l}_2$, with coordination number 6 (5 NH₃ + 1 Cl). (1)
3. (b) $[CoCl(N{H}_3{)}_5]C{l}_2\to [CoCl(N{H}_3{)}_5{]}^{2+}+2C{l}^{-}$: three ions. (1)
4. (c) Mass = 0.2 × 143.5 = 28.7 g. (1)

Answer: (a) [CoCl(NH₃)₅]Cl₂ (b) 3 ions (c) 28.7 g of AgCl

Key terms: ligands, coordination number and oxidation number

NCERT section 5.2 Definitions of Some Important Terms Pertaining to Coordination Compounds

• Central atom or ion: the atom or ion to which a fixed number of ligands are bound, for example Ni²⁺ in $[NiC{l}_2(H_{2}O{)}_4]$. It acts as a Lewis acid.
• Ligands: ions or molecules bound to the central atom. A ligand bonding through one donor atom is unidentate (Cl⁻, H₂O, NH₃); through two, didentate (ethane-1,2-diamine, en; oxalate, C₂O₄²⁻); through several, polydentate (EDTA⁴⁻ is hexadentate).
• Chelate: when a di- or polydentate ligand uses two or more donor atoms to bind one metal ion, forming a ring. Chelate complexes tend to be more stable than similar complexes with unidentate ligands.
• Ambidentate ligand: can bind through either of two different atoms, such as NO₂⁻ (through N, nitrito-N, or through O, nitrito-O) and SCN⁻ (through S, thiocyanato-S, or through N, thiocyanato-N).
• Coordination number (CN): the number of ligand donor atoms directly bonded to the metal. In $[PtC{l}_6{]}^{2-}$ it is 6; in $[Co(en{)}_3{]}^{3+}$ it is also 6 (three didentate ligands).
• Coordination sphere: the central atom and its ligands, written in square brackets.
• Coordination polyhedron: the spatial arrangement of the ligand atoms: octahedral, square planar or tetrahedral.
• Oxidation number: the charge the central atom would carry if all ligands were removed with their electron pairs, written in Roman numerals, e.g. Cu(I) in $[Cu(CN{)}_4{]}^{3-}$.
• Homoleptic complexes have one kind of ligand ($[Co(N{H}_3{)}_6{]}^{3+}$); heteroleptic have more than one ($[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$).

Worked example

For the complex K₃[Fe(C₂O₄)₃], give (a) the oxidation number of iron, (b) the coordination number, (c) the denticity of the ligand and whether the complex is homoleptic or heteroleptic. (2 marks)

Try it first, then show the solutionSolution

1. (a) Let the oxidation number of Fe be x: 3(+1) + x + 3(−2) = 0, so x = +3. (½)
2. (b) Oxalate is didentate; three of them give 3 × 2 = 6 donor atoms, so CN = 6. (½)
3. (c) Oxalate is a didentate (chelating) ligand. Only one kind of ligand is present, so the complex is homoleptic. (1)

Answer: (a) +3 (b) 6 (c) didentate oxalate; homoleptic

Writing formulas and IUPAC names

NCERT section 5.3 Nomenclature of Coordination Compounds

Formulas: the central atom is written first, then ligands in alphabetical order; polydentate ligands are also listed alphabetically, abbreviated ones in brackets. The whole entity goes in square brackets, with the charge outside if it is an ion, and the cation is written before the anion.

Names: the cation is named first, then the anion, whether positive or negative.
1. Name ligands in alphabetical order before the metal. Numerical prefixes (di, tri) are ignored when alphabetising.
2. Anionic ligands end in -o: chlorido (Cl⁻), cyanido (CN⁻), hydroxido (OH⁻), oxalato (C₂O₄²⁻), nitrito-N (NO₂⁻ through N). Neutral ligands keep their names except aqua (H₂O), ammine (NH₃), carbonyl (CO) and nitrosyl (NO).
3. Prefixes di-, tri-, tetra- give the number of simple ligands; bis-, tris-, tetrakis- are used when the ligand name already has a prefix, with the name in brackets, e.g. bis(ethane-1,2-diamine).
4. Give the oxidation state of the metal in Roman numerals in brackets.
5. If the complex is an anion, the metal name ends in -ate: cobaltate, nickelate. Some use Latin names: ferrate (Fe), cuprate (Cu), argentate (Ag), aurate (Au), plumbate (Pb), stannate (Sn).
6. A neutral complex is named like a cation.

Examples: $[Co(N{H}_3{)}_6]C{l}_3$ is hexaamminecobalt(III) chloride; $K_3[Fe(CN{)}_6]$ is potassium hexacyanidoferrate(III); $[Ni(CO{)}_4]$ is tetracarbonylnickel(0).

Worked example

Write the IUPAC names of (a) [CoCl(NH₃)₅]Cl₂ (b) Na₂[Ni(CN)₄] (c) [CoCl₂(en)₂]Cl. (3 marks)

Try it first, then show the solutionSolution

1. (a) Ligands alphabetically: ammine before chlorido; Co is +3: pentaamminechloridocobalt(III) chloride. (1)
2. (b) Complex anion, so nickel becomes nickelate; 2(+1) + x + 4(−1) = 0 gives Ni = +2: sodium tetracyanidonickelate(II). (1)
3. (c) en needs 'bis'; chlorido comes before ethane-1,2-diamine; Co is +3: dichloridobis(ethane-1,2-diamine)cobalt(III) chloride. (1)

Answer: (a) Pentaamminechloridocobalt(III) chloride (b) Sodium tetracyanidonickelate(II) (c) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride

Stereoisomerism: geometrical and optical

NCERT section 5.4.1 Geometric Isomerism; 5.4.2 Optical Isomerism

Stereoisomers have the same formula and bonds but a different arrangement in space.

Geometrical isomerism happens in heteroleptic complexes because ligands can occupy different relative positions.

• Square planar $[M{X}_2{L}_2]$, such as $[Pt(N{H}_3{)}_{2}C{l}_2]$: cis (like groups next to each other) and trans (opposite).
• Octahedral $[M{X}_2{L}_4]$, such as $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$: cis and trans.
• Octahedral $[M{X}_2(L-L{)}_2]$, such as $[CoC{l}_2(en{)}_2{]}^{+}$: cis and trans.
• Octahedral $[M{a}_3{b}_3]$, such as $[Co(N{H}_3{)}_3(N{O}_2{)}_3]$: fac (three identical ligands on one face) and mer (around a meridian).
• Tetrahedral complexes do not show geometrical isomerism, because all four positions are adjacent to one another.

Optical isomerism: optical isomers (enantiomers) are mirror images that cannot be superimposed. They rotate plane-polarised light in opposite directions: dextro (d) to the right and laevo (l) to the left. It is common in octahedral complexes with didentate ligands:

• $[Co(en{)}_3{]}^{3+}$ shows d and l forms.
• In $[PtC{l}_2(en{)}_2{]}^{2+}$ and $[CoC{l}_2(en{)}_2{]}^{+}$, only the cis isomer is optically active; the trans isomer has a plane of symmetry.

Worked example

How many stereoisomers does [CoCl₂(en)₂]⁺ have? Name them and say which is optically active. (2 marks)

Try it first, then show the solutionSolution

1. Geometrical: the two Cl ligands can be cis (90°) or trans (180°). (1)
2. The cis form has no plane of symmetry, so it exists as d and l forms; the trans form is optically inactive. Total = 2 (cis d and l) + 1 (trans) = 3. (1)

Answer: Three: trans, and the d and l forms of the cis isomer; only cis is optically active.

Structural isomerism

NCERT section 5.4.3 Linkage Isomerism; 5.4.4 Coordination Isomerism; 5.4.5 Ionisation Isomerism; 5.4.6 Solvate Isomerism

Structural isomers have different bonds.

• Linkage isomerism: an ambidentate ligand binds through different atoms. $[Co(N{H}_3{)}_5(N{O}_2)]C{l}_2$ (nitrito-N, yellow) and $[Co(N{H}_3{)}_5(ONO)]C{l}_2$ (nitrito-O, red).
• Coordination isomerism: ligands are interchanged between a cationic and an anionic complex: $[Co(N{H}_3{)}_6][Cr(CN{)}_6]$ and $[Cr(N{H}_3{)}_6][Co(CN{)}_6]$.
• Ionisation isomerism: a counter ion is itself a potential ligand and swaps with a ligand, so the compounds give different ions in solution: $[Co(N{H}_3{)}_5(S{O}_4)]Br$ and $[Co(N{H}_3{)}_{5}Br]S{O}_4$. The first gives a cream precipitate with AgNO₃; the second gives a white precipitate with BaCl₂.
• Solvate isomerism (hydrate isomerism when the solvent is water): water is a ligand in one isomer and free solvent in the other. $[Cr(H_{2}O{)}_6]C{l}_3$ (violet) and $[CrCl(H_{2}O{)}_5]C{l}_2\cdot H_{2}O$ (grey-green).

Worked example

Name the type of isomerism shown by each pair and give a chemical test for the third pair: (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)₅(SO₄)]Br and [Co(NH₃)₅Br]SO₄ (3 marks)

Try it first, then show the solutionSolution

1. (a) Linkage isomerism: NO₂⁻ binds through N in one and through O in the other. (1)
2. (b) Coordination isomerism: NH₃ and CN⁻ are exchanged between the two metal ions. (1)
3. (c) Ionisation isomerism. Test: [Co(NH₃)₅(SO₄)]Br gives a pale yellow/cream precipitate of AgBr with AgNO₃, while [Co(NH₃)₅Br]SO₄ gives a white precipitate of BaSO₄ with BaCl₂. (1)

Answer: (a) Linkage (b) Coordination (c) Ionisation; tell them apart with AgNO₃ (Br⁻ free) and BaCl₂ (SO₄²⁻ free).

Valence bond theory: hybridisation, shape and magnetism

NCERT section 5.5.1 Valence Bond Theory; 5.5.2 Magnetic Properties of Coordination Compounds; 5.5.3 Limitations of Valence Bond Theory

In valence bond theory the metal ion uses hybridised orbitals (s, p and d) to accept lone pairs from the ligands. The hybridisation fixes the shape:

• CN 4: sp³ tetrahedral; dsp² square planar
• CN 6: d²sp³ octahedral (inner d orbitals, (n−1)d); sp³d² octahedral (outer d orbitals, nd)

Which one forms depends on whether the ligands force pairing of d electrons.

• $[Co(N{H}_3{)}_6{]}^{3+}$: Co³⁺ is 3d⁶; the six electrons pair up in three 3d orbitals, leaving two 3d orbitals for d²sp³. No unpaired electrons: diamagnetic, an inner orbital or low spin complex.
• $[Co{F}_6{]}^{3-}$: the electrons do not pair; outer 4d orbitals are used, sp³d². Four unpaired electrons: paramagnetic, an outer orbital or high spin complex.
• $[NiC{l}_4{]}^{2-}$: Ni²⁺ is 3d⁸ with two unpaired electrons; sp³, tetrahedral, paramagnetic.
• $[Ni(CN{)}_4{]}^{2-}$: CN⁻ forces pairing; one 3d orbital is free: dsp², square planar, diamagnetic.
• $[Ni(CO{)}_4]$: Ni(0) is 3d⁸4s²; the 4s electrons move into 3d, giving sp³, tetrahedral, diamagnetic.

Magnetic moments are found with $\mu =\sqrt{n(n+2)}$ BM.

Limitations of VBT: it involves many assumptions; gives no quantitative interpretation of magnetic data; does not explain colour; gives no quantitative explanation of thermodynamic or kinetic stability; cannot predict whether a 4-coordinate complex is tetrahedral or square planar; and does not distinguish between weak and strong ligands.

Worked example

Using valence bond theory, predict the hybridisation, shape and magnetic behaviour of [Ni(CN)₄]²⁻ and [NiCl₄]²⁻. (Z of Ni = 28) (3 marks)

Try it first, then show the solutionSolution

1. Ni²⁺: [Ar]3d⁸, with two unpaired electrons in the free ion. (½)
2. [Ni(CN)₄]²⁻: CN⁻ is a strong ligand and pairs the 3d electrons, freeing one 3d orbital: dsp² hybridisation, square planar, no unpaired electron, diamagnetic. (1½)
3. [NiCl₄]²⁻: Cl⁻ is a weak ligand; no pairing, so 4s and 4p are used: sp³, tetrahedral, two unpaired electrons, paramagnetic (μ = 2.83 BM). (1)

Answer: [Ni(CN)₄]²⁻: dsp², square planar, diamagnetic. [NiCl₄]²⁻: sp³, tetrahedral, paramagnetic.

Crystal field theory, spectrochemical series and colour

NCERT section 5.5.4 Crystal Field Theory; 5.5.5 Colour in Coordination Compounds; 5.5.6 Limitations of Crystal Field Theory

Crystal field theory treats the metal–ligand bond as purely electrostatic: ligands are point charges (anions) or point dipoles (neutral molecules). In a free ion the five d orbitals are degenerate; in the electric field of the ligands they split.

Octahedral field: the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point at the ligands and are raised in energy (the e_g set); $d_{xy}$, $d_{yz}$ and $d_{xz}$ point between the ligands and are lowered (the t₂g set). The energy gap is the crystal field splitting, ${\Delta }_o$.

Filling d⁴ and beyond: the fourth electron can enter e_g (no pairing) or pair in t₂g. This depends on ${\Delta }_o$ versus the pairing energy P:

• ${\Delta }_o<P$ (weak field ligands): $t_{2g}^3{e}_g^1$, high spin.
• ${\Delta }_o>P$ (strong field ligands): $t_{2g}^4{e}_g^0$, low spin.

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

Tetrahedral field: the splitting is inverted (e below t₂) and smaller, ${\Delta }_t=\frac{4}{9}{\Delta }_o$, so pairing is not favoured and low spin tetrahedral complexes are rarely seen.

Colour: an electron absorbs light and jumps from t₂g to e_g (a d–d transition); the complex shows the complementary colour. $[Ti(H_{2}O{)}_6{]}^{3+}$ ($t_{2g}^1$) absorbs blue-green light (498 nm) and looks violet. Without ligands there is no splitting, so anhydrous CuSO₄ is white while hydrated CuSO₄·5H₂O is blue. Changing ligands changes ${\Delta }_o$ and so the colour: $[Ni(H_{2}O{)}_6{]}^{2+}$ is green, while adding en gives blue or purple complexes.

Limitations of CFT: ligands are treated as point charges, so anionic ligands should split most, yet they lie at the low end of the series; it ignores the covalent character of bonding.

Worked example

Write the t₂g/e_g electronic configuration of the central metal ion and the number of unpaired electrons in (a) [CoF₆]³⁻ and (b) [Co(NH₃)₆]³⁺. Which is high spin? (Z of Co = 27) (3 marks)

Try it first, then show the solutionSolution

1. Co³⁺ is 3d⁶. (½)
2. (a) F⁻ is a weak field ligand (${\Delta }_o<P$): $t_{2g}^4{e}_g^2$, four unpaired electrons, high spin. (1)
3. (b) NH₃ is a stronger field ligand (${\Delta }_o>P$): $t_{2g}^6{e}_g^0$, no unpaired electrons, low spin. (1)
4. So [CoF₆]³⁻ is the high spin complex (paramagnetic, μ = 4.90 BM). (½)

Answer: [CoF₆]³⁻: t₂g⁴e_g², 4 unpaired, high spin. [Co(NH₃)₆]³⁺: t₂g⁶e_g⁰, 0 unpaired, low spin.

Bonding in metal carbonyls

NCERT section 5.6 Bonding in Metal Carbonyls

Homoleptic carbonyls (only CO ligands) are formed by most transition metals and have simple structures:

• $[Ni(CO{)}_4]$ tetracarbonylnickel(0): tetrahedral
• $[Fe(CO{)}_5]$ pentacarbonyliron(0): trigonal bipyramidal
• $[Cr(CO{)}_6]$ hexacarbonylchromium(0): octahedral
• $[M{n}_2(CO{)}_{10}]$: two square pyramidal Mn(CO)₅ units joined by a Mn–Mn bond
• $[C{o}_2(CO{)}_8]$: a Co–Co bond bridged by two CO groups

Synergic bonding: the metal–carbon bond has both σ and π character.
1. σ bond: the lone pair on the carbon of CO is donated into a vacant orbital of the metal.
2. π back bond: a filled d orbital of the metal donates electrons into the vacant antibonding π* orbital of CO.
Each effect strengthens the other (synergic effect), so the metal–ligand bond is strong. Back bonding puts electrons into the antibonding orbital of CO, which weakens the C–O bond slightly.

Worked example

Explain the synergic bonding in metal carbonyls. Why is the metal–carbon bond in carbonyls stronger than expected from σ donation alone? (2 marks)

Try it first, then show the solutionSolution

1. CO donates its carbon lone pair to an empty metal orbital (σ bond), and the metal donates electrons from a filled d orbital into the empty π* orbital of CO (π back bond). (1)
2. The two effects reinforce each other: σ donation makes the metal more electron-rich and able to back-donate, and back donation reduces the build-up of negative charge on the metal, allowing more σ donation. So the M–C bond has σ and π character and is strong. (1)

Answer: σ donation from CO plus π back donation from metal d orbitals into CO π*; the mutual reinforcement (synergic effect) strengthens the M–C bond.

Importance and applications of coordination compounds

NCERT section 5.7 Importance and Applications of Coordination Compounds

• Analysis: complexes give colour tests and precipitates; hardness of water is estimated by titration with Na₂EDTA, because Ca²⁺ and Mg²⁺ form stable EDTA complexes. Selective estimation is possible because the stability of complexes differs.
• Metallurgy: silver and gold are extracted by forming soluble cyanide complexes, $[Ag(CN{)}_2{]}^{-}$ and $[Au(CN{)}_2{]}^{-}$; nickel can be purified through $[Ni(CO{)}_4]$, which decomposes to give pure nickel.
• Biology: chlorophyll (photosynthesis) is a magnesium complex; haemoglobin (oxygen carrier in blood) is an iron complex; vitamin B₁₂ (cyanocobalamin) is a cobalt complex. Many enzymes are also coordination compounds.
• Catalysis: Wilkinson's catalyst, $[(P{h}_{3}P{)}_{3}RhCl]$, is used for hydrogenation of alkenes.
• Electroplating and photography: articles are plated with silver and gold from $[Ag(CN{)}_2{]}^{-}$ and $[Au(CN{)}_2{]}^{-}$, which give a smoother coating than simple ions. In black and white photography, undecomposed AgBr is dissolved by hypo as the complex $[Ag(S_2{O}_3{)}_2{]}^{3-}$.
• Medicine: chelating agents remove toxic metals: EDTA for lead poisoning, D-penicillamine and desferrioxime B for excess copper and iron. Cisplatin (cis-$[PtC{l}_2(N{H}_3{)}_2]$) and related compounds inhibit the growth of tumours.

Worked example

Name the coordination compound or complex used in each case: (a) estimation of hardness of water, (b) treatment of lead poisoning, (c) inhibiting the growth of tumours, (d) the oxygen carrier in blood. (2 marks)

Try it first, then show the solutionSolution

1. (a) Na₂EDTA, which forms stable complexes with Ca²⁺ and Mg²⁺. (½)
2. (b) EDTA (a chelating agent). (½)
3. (c) Cisplatin, cis-[PtCl₂(NH₃)₂]. (½)
4. (d) Haemoglobin, a complex of iron. (½)

Answer: (a) Na₂EDTA (b) EDTA (c) cisplatin (d) haemoglobin (iron complex)

Chemistry in real life: Coordination Compounds

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• IUPAC Nomenclature of Coordination Compounds
• Isomerism in Coordination Compounds
• Crystal Field Theory

Coordination Compounds formula sheet

Oxidation number of the central atom

$\text{charge of entity}=x+\sum (\text{ligand charges})$

Example: in $[Fe(CN{)}_6{]}^{3-}$, x + 6(−1) = −3, so x = +3.

Coordination number

$\text{CN}=\sum (\text{number of ligands}\times \text{denticity})$

$[Co(en{)}_3{]}^{3+}$: 3 × 2 = 6; $[Ni(CO{)}_4]$: 4.

Spin-only magnetic moment

$\mu =\sqrt{n(n+2)}\text{BM}$

n = number of unpaired electrons.

Octahedral splitting

$e_g(d_{x^2-y^2},d_{z^2})\text{up};t_{2g}(d_{xy},d_{yz},d_{xz})\text{down};\text{gap}={\Delta }_o$

e_g points at the ligands; t₂g points between them.

High spin versus low spin (d⁴ example)

${\Delta }_o<P:t_{2g}^3{e}_g^1;{\Delta }_o>P:t_{2g}^4{e}_g^0$

Weak field ligands give high spin; strong field ligands give low spin.

Tetrahedral splitting

${\Delta }_t=\frac{4}{9}{\Delta }_o$

Too small to force pairing, so tetrahedral complexes are almost always high spin.

Spectrochemical series

$I^{-}<B{r}^{-}<SC{N}^{-}<C{l}^{-}<S^{2-}<F^{-}<O{H}^{-}<C_2{O}_4^{2-}<H_{2}O<NC{S}^{-}<edt{a}^{4-}<N{H}_3<en<C{N}^{-}<CO$

Field strength increases from left to right.

Hybridisation and shape

$s{p}^3:\text{tetrahedral};ds{p}^2:\text{square planar};d^{2}s{p}^3(\text{inner}),s{p}^3{d}^2(\text{outer}):\text{octahedral}$

Inner orbital complexes are low spin; outer orbital complexes are high spin.

Werner's AgCl test

$[CoCl(N{H}_3{)}_5]C{l}_2+2AgN{O}_3\to [CoCl(N{H}_3{)}_5](N{O}_3{)}_2+2AgCl$

Only chloride outside the coordination sphere is precipitated.

Isomer counts to remember

$[M{X}_2{L}_2](\text{sq. planar}):2;[M{X}_2{L}_4]:2;[M{a}_3{b}_3]:2(fac,mer);[M{X}_2(AA{)}_2]:3;[M(AA{)}_3]:2$

The 3 for [MX₂(AA)₂] is trans plus d and l cis.

Coordination Compounds: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Coordination Compounds

The question patterns to expect, and the method that works for each.

Tricks to remember Coordination Compounds

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Ligands alphabetically, metal last, anion '-ate'

  Name = cation first; inside the complex: ligands A to Z (ignore di/tri), then metal, then (oxidation state). If the whole complex is negative, add -ate.

  Use it when: Every IUPAC naming question.

• Spectrochemical series as a sentence

  'I Bought Some Cool Shoes For Old Ox, Water Not Easy, Need Energetic Cyanide Company' = I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO.

  Use it when: Deciding weak or strong field.

• Inner = Pairs = Low spin

  Inner orbital (d²sp³) complexes have paired electrons and are low spin; outer orbital (sp³d²) complexes are high spin.

  Use it when: Linking VBT and CFT answers.

• Tetrahedral never cis

  In a tetrahedron every corner touches every other, so there is no cis/trans. Square planar and octahedral can have it.

  Use it when: Assertion-reason on geometrical isomerism.

• Back-bonding is the 'give and take' bond

  CO gives σ to the metal; the metal gives back π into CO's π*. Give and take makes the bond stronger.

  Use it when: Synergic bonding questions.

Coordination Compounds: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Coordination Compounds

The slips examiners see most often, and what to write instead.

• ✗Mistake: Wrong order of ligands in names

  Wrong working

  Chloridopentaamminecobalt(III) chloride

  Right working

  Pentaamminechloridocobalt(III) chloride (a for ammine comes before c for chlorido; 'penta' is ignored)

  ✓Do this: Alphabetise by the ligand name, not the prefix.

  Usually costs: The whole naming mark (½ to 1).

• ✗Mistake: Forgetting '-ate' for complex anions

  Wrong working

  Potassium hexacyanidoiron(III)

  Right working

  Potassium hexacyanidoferrate(III)

  ✓Do this: If the complex carries a negative charge, the metal name ends in -ate, using Latin roots for Fe, Cu, Ag, Au, Pb, Sn.

  Usually costs: ½ to 1 mark.

• ✗Mistake: Using the oxidation state of the free atom for the d count

  Wrong working

  [Co(NH₃)₆]³⁺: Co is 3d⁷4s², so it is paramagnetic

  Right working

  Co³⁺ is 3d⁶; with NH₃ the six electrons pair (t₂g⁶), so it is diamagnetic

  ✓Do this: Always find the oxidation state, then write the ion's d configuration.

  Usually costs: Hybridisation and magnetism marks.

• ✗Mistake: Saying trans-[CoCl₂(en)₂]⁺ is optically active

  ✓Do this: The trans form has a mirror plane through the metal and both en ligands; only the cis form is chiral.

  Usually costs: 1 mark in an isomer-count question.

• ✗Mistake: Counting ligands instead of donor atoms for coordination number

  Wrong working

  [Co(en)₃]³⁺ has CN = 3

  Right working

  [Co(en)₃]³⁺ has CN = 6, because each en binds through two N atoms

  ✓Do this: Multiply each ligand by its denticity.

  Usually costs: ½ to 1 mark.

NCERT exercise guide: Coordination Compounds

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Coordination Compounds: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 29 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Coordination Compounds

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Coordination Compounds question, by year

Practice questions on Coordination Compounds

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of the following is an ambidentate ligand?
   (a) NH₃ (b) en (c) SCN⁻ (d) C₂O₄²⁻

   Hint

   It can bind through two different atoms.

   Answer

   (c) SCN⁻, which can bind through S or N.

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Tetrahedral complexes do not show geometrical isomerism.
   Reason (R): In a tetrahedron, all four positions are adjacent to one another.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Can two ligands ever be 'opposite' in a tetrahedron?

   Answer

   (a)

3. Q3 · 2 marks · Very short answerBoard level

   What is meant by (a) a homoleptic complex and (b) a chelate? Give one example of each. (2 marks)

   Hint

   One kind of ligand; a ring.

   Answer

   (a) A complex with only one kind of ligand, e.g. [Co(NH₃)₆]³⁺ (1). (b) A ring complex formed when a di- or polydentate ligand binds one metal ion through two or more donor atoms, e.g. [Co(en)₃]³⁺ or [Ca(EDTA)]²⁻ (1).

4. Q4 · 3 marks · Short answerBoard level

   A complex has the formula CoCl₃·4NH₃. One mole of it gives one mole of AgCl with excess AgNO₃. (a) Write its structural formula. (b) Draw (in words) its two geometrical isomers. (c) How many ions does it give in solution? (3 marks)

   Hint

   Only one Cl is outside the sphere.

   Answer

   (a) [CoCl₂(NH₃)₄]Cl (1). (b) Cis: two Cl adjacent on the octahedron; trans: two Cl opposite (1). (c) Two ions: [CoCl₂(NH₃)₄]⁺ and Cl⁻ (1).

5. Q5 · 3 marks · Short answerHarder board level

   Write the formulas of (a) tetraamminediaquacobalt(III) chloride, (b) potassium trioxalatochromate(III), (c) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate. (3 marks)

   Hint

   Balance the charge of the complex with counter ions.

   Answer

   (a) [Co(NH₃)₄(H₂O)₂]Cl₃ (1) (b) K₃[Cr(C₂O₄)₃] (1) (c) [PtCl₂(en)₂](NO₃)₂ (1).

6. Q6 · 3 marks · Short answerHarder board level

   Give the hybridisation, shape and magnetic behaviour of [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺. (Z of Fe = 26) (3 marks)

   Hint

   Fe²⁺ is 3d⁶ in both; compare ligand strengths.

   Answer

   [Fe(CN)₆]⁴⁻: CN⁻ pairs the 3d⁶ electrons (t₂g⁶), d²sp³, octahedral, diamagnetic (1½). [Fe(H₂O)₆]²⁺: H₂O is weak, no pairing (t₂g⁴e_g²), sp³d², octahedral, 4 unpaired electrons, paramagnetic, μ = 4.90 BM (1½).

7. Q7 · 5 marks · Long answerHarder board level

   (a) Draw the octahedral crystal field splitting diagram. (b) Write the configuration of a d⁶ ion with a weak field and a strong field ligand and give the number of unpaired electrons. (c) Why are low spin tetrahedral complexes rarely observed? (d) Name the ligand at the strong end of the spectrochemical series. (5 marks)

   Hint

   Δt = (4/9)Δo.

   Answer

   (a) Five degenerate levels splitting into lower t₂g (three) and upper e_g (two), gap Δo (1½). (b) Weak field: t₂g⁴e_g², 4 unpaired; strong field: t₂g⁶e_g⁰, 0 unpaired (1½). (c) Δt = (4/9)Δo is small and usually less than the pairing energy, so electrons do not pair (1). (d) CO (1).

8. Q8 · 2 marks · Very short answerHarder board level

   Write the type of isomerism shown by [Cr(H₂O)₆]Cl₃ and [CrCl(H₂O)₅]Cl₂·H₂O. How would the number of moles of AgCl precipitated per mole differ? (2 marks)

   Hint

   Count chloride ions outside the brackets.

   Answer

   Solvate (hydrate) isomerism (1). [Cr(H₂O)₆]Cl₃ gives 3 mol AgCl; [CrCl(H₂O)₅]Cl₂·H₂O gives 2 mol AgCl (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose a chemist finds that a solution of [Ni(H₂O)₆]²⁺ is green, but turns violet when ethane-1,2-diamine is added in excess. Explain the colour change using crystal field theory and predict the effect on the magnetic moment. (3 marks)

   Hint

   en is higher than H₂O in the spectrochemical series; Ni²⁺ is d⁸.

   Answer

   en replaces H₂O to form [Ni(en)₃]²⁺; en is a stronger field ligand, so Δo increases (1). The complex now absorbs light of higher energy (shorter wavelength), so a different complementary colour is seen (1). Ni²⁺ is d⁸ (t₂g⁶e_g²) in both, with 2 unpaired electrons, so the spin-only moment stays 2.83 BM (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    Explain why the C–O bond in a metal carbonyl is slightly longer than in free CO, while the metal–carbon bond is strong. (2 marks)

    Hint

    Where do the back-donated electrons go?

    Answer

    π back donation from metal d orbitals fills the antibonding π* orbital of CO, which weakens and lengthens the C–O bond (1). The same back donation adds π character to the M–C bond, and σ donation plus π back donation reinforce each other (synergic), so the M–C bond is strong (1).

11. Q11 · 2 marks · Very short answerBoard level

    Give two limitations each of valence bond theory and crystal field theory. (2 marks)

    Hint

    Think colour, magnetism and the nature of bonding.

    Answer

    VBT: does not explain colour; cannot distinguish weak and strong ligands or predict tetrahedral versus square planar (any two) (1). CFT: treats ligands as point charges, so anionic ligands should split most but are at the weak end of the series; ignores covalent character of the bond (1).

12. Q12 · 3 marks · Short answerHOTS / competency

    Suppose you are given three square planar platinum(II) complexes, [PtCl₂(NH₃)₂], [Pt(NH₃)₄]Cl₂ and K₂[PtCl₄]. (a) Which shows geometrical isomerism? (b) Arrange them in increasing order of molar conductivity in water, with reason. (c) Name K₂[PtCl₄]. (3 marks)

    Hint

    Count the ions each gives in solution.

    Answer

    (a) [PtCl₂(NH₃)₂] (MX₂L₂: cis and trans) (1). (b) [PtCl₂(NH₃)₂] (0 ions, neutral) < [Pt(NH₃)₄]Cl₂ ≈ K₂[PtCl₄] (3 ions each); conductivity rises with the number of ions (1). (c) Potassium tetrachloridoplatinate(II) (1).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the difference between primary and secondary valence in Werner's theory.Show answer

   Primary valence is ionisable, satisfied by negative ions and equals the oxidation state; secondary valence is non-ionisable, satisfied by neutral molecules or anions and equals the coordination number.

2. 2.What is an ambidentate ligand? Give two examples.Show answer

   A ligand that can bind through either of two different donor atoms: NO₂⁻ (N or O) and SCN⁻ (S or N).

3. 3.What is a chelate, and why are chelates more stable?Show answer

   A ring complex formed when a di- or polydentate ligand binds one metal ion through two or more donor atoms; ring formation makes them more stable than similar unidentate complexes.

4. 4.Name [Ni(CO)₄] and give its shape.Show answer

   Tetracarbonylnickel(0); tetrahedral (sp³).

5. 5.Why don't tetrahedral complexes show geometrical isomerism?Show answer

   All four positions are adjacent to each other, so the relative positions of ligands cannot differ.

6. 6.What is the hybridisation and magnetic nature of [Co(NH₃)₆]³⁺?Show answer

   d²sp³; diamagnetic (inner orbital, low spin).

7. 7.What is Δt in terms of Δo?Show answer

   Δt = (4/9) Δo.

8. 8.Why is [Ti(H₂O)₆]³⁺ violet?Show answer

   Its t₂g¹ electron absorbs blue-green light (about 498 nm) to jump to e_g; the complementary colour, violet, is seen.

9. 9.What is meant by synergic bonding in carbonyls?Show answer

   σ donation from CO to the metal plus π back donation from metal d orbitals to CO π*, each strengthening the other.

10. 10.Name Wilkinson's catalyst and its use.Show answer

    [(Ph₃P)₃RhCl]; hydrogenation of alkenes.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you predict the moles of AgCl precipitated by any cobalt ammine chloride?
• ☐Can you find the oxidation number and coordination number in K₃[Fe(C₂O₄)₃] and [Co(en)₂Cl₂]⁺?
• ☐Can you name [Pt(NH₃)₂Cl(NO₂)] and write the formula of potassium tetracyanidonickelate(II)?
• ☐Can you draw cis and trans [Co(NH₃)₄Cl₂]⁺ and fac and mer [Co(NH₃)₃(NO₂)₃]?
• ☐Can you explain why cis-[CoCl₂(en)₂]⁺ is optically active?
• ☐Can you give hybridisation, shape and magnetism for [Ni(CN)₄]²⁻, [NiCl₄]²⁻, [CoF₆]³⁻ and [Co(NH₃)₆]³⁺?
• ☐Can you draw octahedral splitting and fill d⁴ to d⁷ for weak and strong ligands?
• ☐Can you explain synergic bonding and list five applications of complexes?

Coordination Compounds: frequently asked questions