Coordination Compounds ISC Class 12 Chemistry: Notes

What Coordination Compounds is about

Weightage: Coordination Compounds is part of the Inorganic Chemistry group, which carries 14 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Complexes, ligands and coordination number

Complex vs double salt: a complex ion persists in solution (K₄[Fe(CN)₆] gives no test for Fe²⁺ or CN⁻); a double salt gives tests for all its ions (potash alum gives K⁺, Al³⁺ and SO₄²⁻). Ligand: an ion or molecule that donates a lone pair to the central metal. Denticity: number of donor atoms in one ligand. Monodentate: Cl⁻, H₂O, NH₃, CN⁻, CO. Bidentate: ethane-1,2-diamine (en), oxalate ($C_2{O}_4^{2-}$, ox). Tridentate: diethylenetriamine (dien). Tetradentate: triethylenetetramine (trien). Pentadentate: ethylenediaminetriacetate. Hexadentate: ethylenediaminetetraacetate (EDTA⁴⁻, two N and four O donors). Chelating ligand: a polydentate ligand that forms a ring with the metal; chelate complexes are especially stable (chelate effect). Ambidentate ligand: can bind through either of two atoms: $N{O}_2^{-}$ (nitrito-N) / $ON{O}^{-}$ (nitrito-O); $SC{N}^{-}$ / $NC{S}^{-}$. Coordination number (CN): the number of donor atoms bonded to the metal (not the number of ligands). Coordination sphere: the metal and ligands in square brackets; ions outside are counter ions.

Worked example

Find the coordination number and the oxidation state of the metal in (a) $[Co(en{)}_{2}C{l}_2]N{O}_3$, (b) $K_3[Cr(C_2{O}_4{)}_3]$, (c) $[Ni(CO{)}_4]$.

Try it first, then show the solutionSolution

1. (a) Two en (2 donors each) + 2 Cl⁻ = CN 6; x + 0 + 2(−1) = +1 (balances one NO₃⁻): Co +3.
2. (b) Three bidentate oxalates: CN 6; 3(+1) + x + 3(−2) = 0: Cr +3.
3. (c) Four CO: CN 4; CO is neutral: Ni 0.

Answer: (a) 6, +3 (b) 6, +3 (c) 4, 0

IUPAC nomenclature of mononuclear complexes

Rules: (1) Name the cation first, then the anion (as for any salt). (2) Within the complex, name ligands first in alphabetical order (ignore multiplying prefixes), then the metal. (3) Prefixes di-, tri-, tetra- give the number of simple ligands; bis-, tris-, tetrakis- are used for ligands whose names already contain a number, e.g. bis(ethane-1,2-diamine). (4) Anionic ligands end in -o: chlorido, cyanido, hydroxido, oxalato, nitrito-N. Neutral ligands keep their names except aqua (H₂O), ammine (NH₃, two m's), carbonyl (CO), nitrosyl (NO). (5) Give the metal's oxidation state in Roman numerals in brackets, with no space. (6) If the complex is an anion, the metal name ends in -ate (cobaltate, nickelate, ferrate for Fe, cuprate for Cu, argentate for Ag, aurate for Au). (7) Do not write 'ion' after a neutral or salt name. Examples: $[Co(N{H}_3{)}_6]C{l}_3$ hexaamminecobalt(III) chloride; $K_4[Fe(CN{)}_6]$ potassium hexacyanidoferrate(II); $[Pt(N{H}_3{)}_{2}C{l}_2]$ diamminedichloridoplatinum(II); $[Co(en{)}_3{]}^{3+}$ tris(ethane-1,2-diamine)cobalt(III) ion.

Worked example

Write IUPAC names of (a) $[Co(N{H}_3{)}_{4}C{l}_2]Cl$, (b) $K_2[NiC{l}_4]$, (c) $[Cr(H_{2}O{)}_{4}C{l}_2]N{O}_3$.

Try it first, then show the solutionSolution

1. (a) ammine before chlorido; Co: x − 2 = +1, x = +3: tetraamminedichloridocobalt(III) chloride
2. (b) anion: x − 4 = −2, x = +2: potassium tetrachloridonickelate(II)
3. (c) aqua before chlorido; x − 2 = +1: tetraaquadichloridochromium(III) nitrate

Answer: As above

Werner's theory

Werner (1893) proposed: (1) a metal shows two kinds of valency. Primary valency is ionisable, equals the oxidation state, and is satisfied by negative ions (shown by dotted lines). Secondary valency is non-ionisable, equals the coordination number, and is satisfied by ligands (neutral or negative) shown by solid lines. (2) Secondary valencies are directed in space, giving definite geometries (octahedral for CN 6, tetrahedral or square planar for CN 4). (3) Every metal tends to satisfy both. Evidence from the cobalt(III) chloride-ammonia series: the number of Cl⁻ precipitated by AgNO₃ and the number of ions from molar conductance tell which Cl are outside the sphere. CoCl₃·6NH₃: 3 AgCl, 4 ions, $[Co(N{H}_3{)}_6]C{l}_3$. CoCl₃·5NH₃: 2 AgCl, 3 ions, $[Co(N{H}_3{)}_{5}Cl]C{l}_2$. CoCl₃·4NH₃: 1 AgCl, 2 ions, $[Co(N{H}_3{)}_{4}C{l}_2]Cl$ (exists as green trans and violet cis isomers). CoCl₃·3NH₃: no AgCl, non-electrolyte, $[Co(N{H}_3{)}_{3}C{l}_3]$. Limitations: it does not explain why only certain elements form complexes, the directional nature of bonds, or colour and magnetism.

Worked example

PtCl₄·2NH₃ gives no precipitate with AgNO₃ and does not conduct electricity. Write its formula according to Werner and give its name.

Try it first, then show the solutionSolution

1. No ionisable Cl and no ions: all four Cl and both NH₃ are in the coordination sphere (CN 6).
2. Formula $[Pt(N{H}_3{)}_{2}C{l}_4]$; Pt: x − 4 = 0, so +4.

Answer: $[Pt(N{H}_3{)}_{2}C{l}_4]$, diamminetetrachloridoplatinum(IV)

Isomerism in coordination compounds

Structural isomerism: (1) Ionisation: an ion inside the sphere swaps with a counter ion: $[Co(N{H}_3{)}_{5}Br]S{O}_4$ (red-violet, gives BaSO₄ with BaCl₂) and $[Co(N{H}_3{)}_{5}S{O}_4]Br$ (red, gives pale yellow AgBr with AgNO₃). (2) Linkage: 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). (3) Coordination: ligands exchange between cationic and anionic complexes: $[Co(N{H}_3{)}_6][Cr(CN{)}_6]$ and $[Cr(N{H}_3{)}_6][Co(CN{)}_6]$. (4) Hydrate (solvate): water inside vs outside: $[Cr(H_{2}O{)}_6]C{l}_3$ (violet), $[Cr(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$ (grey-green), $[Cr(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$ (dark green). Stereoisomerism: (1) Geometrical (cis-trans): in square planar $[M{A}_2{B}_2]$ such as $[Pt(N{H}_3{)}_{2}C{l}_2]$ (cis and trans); in octahedral $[M{A}_4{B}_2]$ such as $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$ (cis: Cl at 90°; trans: Cl at 180°); also fac-mer in $[M{A}_3{B}_3]$. Tetrahedral complexes show no geometrical isomerism. (2) Optical: non-superimposable mirror images (d and l), common for octahedral complexes with bidentate ligands: $[Co(en{)}_3{]}^{3+}$ and cis-$[Co(en{)}_{2}C{l}_2{]}^{+}$ (trans is optically inactive).

Worked example

How would you distinguish chemically between the ionisation isomers $[Pt(N{H}_3{)}_{4}C{l}_2]B{r}_2$ and $[Pt(N{H}_3{)}_{4}B{r}_2]C{l}_2$?

Try it first, then show the solutionSolution

1. Add AgNO₃ solution to each: only the counter ions outside the bracket are free to react.
2. $[Pt(N{H}_3{)}_{4}C{l}_2]B{r}_2$ has free Br⁻ and gives a pale yellow precipitate of AgBr (sparingly soluble in NH₃); $[Pt(N{H}_3{)}_{4}B{r}_2]C{l}_2$ has free Cl⁻ and gives a white precipitate of AgCl (readily soluble in NH₃).

Answer: AgNO₃: pale yellow AgBr from the first, white AgCl from the second

Valence bond theory

The metal ion provides vacant hybrid orbitals equal in number to its coordination number; each ligand donates a lone pair into one. CN 6, octahedral: **$d^{2}s{p}^3$ uses inner (n−1)d orbitals: an inner-orbital, low-spin** complex, formed with strong field ligands (CN⁻, CO, NH₃ with Co³⁺) which force electrons to pair. Example: $[Co(N{H}_3{)}_6{]}^{3+}$, Co³⁺ $3d^6$ paired into three orbitals, two 3d + 4s + three 4p hybridise: diamagnetic. **$s{p}^3{d}^2$ uses outer nd orbitals: an outer-orbital, high-spin** complex with weak field ligands (F⁻, H₂O): $[Co{F}_6{]}^{3-}$, four unpaired electrons, paramagnetic (4.90 BM). CN 4: **$ds{p}^2$, square planar**, e.g. $[Ni(CN{)}_4{]}^{2-}$: Ni²⁺ $3d^8$ pairs up, freeing one 3d orbital: diamagnetic. **$s{p}^3$, tetrahedral**, e.g. $[NiC{l}_4{]}^{2-}$ (two unpaired, paramagnetic) and $[Ni(CO{)}_4]$ (Ni 0: $3d^{8}4s^2$ rearranges to $3d^{10}$, diamagnetic). Limitations: it does not explain colour, gives no quantitative picture of energies, cannot reliably predict high or low spin, and treats bonding as purely covalent.

Worked example

Using VBT, predict the hybridisation, shape and magnetic behaviour of $[Fe(CN{)}_6{]}^{3-}$ (Fe Z = 26).

Try it first, then show the solutionSolution

1. Fe³⁺: $3d^5$. CN⁻ is a strong field ligand, so electrons pair into three 3d orbitals: one unpaired electron remains.
2. Two vacant 3d + one 4s + three 4p → $d^{2}s{p}^3$ hybrid orbitals accept six CN⁻ lone pairs: octahedral, inner-orbital.
3. One unpaired electron: paramagnetic, μ = √3 = 1.73 BM (low spin).

Answer: $d^{2}s{p}^3$, octahedral, paramagnetic (1 unpaired electron)

Crystal field theory: splitting, spin and magnetism

CFT treats ligands as point charges (or dipoles) whose approach raises the energies of the metal's d orbitals unequally. Octahedral field: ligands lie along the axes, so $d_{x^2-y^2}$ and $d_{z^2}$ (the **$e_g$** set) are raised by $+0.6{\Delta }_o$, and $d_{xy}$, $d_{yz}$, $d_{zx}$ (the **$t_{2g}$** set) are lowered by $-0.4{\Delta }_o$. ${\Delta }_o$ is the crystal field splitting energy. **Filling for $d^4$-$d^7$:** if ${\Delta }_o$ < pairing energy P (weak field ligands), the 4th electron enters $e_g$: high spin (e.g. $d^6$ as $t_{2g}^4{e}_g^2$, 4 unpaired). If ${\Delta }_o$ > P (strong field ligands), electrons pair in $t_{2g}$: low spin ($d^6$ as $t_{2g}^6{e}_g^0$, 0 unpaired). Spectrochemical series (increasing field strength): $I^{-}<B{r}^{-}<SC{N}^{-}<C{l}^{-}<F^{-}<O{H}^{-}<C_2{O}_4^{2-}<H_{2}O<NC{S}^{-}<N{H}_3<en<N{O}_2^{-}<C{N}^{-}<CO$. Tetrahedral field: ligands lie between axes, so the order is inverted: $e$ set lower, $t_2$ higher, with ${\Delta }_t=\frac{4}{9}{\Delta }_o$. Because ${\Delta }_t$ is small, tetrahedral complexes are almost always high spin.

Worked example

Two octahedral $d^5$ complexes are $[Fe(H_{2}O{)}_6{]}^{3+}$ and $[Fe(CN{)}_6{]}^{3-}$. Write the $t_{2g}$/$e_g$ configuration of each, the number of unpaired electrons and the spin-only magnetic moment.

Try it first, then show the solutionSolution

1. H₂O is a weak field ligand (${\Delta }_o$ < P): the five electrons occupy all five orbitals singly, $t_{2g}^3{e}_g^2$, high spin, 5 unpaired, μ = √35 = 5.92 BM.
2. CN⁻ is a strong field ligand (${\Delta }_o$ > P): electrons pair in $t_{2g}$, $t_{2g}^5{e}_g^0$, low spin, 1 unpaired, μ = √3 = 1.73 BM.

Answer: $[Fe(H_{2}O{)}_6{]}^{3+}$: $t_{2g}^3{e}_g^2$, 5 unpaired, 5.92 BM; $[Fe(CN{)}_6{]}^{3-}$: $t_{2g}^5{e}_g^0$, 1 unpaired, 1.73 BM

Colour of complexes

In a complex with a partly filled d subshell, an electron can absorb visible light of energy $\Delta$ and jump from $t_{2g}$ to $e_g$ (d-d transition). The complex shows the complementary colour of the light absorbed. $[Ti(H_{2}O{)}_6{]}^{3+}$ ($d^1$) absorbs green-blue light near 498 nm and looks purple. Changing ligands changes Δ and hence colour: $[Ni(H_{2}O{)}_6{]}^{2+}$ is green, $[Ni(en{)}_3{]}^{2+}$ violet (stronger field, larger Δ, absorbs higher-energy light). Anhydrous CuSO₄ is white and hydrated CuSO₄ blue: without water ligands there is no crystal field splitting in the visible range. $d^0$ and $d^{10}$ complexes ($[Sc(H_{2}O{)}_6{]}^{3+}$, $[Zn(H_{2}O{)}_6{]}^{2+}$) are colourless. Relation: $\Delta =\frac{hc{N}_A}{\lambda }$ per mole.

Worked example

$[Ti(H_{2}O{)}_6{]}^{3+}$ absorbs most strongly at 498 nm. Calculate ${\Delta }_o$ in kJ mol⁻¹. (h = $6.626\times 10^{-34}$ J s, c = $3\times 10^8$ m s⁻¹, $N_A=6.022\times 10^{23}$)

Try it first, then show the solutionSolution

1. Energy per ion $=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{498\times 10^{-9}}=3.99\times 10^{-19}$ J
2. Per mole: $3.99\times 10^{-19}\times 6.022\times 10^{23}=2.40\times 10^5$ J = 240 kJ mol⁻¹

Answer: About 240 kJ mol⁻¹

Stability of complexes and importance

Stability constant: for $M+nL\rightleftharpoons M{L}_n$, ${\beta }_n=\frac{[M{L}_n]}{[M][L{]}^n}$. The larger K (β), the more stable the complex. E.g. $\beta$ for $[Cu(N{H}_3{)}_4{]}^{2+}$ ≈ $4.5\times 10^{11}$ and for $[Cu(CN{)}_4{]}^{3-}$ much larger. Stability increases with higher charge and smaller size of the metal ion, more basic ligands, and chelation (chelate effect). Importance: (1) Qualitative analysis: Ni²⁺ gives a red precipitate with dimethylglyoxime; Cu²⁺ gives deep blue $[Cu(N{H}_3{)}_4{]}^{2+}$ with excess NH₃; AgCl dissolves in ammonia: $AgCl+2N{H}_3\to [Ag(N{H}_3{)}_2]Cl$; Fe³⁺ gives blood-red $[Fe(SCN){]}^{2+}$. Hardness of water is estimated by titration with EDTA. (2) Extraction of metals: cyanide process for gold and silver: $4Au+8C{N}^{-}+2H_{2}O+O_2\to 4[Au(CN{)}_2{]}^{-}+4O{H}^{-}$, then $2[Au(CN{)}_2{]}^{-}+Zn\to [Zn(CN{)}_4{]}^{2-}+2Au$; Mond process for nickel: $Ni+4CO\to Ni(CO{)}_4$, decomposed at 450-470 K to pure Ni. (3) Biological systems: haemoglobin (Fe²⁺ complex, carries O₂), chlorophyll (Mg²⁺), vitamin B₁₂ (cyanocobalamin, Co). (4) Medicine: EDTA (as its calcium salt) removes lead in lead poisoning; cisplatin cis-$[Pt(N{H}_3{)}_{2}C{l}_2]$ treats cancer; desferrioxamine removes excess iron.

Worked example

Why does AgCl, insoluble in water, dissolve in aqueous ammonia, and what is the use of this in analysis?

Try it first, then show the solutionSolution

1. Ag⁺ forms the stable complex diamminesilver(I): $AgCl+2N{H}_3\to [Ag(N{H}_3{)}_2]Cl$.
2. Removing Ag⁺ from solution as the complex shifts the solubility equilibrium of AgCl, so it dissolves. This distinguishes AgCl (dissolves readily) from AgBr and AgI.

Answer: Formation of soluble $[Ag(N{H}_3{)}_2{]}^{+}$; used in testing for halide ions

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

• Isomerism in Coordination Compounds
• Valence Bond Theory of Coordination Compounds
• Crystal Field Theory

Coordination Compounds formula sheet

Oxidation number of the metal

$x+\sum (\text{ligand charges})=\text{charge on the complex}$

Neutral ligands: H₂O, NH₃, CO, en count 0.

Spin-only magnetic moment

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

n = unpaired electrons from VBT or CFT.

Octahedral splitting

$E(e_g)=+0.6{\Delta }_o,E(t_{2g})=-0.4{\Delta }_o$

High spin if ${\Delta }_o$ < P; low spin if ${\Delta }_o$ > P.

Tetrahedral splitting

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

Order inverted; almost always high spin.

Splitting energy from absorption

$\Delta =\frac{hc{N}_A}{\lambda }$

Colour seen is complementary to the light absorbed.

Overall stability constant

${\beta }_n=\frac{[M{L}_n]}{[M][L{]}^n}$

Larger β, more stable complex.

AgCl in ammonia

$AgCl+2N{H}_3\to [Ag(N{H}_3{)}_2]Cl$

Diamminesilver(I) chloride.

Cyanide process (leaching)

$4Au+8C{N}^{-}+2H_{2}O+O_2\to 4[Au(CN{)}_2{]}^{-}+4O{H}^{-}$

Then displaced by zinc.

Cyanide process (recovery)

$2[Au(CN{)}_2{]}^{-}+Zn\to [Zn(CN{)}_4{]}^{2-}+2Au$

Zinc is the reducing agent.

Mond process

$Ni+4CO\to Ni(CO{)}_4$

Formed at about 330-350 K, decomposed at 450-470 K.

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.

• Count donors, not ligands

  CN = number of donor atoms: en and ox count 2 each, EDTA 6.

  Use it when: Coordination number questions.

• Strong ligands squeeze pairs

  Strong field (CN⁻, CO) → pairing → low spin → inner $d^{2}s{p}^3$. Weak field (F⁻, H₂O) → high spin → outer $s{p}^3{d}^2$.

  Use it when: VBT and CFT magnetism.

• Spectrochemical order

  'I Bought Some Clean Fresh Oranges On Wednesday, Nine Eggs, New Cartons, Clean Cartons': I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < ox < H₂O < NCS⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO.

  Use it when: Weak/strong field decisions.

• BaCl₂ finds sulphate, AgNO₃ finds chloride

  Test the counter ion outside the bracket.

  Use it when: Ionisation isomer tests.

• EDTA lead, platinum cancer

  EDTA for lead poisoning; cisPlatin for Proliferating cells (cancer).

  Use it when: Uses fill-in-the-blanks.

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 5-mark question split into 1 + 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 Werner formula and isomer drawings

  CISCE examiners' comments, 2024: for CoCl₃·4NH₃ giving two ions, candidates wrote [Co(NH₃)₄]Cl₃ or [Co(NH₃)Cl]Cl₂ instead of [Co(NH₃)₄Cl₂]Cl, drew three chlorine ligands in the complex and confused cis and trans.

  Wrong working

  $[Co(N{H}_3{)}_4]C{l}_3$ with three Cl drawn on the octahedron.

  Right working

  $[Co(N{H}_3{)}_{4}C{l}_2]Cl$: only two Cl inside; cis = 90°, trans = 180°.

  ✓Do this: Count the ions first; the number outside the bracket = AgCl precipitated.

  Usually costs: 1-2 marks.

• ✗Mistake: Spelling and oxidation-state errors in IUPAC names

  CISCE examiners' comments, 2024 and 2025: candidates wrote 'tetramine' and 'triamine' for tetraammine and triammine, 'ethylenediammine' for ethylenediamine, gave the wrong oxidation state of the metal and added 'ion' after the name of a salt.

  Wrong working

  tetraminedichlorocobalt(II) chloride ion

  Right working

  tetraamminedichloridocobalt(III) chloride

  ✓Do this: Double 'a' and double 'm' in ammine; calculate the oxidation state from charges; no 'ion' for a salt.

  Usually costs: 1 mark.

• ✗Mistake: Wrong reagent for ionisation isomers

  CISCE examiners' comments, 2024 and 2025: candidates used BaSO₄, AgCl or AgNO₃ as the reagent for a sulphate test, did not state the observation, or described a light yellow precipitate where a white one forms.

  Wrong working

  Add AgNO₃ to detect the sulphate isomer.

  Right working

  BaCl₂ gives a white precipitate of BaSO₄ with free sulphate; AgNO₃ gives a white precipitate of AgCl with free chloride.

  ✓Do this: Name reagent, observation and colour every time.

  Usually costs: 1 mark.

• ✗Mistake: Confusing high spin, low spin and field strength

  CISCE examiners' comments, 2024 and 2025: candidates named $t_{2g}^4{e}_g^2$ as the low-spin complex, gave wrong unpaired-electron counts (2 or 0 for A, 2 or 4 for B), confused strong and weak field ligands, chose diamagnetic complexes as paramagnetic, and called [Ni(CO)₄] square planar instead of tetrahedral.

  Wrong working

  $t_{2g}^4{e}_g^2$: low spin, 2 unpaired.

  Right working

  $t_{2g}^4{e}_g^2$: high spin, 4 unpaired, weak field; $t_{2g}^6{e}_g^0$: low spin, 0 unpaired, strong field.

  ✓Do this: Draw the boxes and count arrows; never guess.

  Usually costs: 1 mark each.

• ✗Mistake: Interchanging uses of complexes

  CISCE examiners' comments, 2024: candidates interchanged the answers, giving cancer for EDTA and lead poisoning for cisplatin.

  Wrong working

  EDTA treats cancer.

  Right working

  EDTA treats lead poisoning; cisplatin treats cancer.

  ✓Do this: Use the mnemonic 'EDTA lead, platinum cancer'.

  Usually costs: 1 mark.

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

   The coordination number of Fe in $[Fe(C_2{O}_4{)}_3{]}^{3-}$ is:
   (a) 3
   (b) 4
   (c) 6
   (d) 9

   Hint

   Oxalate is bidentate.

   Answer

   (c) 6

2. Q2 · 1 mark · MCQBoard level

   Which ligand is ambidentate?
   (a) NH₃
   (b) en
   (c) SCN⁻
   (d) H₂O

   Hint

   Two possible donor atoms.

   Answer

   (c) SCN⁻ (through S or N)

3. Q3 · 1 mark · Very short answerBoard level

   Fill in the blanks: $[Co(N{H}_3{)}_6{]}^{3+}$ is an ______-orbital complex and is ______.

   Hint

   $d^{2}s{p}^3$.

   Answer

   inner; diamagnetic

4. Q4 · 1 mark · Very short answerBoard level

   What is a chelating ligand? Give an example.

   Hint

   Ring formation.

   Answer

   A polydentate ligand that binds through two or more donor atoms forming a ring with the metal, e.g. ethane-1,2-diamine or EDTA.

5. Q5 · 2 marks · Short answerHarder board level

   Write the formulae of (a) potassium hexacyanidoferrate(III) and (b) diamminesilver(I) dicyanidoargentate(I).

   Hint

   Balance charges.

   Answer

   (a) $K_3[Fe(CN{)}_6]$ (b) $[Ag(N{H}_3{)}_2][Ag(CN{)}_2]$

6. Q6 · 2 marks · Short answerBoard level

   How many geometrical isomers does $[Pt(N{H}_3{)}_{2}C{l}_2]$ have? Which is the anticancer drug?

   Hint

   Square planar.

   Answer

   Two (cis and trans); cis-diamminedichloridoplatinum(II), cisplatin.

7. Q7 · 3 marks · Short answerHarder board level

   Draw the optical isomers of $[Co(en{)}_3{]}^{3+}$ and explain why trans-$[Co(en{)}_{2}C{l}_2{]}^{+}$ is optically inactive.

   Hint

   Plane of symmetry.

   Answer

   Two non-superimposable mirror-image octahedra, each with three en spanning cis positions (d and l). trans-$[Co(en{)}_{2}C{l}_2{]}^{+}$ has a plane of symmetry containing the two en rings, so its mirror image is superimposable.

8. Q8 · 3 marks · Short answerHarder board level

   Predict the number of unpaired electrons and the magnetic moment of $[MnC{l}_4{]}^{2-}$ (Mn Z = 25) and $[Mn(CN{)}_6{]}^{4-}$.

   Hint

   Mn²⁺ $d^5$.

   Answer

   $[MnC{l}_4{]}^{2-}$: tetrahedral, high spin, 5 unpaired, 5.92 BM. $[Mn(CN{)}_6{]}^{4-}$: octahedral, strong field, low spin $t_{2g}^5$, 1 unpaired, 1.73 BM.

9. Q9 · 2 marks · Short answerHarder board level

   A complex absorbs light of 600 nm. Calculate Δ in kJ mol⁻¹ and state the approximate colour seen. (h = $6.626\times 10^{-34}$ J s, c = $3\times 10^8$ m s⁻¹, $N_A=6.022\times 10^{23}$)

   Hint

   Δ = hcN_A/λ.

   Answer

   Δ = 6.626 × 10⁻³⁴ × 3 × 10⁸ × 6.022 × 10²³/600 × 10⁻⁹ = 1.99 × 10⁵ J ≈ 199.5 kJ mol⁻¹. It absorbs orange light and appears blue (the complementary colour).

10. Q10 · 2 marks · Short answerHarder board level

    Arrange in increasing order of crystal field splitting and explain: $[Co(H_{2}O{)}_6{]}^{3+}$, $[Co(CN{)}_6{]}^{3-}$, $[Co{F}_6{]}^{3-}$, $[Co(N{H}_3{)}_6{]}^{3+}$.

    Hint

    Spectrochemical series.

    Answer

    $[Co{F}_6{]}^{3-}$ < $[Co(H_{2}O{)}_6{]}^{3+}$ < $[Co(N{H}_3{)}_6{]}^{3+}$ < $[Co(CN{)}_6{]}^{3-}$, following F⁻ < H₂O < NH₃ < CN⁻ in the spectrochemical series.

11. Q11 · 5 marks · Long answerHarder board level

    (a) State two postulates of Werner's theory. [2]
    (b) Write the Werner formula and the number of ions for CoCl₃·6NH₃ and CoCl₃·3NH₃. [2]
    (c) Give one limitation of Werner's theory. [1]

    Hint

    Primary and secondary valencies.

    Answer

    (a) Metals have primary (ionisable, = oxidation state, satisfied by anions) and secondary (non-ionisable, = CN, satisfied by ligands) valencies; secondary valencies are directed in space giving definite geometry. (b) $[Co(N{H}_3{)}_6]C{l}_3$, 4 ions; $[Co(N{H}_3{)}_{3}C{l}_3]$, non-electrolyte (0 ions). (c) It cannot explain colour, magnetic properties or why only some metals form complexes.

12. Q12 · 2 marks · Short answerHOTS / competency

    Hydrated CuSO₄ is blue but anhydrous CuSO₄ is white. Explain using crystal field theory.

    Hint

    What splits the d orbitals?

    Answer

    In the hydrate, water ligands around Cu²⁺ ($d^9$) split the d orbitals, so d-d transitions absorb red-orange light and blue is seen. Without water ligands the splitting is absent (or falls outside the visible range), so no visible light is absorbed: white.

13. Q13 · 2 marks · Short answerHOTS / competency

    Explain why $[Fe(CN{)}_6{]}^{4-}$ is more stable than $[Fe(H_{2}O{)}_6{]}^{2+}$, and why EDTA complexes are especially stable.

    Hint

    Field strength; chelate effect.

    Answer

    CN⁻ is a stronger, more basic ligand forming low-spin $t_{2g}^6$ with large CFSE, so the stability constant is very high. EDTA is hexadentate: chelation forms several stable rings and releases many water molecules (entropy gain), giving very large K.

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.Define coordination number.Show answer

   The number of ligand donor atoms directly bonded to the central metal.

2. 2.Name a hexadentate ligand.Show answer

   EDTA⁴⁻ (ethylenediaminetetraacetate).

3. 3.IUPAC name of $K_4[Fe(CN{)}_6]$?Show answer

   Potassium hexacyanidoferrate(II).

4. 4.Hybridisation and shape of $[Ni(CN{)}_4{]}^{2-}$?Show answer

   $ds{p}^2$, square planar, diamagnetic.

5. 5.What is ${\Delta }_t$ in terms of ${\Delta }_o$?Show answer

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

6. 6.Which metal is in chlorophyll and in vitamin B₁₂?Show answer

   Mg and Co.

7. 7.Give an example of linkage isomers.Show answer

   $[Co(N{H}_3{)}_5(N{O}_2)]C{l}_2$ and $[Co(N{H}_3{)}_5(ONO)]C{l}_2$.

8. 8.Which process purifies nickel using CO?Show answer

   Mond process.

Quick revision checklist

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

• ☐Can you find CN and oxidation state for any complex?
• ☐Can you name and write formulae of complexes without spelling errors?
• ☐Can you deduce a Werner formula from AgCl or conductance data?
• ☐Can you draw cis/trans and d/l isomers?
• ☐Can you distinguish ionisation isomers with BaCl₂ and AgNO₃?
• ☐Can you draw VBT diagrams for $[Co(N{H}_3{)}_6{]}^{3+}$, $[Co{F}_6{]}^{3-}$, $[Ni(CN{)}_4{]}^{2-}$, $[NiC{l}_4{]}^{2-}$?
• ☐Can you write $t_{2g}$/$e_g$ configurations for high- and low-spin $d^4$-$d^7$?
• ☐Can you give uses: EDTA, cisplatin, cyanide process, Mond process, haemoglobin?

Coordination Compounds: frequently asked questions