Take a breath. If Coordination Compounds has been sitting at the bottom of your Class 12 Chemistry to-do list because the formulas look like alphabet soup in square brackets, you are in very ordinary company. Almost every student meets [Co(NH3)6]Cl3 for the first time and quietly panics. The good news is that this chapter is one of the most rule-driven topics in the whole syllabus. Once you own about eight rules, the questions stop being surprises.
This page is a full walkthrough of NCERT Chapter 5 for the CBSE 2026-27 session — the seven headings CBSE actually names, taught from zero. You will get coordination compounds class 12 chemistry important questions answered step by step, a coordination compounds formula and ligand list you can revise in five minutes, worked nomenclature drills, the full isomerism family, both bonding theories, and a practice worksheet with answers hidden until you want them. Nothing here assumes you remember anything beyond “atoms have electrons”.
One idea runs through the whole page, and it is the reason this chapter will finally click: treat the central metal atom as a host, and the ligands as guests at its dinner party. The host has empty seats. Each guest arrives carrying a lone pair as a gift. Some guests hold hands and arrive as a pair. Some guests can shake with either hand. Some cling to the host so hard that the host has to rearrange its own furniture to make room. Every single definition in this chapter — coordination number, denticity, chelation, strong field, low spin — is a sentence about that dinner party. Keep the party in your head and the vocabulary stops feeling foreign.
Meet Your Tutor
Coordination chemistry is rule-driven once the formula is read in a fixed order. I will help you identify the metal, oxidation state, ligands, coordination number and geometry before naming or predicting magnetism, so each answer grows from the complex rather than from memory alone.
What You’ll Learn
- What Coordination Compounds Are, and Werner’s Big Idea
- The Host-and-Guest Vocabulary: Sphere, Seats and Counter Ions
- Types of Ligands: Mono, Bi, Poly, Ambidentate and Chelating
- Oxidation Number and Effective Atomic Number
- IUPAC Nomenclature of Mononuclear Coordination Compounds
- Isomerism in Coordination Compounds
- Bonding I: Valence Bond Theory and the Inner/Outer Decision
- The Exam Decision Flow: Formula to Magnetic Moment
- Bonding II: Crystal Field Theory and d-Orbital Splitting
- Colour, CFSE and the Spectrochemical Series
- Bonding in Metal Carbonyls: Synergic Bonding
- Stability of Coordination Compounds
- Importance and Applications of Coordination Compounds
Your Game Plan
- Learn to split a formula into inside the brackets and outside the brackets. Ten minutes. Everything else stands on this.
- Drill oxidation numbers until they take five seconds. This one skill unlocks nomenclature, VBT and CFT.
- Memorise the ligand name table and the naming order. Write out fifteen names by hand, not by reading.
- Learn the isomerism tree as a tree, not as a list — four structural, two stereo.
- Run the HOST → μ decision flow on ten complexes until you never have to think about it again.
- Add crystal field theory last. It answers “why is it coloured?” and “why low spin?” once VBT has given you the vocabulary.
- Finish with the worksheet at the bottom, then a timed run at the official CBSE Class 12 Chemistry sample paper and marking scheme so you see how the 7 marks are actually distributed.
Study Notes
What Coordination Compounds Are, and Werner’s Big Idea
Start with something that genuinely puzzled chemists in the 1890s. Cobalt(III) chloride is CoCl3. Ammonia is NH3. Both are perfectly stable, “finished” compounds — every valency satisfied. And yet if you mix them, they combine, and they combine in several different fixed ratios: CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3. Two saturated compounds should not stick together at all, let alone in four different flavours with four different colours.
Alfred Werner’s answer, which won him the 1913 Nobel Prize, was that a metal has two kinds of valency, not one.
In the dinner-party picture: primary valency is the guest list the host owes money to (the counter ions, waiting in the corridor). Secondary valency is the number of chairs actually at the table. A host can owe three people money and still have only six chairs.
Werner had no X-ray crystallography. He proved all of this with a silver nitrate bottle and a conductivity meter. Add excess AgNO3 to the solution: only chloride that is genuinely free in solution precipitates as AgCl. Chloride that is bonded to cobalt inside the coordination sphere is not free and does not precipitate. Count the moles of AgCl and you have counted the primary valencies.
CoCl3·6NH3 gives 3 mol AgCl. All three chlorides are free → all six ammonias are at the table → [Co(NH3)6]Cl3. It gives 4 ions in solution (1 complex cation + 3 Cl−).
CoCl3·5NH3 gives 2 mol AgCl. One chloride has taken the empty sixth chair → [Co(NH3)5Cl]Cl2, 3 ions.
CoCl3·4NH3 gives 1 mol AgCl. Two chlorides are inside → [Co(NH3)4Cl2]Cl, 2 ions.
CoCl3·3NH3 gives no AgCl. → [Co(NH3)3Cl3], a neutral non-electrolyte, 0 ions.
Notice the pattern: the secondary valency stayed at 6 in all four. Only the primary valency changed. That constancy is the whole of Werner’s theory in one column of data.
No AgCl means zero free chloride, so all six chlorides are inside the sphere. Three ions with two potassiums outside means the third ion is a single complex anion carrying 2− charge. So the formula is K2[PtCl6]: 2 K+ + 1 [PtCl6]2− = 3 ions. Check the platinum: charge on ion = x + 6(−1) = −2, so x = +4. Consistent with PtCl4.
One more distinction that examiners love. A double salt such as Mohr’s salt, FeSO4·(NH4)2SO4·6H2O, dissolves and completely falls apart into its simple ions — you can detect Fe2+ with an ordinary test. A coordination compound such as K4[Fe(CN)6] dissolves and keeps the [Fe(CN)6]4− unit intact — you cannot detect free Fe2+. The party survives being poured into water; the double salt does not.
The Host-and-Guest Vocabulary: Sphere, Seats and Counter Ions
This section is pure vocabulary, and it is worth slowing down for, because every later section reuses these six words. I will give each one its dinner-party translation.
| Term | Strict definition | Dinner-party translation |
|---|---|---|
| Central atom / ion | The metal atom or ion that accepts lone pairs; a Lewis acid. | The host. |
| Ligand | An ion or molecule with at least one lone pair that it donates to the metal; a Lewis base. | A guest, arriving with a gift. |
| Coordination number (CN) | The number of ligand donor atoms bonded directly to the central atom — count donor atoms, not ligand molecules. | How many chairs are occupied at the table. |
| Coordination sphere | The central atom plus its ligands, written inside square brackets. Behaves as one unit and carries one overall charge. | The dining room. |
| Counter ion | The ion written outside the brackets, which ionises freely in water. | Someone waiting in the corridor, not at the table. |
| Coordination polyhedron | The geometry traced out by the donor atoms — octahedral, tetrahedral, square planar. | The shape of the table. |
| Oxidation number of the metal | The charge the central atom would carry if every ligand were removed along with its lone pairs. | What the host is worth once every guest takes their gift back. |
The single most useful habit in this chapter: before you do anything with a formula, physically draw a line separating what is inside the square brackets from what is outside. Everything inside is one particle. Everything outside is a free ion.
(b) K3[Fe(C2O4)3]. Inside: Fe with three oxalate ions. Oxalate is bidentate — each one occupies two chairs. Donor atoms = 3 × 2 = CN 6, even though there are only three ligands. Counter ions = 3 K+. Ions = 4.
(c) [Co(en)2Cl2]+. “en” is ethane-1,2-diamine, bidentate. Donor atoms = (2 × 2 N) + 2 Cl = CN 6. Two ligands, four chairs, plus two more — the arithmetic only works if you count donor atoms.
Types of Ligands: Mono, Bi, Poly, Ambidentate and Chelating
A ligand is classified by denticity — from the Latin for tooth. How many teeth does the guest sink into the host? Count the donor atoms that one ligand molecule can attach with at the same time.
| Class | Donor atoms used | Examples | Party translation |
|---|---|---|---|
| Monodentate (unidentate) | 1 | Cl−, Br−, CN−, OH−, H2O, NH3, CO, NO2− | A guest who takes one chair. |
| Bidentate | 2 | ethane-1,2-diamine (en, 2 N), oxalate C2O42− (2 O), glycinate (1 N + 1 O) | Two guests holding hands — two chairs, one arrival. |
| Polydentate | 3 or more | EDTA4− (hexadentate: 2 N + 4 O) | A whole family that fills half the table. |
| Ambidentate | 1 at a time, but a choice of which | NO2− (bonds through N or O), SCN− (through S or N), CN− (through C or N) | A guest who can shake with either hand — only one hand at a time. |
| Chelating | 2 or more, forming a ring with the metal | en, oxalate, EDTA4− | Guests who link arms around the host and will not let go. |
Chelation deserves its own paragraph because it produces one of the most exam-friendly facts in the chapter. When a bidentate or polydentate ligand grips a metal, the metal plus the ligand’s backbone closes a ring — typically a five- or six-membered ring. That ring is called a chelate (from the Greek for a crab’s claw). Chelated complexes are dramatically more stable than the same metal holding the equivalent number of separate monodentate ligands. This is the chelate effect, and the reason is largely entropy: one EDTA4− molecule replaces six water molecules, so one particle goes in and six come out. The disorder of the system rises, and reactions that increase disorder are favoured.
(a) en is bidentate. 3 × 2 = CN 6. Three chelate rings, so this ion is notably stable.
(b) EDTA4− is hexadentate (2 nitrogen + 4 carboxylate oxygen). 1 × 6 = CN 6 from a single ligand. Five chelate rings.
(c) NH3 is monodentate; SCN− is ambidentate but binds through only one atom. 5 + 1 = CN 5 + 1 = 6. Because SCN− may bind through S (thiocyanato-S) or through N (thiocyanato-N), this complex has linkage isomers.
Model answer. EDTA4− is a hexadentate chelating ligand: it binds calcium through six donor atoms at once, closing five stable rings. Replacing six separate water ligands with one EDTA ion produces more particles in solution than it consumes, so the entropy change is strongly positive and the formation constant is very large. This is the chelate effect. Monodentate water forms no rings and gives no such entropy gain.
Marking note: say the words “hexadentate”, “chelate ring” and “entropy”. Those are the three marking points.
Oxidation Number and Effective Atomic Number
The oxidation number of the central metal is the single most-used quantity in this chapter. If you get it wrong, the name is wrong, the d-electron count is wrong, the hybridisation is wrong and the magnetic moment is wrong — a four-mark answer collapses from one slip. So make it mechanical.
[Co(NH3)5Cl]Cl2 → two Cl− outside, sphere is 2+. x + 5(0) + 1(−1) = +2 → x = +3.
[Cr(H2O)4Cl2]Cl → sphere is 1+. x + 4(0) + 2(−1) = +1 → x = +3.
K3[Fe(C2O4)3] → sphere is 3−. x + 3(−2) = −3 → x = +3.
[Ni(CO)4] → neutral, CO is neutral. x + 0 = 0 → x = 0. Zero is a perfectly legal oxidation state.
[Fe(H2O)5NO]SO4 (the brown ring) → SO42− outside, sphere is 2+. Taking NO as NO+: x + 5(0) + (+1) = +2 → x = +1. Iron(I) is genuinely unusual, and that is exactly why examiners ask it.
Once you have the oxidation number, the d-electron count follows in one subtraction. For a first-row transition metal, dn = (group number, counting the 4s electrons) − (oxidation number). Practically: Sc 3, Ti 4, V 5, Cr 6, Mn 7, Fe 8, Co 9, Ni 10, Cu 11, Zn 12.
| Ion | Arithmetic | d-configuration |
|---|---|---|
| Ti3+ | 4 − 3 | d1 |
| V3+ | 5 − 3 | d2 |
| Cr3+ | 6 − 3 | d3 |
| Mn2+ | 7 − 2 | d5 |
| Fe3+ | 8 − 3 | d5 |
| Fe2+ | 8 − 2 | d6 |
| Co3+ | 9 − 3 | d6 |
| Co2+ | 9 − 2 | d7 |
| Ni2+ | 10 − 2 | d8 |
| Cu2+ | 11 − 2 | d9 |
| Zn2+ | 12 − 2 | d10 |
The Effective Atomic Number (EAN) is Sidgwick’s old rule of thumb, and it is a lovely one-line stability check. Count the electrons the metal ends up “feeling”: its own atomic number, minus the electrons it lost on becoming an ion, plus two for every lone pair donated by ligands.
[Co(NH3)6]3+: Z(Co) = 27, +3, 6 donors. EAN = 27 − 3 + 12 = 36.
[Ni(CO)4]: Z(Ni) = 28, 0, 4 donors. EAN = 28 − 0 + 8 = 36. This is why nickel forms a tetracarbonyl and not a pentacarbonyl.
[Fe(CO)5]: 26 − 0 + 10 = 36. [Cr(CO)6]: 24 − 0 + 12 = 36. Three different metals, three different carbonyl formulas, one identical answer — that is the rule earning its keep.
IUPAC Nomenclature of Mononuclear Coordination Compounds
Naming is where most of the easy marks live, and it is completely algorithmic. There is no understanding required — only a fixed order, obeyed every time. Here is the order, in the sequence you actually write the name.
- Cation first, anion second — exactly as in NaCl, whichever one is the complex.
- Inside the sphere, name the ligands first, alphabetically, then the metal. Yes, ligands before metal, even though the formula writes the metal first.
- Alphabetise by the ligand name, ignoring the multiplying prefix. Ammine comes before aqua comes before chlorido comes before cyanido.
- Use di, tri, tetra, penta, hexa for simple ligands; use bis, tris, tetrakis with brackets when the ligand name already contains a number-word or would be ambiguous — bis(ethane-1,2-diamine), tris(oxalato).
- Anionic ligands end in -o (chlorido, cyanido, oxalato); neutral ligands keep their name, with the special cases aqua, ammine, carbonyl, nitrosyl.
- Give the metal’s oxidation number as a Roman numeral in round brackets, with no space before the bracket.
- If the coordination sphere is an anion, the metal name takes the suffix -ate — and several metals switch to their Latin stem.
- Write the whole name as one word, with the counter ion as a separate word.
| Metal | In a cationic or neutral sphere | In an anionic sphere |
|---|---|---|
| Iron | iron | ferrate |
| Copper | copper | cuprate |
| Silver | silver | argentate |
| Gold | gold | aurate |
| Lead | lead | plumbate |
| Tin | tin | stannate |
| Chromium | chromium | chromate |
| Cobalt | cobalt | cobaltate |
| Nickel | nickel | nickelate |
| Manganese | manganese | manganate |
| Zinc | zinc | zincate |
| Platinum | platinum | platinate |
(b) K3[Fe(C2O4)3]. Fe is +3, sphere is an anion → ferrate. → potassium trioxalatoferrate(III). (Strict current IUPAC would write tris(oxalato)ferrate(III); CBSE accepts the NCERT form, and either is safe if you are consistent.)
(c) [Cr(NH3)3(H2O)3]Cl3. Cr = +3. Ammine before aqua. → triamminetriaquachromium(III) chloride.
(d) [NiCl4]2−. Ni: x + 4(−1) = −2, so +2. Anionic sphere → nickelate. → tetrachloridonickelate(II) ion.
(e) [CoCl2(en)2]+. Co: x + 2(−1) + 0 = +1, so +3. Alphabetise chlorido before ethane-1,2-diamine; en needs bis(…). → dichloridobis(ethane-1,2-diamine)cobalt(III) ion.
(b) Hexaamminecobalt(III) sulphate. Sphere = [Co(NH3)6]3+. Balancing 3+ against SO42− needs 2 and 3. → [Co(NH3)6]2(SO4)3.
(c) Tetracarbonylnickel(0). Neutral CO, metal oxidation state zero. → [Ni(CO)4].
(d) Diamminechloridonitrito-N-platinum(II). Pt = +2; ligands 2 NH3 (0), 1 Cl−, 1 NO2− bonded through nitrogen. Total charge = +2 −1 −1 = 0, a neutral complex. → [Pt(NH3)2Cl(NO2)].
Isomerism in Coordination Compounds
Isomers are compounds with the same molecular formula but a different arrangement of atoms. In coordination chemistry the family splits cleanly into two: structural isomers, where the atoms are actually joined in a different order, and stereoisomers, where the same atoms are joined in the same order but positioned differently in space.
Structural isomerism has four types in the CBSE scope, and there is a neat way to remember them: “I Have Lost Count” — Ionisation, Hydrate, Linkage, Coordination.
| Type | What swaps | Worked pair | How to tell them apart in the lab |
|---|---|---|---|
| Ionisation | A ligand inside the sphere trades places with the counter ion outside | [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 | Add BaCl2: only the second gives a white BaSO4 precipitate. Add AgNO3: only the first gives pale-yellow AgBr. |
| Hydrate (solvate) | Water moves between the coordination sphere and the crystal lattice | [Cr(H2O)6]Cl3 (violet), [Cr(H2O)5Cl]Cl2·H2O (blue-green), [Cr(H2O)4Cl2]Cl·2H2O (dark green) | Count the moles of AgCl with excess AgNO3: 3, 2 and 1 respectively. The colours differ too. |
| Linkage | An ambidentate ligand switches which atom it donates through | [Co(NH3)5(NO2)]2+ nitrito-N (yellow) and [Co(NH3)5(ONO)]2+ nitrito-O (red) | Colour, and the infrared spectrum. Only ambidentate ligands (NO2−, SCN−, CN−) can do this. |
| Coordination | Ligands are shared out differently between two complex ions in the same salt | [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] | Needs both the cation and the anion to be complex. If one ion is simple, this isomerism is impossible. |
A white precipitate with BaCl2 means free sulphate ions in solution, so sulphate must be the counter ion and bromide must be inside the sphere. → A = [Co(NH3)5Br]SO4.
B gives AgBr with AgNO3, so bromide is free and sulphate is coordinated. → B = [Co(NH3)5SO4]Br.
Cobalt is +3 in both: for A, x + 5(0) + (−1) = +2; for B, x + 5(0) + (−2) = +1. Both give x = +3, as they must — isomers share a formula.
Stereoisomerism splits into geometrical (cis/trans, and fac/mer) and optical (mirror-image forms that rotate plane-polarised light in opposite directions). Which ones are possible depends entirely on the geometry and on how the ligands are distributed.
| Complex type | Geometry | Geometrical isomers | Optically active? |
|---|---|---|---|
| [Ma2b2], e.g. [Pt(NH3)2Cl2] | Square planar | 2 — cis and trans | No. Both forms have a plane of symmetry. |
| [Ma2b2], e.g. [NiCl2(CO)2] | Tetrahedral | None — in a tetrahedron every position is adjacent to every other | No (unless all four ligands differ) |
| [Ma4b2], e.g. [Co(NH3)4Cl2]+ | Octahedral | 2 — cis and trans | No |
| [Ma3b3], e.g. [Co(NH3)3Cl3] | Octahedral | 2 — facial (fac) and meridional (mer) | No |
| [M(AA)2b2], e.g. [Co(en)2Cl2]+ | Octahedral | 2 — cis and trans | Yes, but only the cis form. The trans form has a plane of symmetry. |
| [M(AA)3], e.g. [Co(en)3]3+ | Octahedral | None | Yes — a d and an l form, a classic chiral complex |
(i) Square planar. In the cis form the two chlorides occupy adjacent corners (90° apart); in the trans form they are opposite (180° apart). The cis form is the anticancer drug cisplatin; the trans form is therapeutically inactive — a real reminder that geometry is not decoration.
(ii) Only the cis isomer. Its mirror image cannot be superimposed on it, so it exists as d and l forms. The trans isomer contains a plane of symmetry and is optically inactive.
(iii) Two. In the fac (facial) isomer the three chlorides occupy one triangular face of the octahedron; in the mer (meridional) isomer they lie along a meridian. Neither is optically active.
Illustrator note: the three-dimensional wedge-and-dash drawings of the octahedral fac/mer pair are worth sketching by hand in your own notes — drawing them yourself is what makes them stick.
Bonding I: Valence Bond Theory and the Inner/Outer Decision
You now know what a complex is. Valence Bond Theory answers why the bonds exist and why the shape is what it is. Linus Pauling’s picture is delightfully concrete: the metal ion empties out some of its orbitals, mixes them into a set of identical hybrid orbitals pointing in specific directions, and each ligand lone pair slots into one of them. A coordinate bond is nothing more exotic than a lone pair moving into an empty orbital.
The only genuinely tricky judgement is which six-orbital set the metal uses, and here is the way to hold it in your head. In d2sp3, the two d orbitals come from the inner (n−1)d shell — the host had to shuffle its own electrons together to free those orbitals up. In sp3d2, the two d orbitals come from the outer nd shell — the host left its own electrons alone and used spare rooms upstairs.
One sentence to keep: strong guests rearrange the host’s own furniture; weak guests use the spare room upstairs.
Magnetic behaviour is how we check the prediction experimentally. Unpaired electrons make a substance paramagnetic — it is drawn into a magnetic field. All-paired means diamagnetic. The spin-only magnetic moment connects the two worlds:
μ = √[ n(n + 2) ] BM
where n is the number of unpaired electrons and BM is the Bohr magneton
| Unpaired electrons, n | μ (spin-only), computed to 2 dp | Typical example |
|---|---|---|
| 0 | 0.00 BM — diamagnetic | [Co(NH3)6]3+, [Ni(CN)4]2−, [Ni(CO)4] |
| 1 | 1.73 BM | [Fe(CN)6]3−, [Cu(H2O)6]2+ |
| 2 | 2.83 BM | [NiCl4]2−, [Ni(H2O)6]2+ |
| 3 | 3.87 BM | [Cr(NH3)6]3+, [Fe(H2O)5NO]2+ |
| 4 | 4.90 BM | [CoF6]3−, [Fe(H2O)6]2+ |
| 5 | 5.92 BM | [FeF6]3−, [Mn(H2O)6]2+ |
[Co(NH3)6]3+: NH3 is a strong-field ligand. The six d electrons pair into three of the five 3d orbitals, leaving two 3d orbitals empty. Hybridisation = two 3d + one 4s + three 4p = d2sp3. Geometry octahedral, inner orbital, low spin. n = 0, so μ = 0 BM, diamagnetic.
[CoF6]3−: F− is a weak-field ligand. The six d electrons stay as t2g4eg2-style maximum spin — 4 unpaired. No 3d orbital is free, so cobalt uses 4d orbitals: sp3d2, outer orbital, high spin. μ = √[4 × 6] = √24 = 4.90 BM, paramagnetic.
[Ni(CN)4]2−: CN− is strongly field-splitting. The eight d electrons squeeze into four 3d orbitals, freeing one. Hybridisation = one 3d + one 4s + two 4p = dsp2 → square planar, n = 0, μ = 0 BM, diamagnetic.
[NiCl4]2−: Cl− is weak-field. The 3d8 arrangement is untouched and keeps 2 unpaired electrons. Nickel uses sp3 → tetrahedral. μ = √[2 × 4] = √8 = 2.83 BM, paramagnetic.
This pair is the single most-asked VBT question in the chapter. Learn it as a pair and you have answered half the bonding questions ever set.
Both: x + 6(−1) = −3 → Fe is +3 → 3d5.
[Fe(CN)6]3−: CN− strong field → five electrons crowd into three orbitals as two pairs plus one lone electron → n = 1. Two 3d orbitals freed → d2sp3, octahedral, inner orbital. μ = √[1 × 3] = 1.73 BM.
[FeF6]3−: F− weak field → all five d electrons stay unpaired, n = 5. Uses 4d → sp3d2, octahedral, outer orbital. μ = √[5 × 7] = √35 = 5.92 BM.
Note the shape is octahedral for both. Hybridisation set changes; geometry does not. Students often assume sp3d2 means a different shape — it does not.
The Exam Decision Flow: From Formula to Magnetic Moment
Here is the promise of this chapter. Almost every four-mark bonding question in this unit is the same question wearing a different formula. Rather than learning a hundred complexes, learn one route and walk it. I call it the HOST → μ check, which also happens to be exactly what you are doing — interrogating the host.
- H — How much charge? Find the metal’s oxidation number with the three-line method.
- O — Occupancy of d. dn = group electron count − oxidation number.
- S — Strength of the guests. Look the ligand up on the spectrochemical series: strong field or weak field?
- T — Type of hybridisation. Strong + CN 6 → d2sp3. Weak + CN 6 → sp3d2. Strong + CN 4 with d8 → dsp2. Weak + CN 4 → sp3.
- → geometry follows straight from the hybridisation.
- → μ. Count the unpaired electrons n and evaluate μ = √[n(n+2)] BM.
O: Mn sits at group count 7, so dn = 7 − 2 = d5.
S: H2O sits in the middle-to-weak part of the series — for Mn(II) it behaves as a weak-field ligand.
T: weak field with CN 6 → no 3d orbital is vacated → sp3d2, outer orbital complex.
Geometry: octahedral.
μ: five unpaired electrons → μ = √[5 × 7] = √35 = 5.92 BM, strongly paramagnetic.
And a bonus prediction for free: a d5 high-spin ion has all d-d transitions spin-forbidden, which is why manganese(II) salts are famously pale pink rather than deeply coloured.
[Co(en)3]3+. H: en is neutral and bidentate, so x = +3; CN = 3 × 2 = 6. O: 9 − 3 = d6. S: en is a strong-field chelating ligand. T: pairing gives t2g6, two 3d free → d2sp3, octahedral, inner orbital. n = 0 → μ = 0 BM, diamagnetic. It is also chiral, as we saw in the isomerism section — d and l forms.
Notice how, for d1–d3 and d8–d10 octahedral ions, the strong/weak distinction makes no difference to the number of unpaired electrons. The high-spin/low-spin question only bites for d4, d5, d6 and d7. That alone saves you a lot of worry.
Bonding II: Crystal Field Theory and d-Orbital Splitting
Crystal Field Theory throws away the idea of orbital overlap entirely and replaces it with something almost embarrassingly simple: treat the metal–ligand interaction as purely electrostatic attraction between a positive metal ion and negative ions (or the negative ends of dipoles). If you are comfortable with the idea of a field created by charges — the same idea you met in Electric Charges and Fields in Class 12 Physics — you already have the physics. All that is new is what that field does to five d orbitals.
In a free metal ion, the five d orbitals are degenerate — identical in energy. Bring six ligands up along the x, y and z axes, and that stops being true. Two of the d orbitals (dx²−y² and dz², called the eg set) point straight at the incoming ligands. An electron sitting in one of those feels strong repulsion, so those orbitals go up in energy. The other three (dxy, dyz, dzx, the t2g set) point between the axes, dodging the ligands, so they go down.
In a tetrahedral field everything flips: t2 rises by +0.4 Δt (3 orbitals on the balcony) and e falls by −0.6 Δt (2 orbitals in the basement). Check: 3(+0.4) + 2(−0.6) = 0.
Memory line: “Octahedral is 2 up, 3 down. Tetrahedral is 3 up, 2 down.” The number of orbitals and the fraction always swap together.
The tetrahedral gap is also much smaller. There are only four ligands instead of six, and none of them points directly at a d orbital — they approach through the faces. Working it through gives Δt = (4/9) Δo ≈ 0.44 Δo. That single number has a large consequence: Δt is almost never bigger than the pairing energy, so tetrahedral complexes are essentially always high spin. You will not be asked to justify a low-spin tetrahedral complex, because in practice there is nothing to justify.
| Octahedral dn | High spin (t2g, eg) | n unpaired, μ / BM | Low spin (t2g, eg) | n unpaired, μ / BM |
|---|---|---|---|---|
| d1 | (1, 0) | 1, 1.73 | (1, 0) | 1, 1.73 |
| d2 | (2, 0) | 2, 2.83 | (2, 0) | 2, 2.83 |
| d3 | (3, 0) | 3, 3.87 | (3, 0) | 3, 3.87 |
| d4 | (3, 1) | 4, 4.90 | (4, 0) | 2, 2.83 |
| d5 | (3, 2) | 5, 5.92 | (5, 0) | 1, 1.73 |
| d6 | (4, 2) | 4, 4.90 | (6, 0) | 0, 0.00 |
| d7 | (5, 2) | 3, 3.87 | (6, 1) | 1, 1.73 |
| d8 | (6, 2) | 2, 2.83 | (6, 2) | 2, 2.83 |
| d9 | (6, 3) | 1, 1.73 | (6, 3) | 1, 1.73 |
| d10 | (6, 4) | 0, 0.00 | (6, 4) | 0, 0.00 |
(a) [Fe(H2O)6]2+: Δo < P, so fill all five orbitals singly first, then the sixth electron pairs in t2g. Arrangement t2g4 eg2, n = 4, μ = √24 = 4.90 BM, high spin, paramagnetic.
(b) [Fe(CN)6]4−: Δo > P, so all six pack into t2g. Arrangement t2g6 eg0, n = 0, μ = 0 BM, low spin, diamagnetic.
Same ion, same charge, same coordination number, same shape. Only the guests changed, and the magnetism flipped completely. That is the whole point of the theory.
Colour, CFSE and the Spectrochemical Series
Now for the payoff that Valence Bond Theory could never deliver: why are these compounds coloured at all?
The splitting gap Δo happens to be about the size of a quantum of visible light. When white light passes through a solution of a complex that has at least one electron in t2g and at least one vacancy in eg, an electron absorbs a photon of exactly the right energy and jumps up — a d–d transition. That wavelength is removed from the beam. What reaches your eye is white light minus one colour, and what you perceive is the complementary colour of the light absorbed.
Ti: x + 6(0) = +3 → Ti3+, and 4 − 3 = d1. That one electron sits in t2g. Absorbing a photon of about 500 nm — green-yellow light — promotes it to eg (t2g1eg0 → t2g0eg1). The transmitted light is white minus green-yellow, which the eye reads as violet/purple.
On heating, the aqua ligands are driven off. With no ligands there is no crystal field, no splitting, no d–d transition — and the solid turns colourless. The same argument explains why hydrated CuSO4·5H2O is blue while anhydrous CuSO4 is white.
Sc3+: 3 − 3 = d0. There is no d electron to promote, so no d–d transition is possible.
Zn2+: 12 − 2 = d10. Every d orbital is full, so there is no vacancy to promote into.
A complex is coloured (by the d–d mechanism) only when the d shell is partly filled — d1 to d9. Empty or full means colourless. Cu(I) compounds being white while Cu(II) compounds are blue is the same story: d10 versus d9.
Which ligands split the d orbitals most? That order, measured experimentally from absorption spectra, is the spectrochemical series. Learning it in order is genuinely useful and genuinely annoying, so here is a mnemonic built for this chapter’s analogy — the Wedding Guest List, running from the guests who barely engage with the host to the ones who will not let go.
“I Bring Some Cold Sandwiches For Our Outdoor Wedding, No Extra Napkins — Everyone Complains Constantly.”
I = I− · Bring = Br− · Some = SCN− · Cold = Cl− · Sandwiches = S2− · For = F− · Our = OH− · Outdoor = oxalate C2O42− · Wedding = water H2O · No = NCS− · Extra = EDTA4− · Napkins = NH3 · Everyone = en · Complains = CN− · Constantly = CO
Fifteen words, fifteen ligands, in exactly the NCERT order. The two Cs at the end — CN− and CO — are the strongest field ligands you will meet, and they are the ones who complain constantly: they never leave the host alone.
The Crystal Field Stabilisation Energy (CFSE) is simply how much energy the complex gains by splitting the levels rather than leaving them degenerate. Add up the contributions: each t2g electron gives −0.4Δo, each eg electron gives +0.6Δo.
| dn | High-spin arrangement | High-spin CFSE / Δo | Low-spin arrangement | Low-spin CFSE / Δo |
|---|---|---|---|---|
| d1 | t2g1 | −0.4 | same | −0.4 |
| d2 | t2g2 | −0.8 | same | −0.8 |
| d3 | t2g3 | −1.2 | same | −1.2 |
| d4 | t2g3eg1 | −0.6 | t2g4 | −1.6 |
| d5 | t2g3eg2 | 0 | t2g5 | −2.0 |
| d6 | t2g4eg2 | −0.4 | t2g6 | −2.4 |
| d7 | t2g5eg2 | −0.8 | t2g6eg1 | −1.8 |
| d8 | t2g6eg2 | −1.2 | same | −1.2 |
| d9 | t2g6eg3 | −0.6 | same | −0.6 |
| d10 | t2g6eg4 | 0 | same | 0 |
(b) [Cr(H2O)6]3+, d3: t2g3. CFSE = 3 × (−0.4Δo) = −1.2 Δo. This is the largest stabilisation available without pairing, which is a big part of why Cr(III) complexes are so stubbornly stable.
(c) [Fe(CN)6]4−, low-spin d6: t2g6. CFSE = 6 × (−0.4Δo) = −2.4 Δo — the largest value in the whole table, and the reason the ferrocyanide ion is so hard to break apart.
Note on rigour: these are crystal-field terms only. A complete energy balance would also add the extra pairing energy paid in the low-spin cases. CBSE questions ask for the Δo term, so quote it and say explicitly that pairing energy is being ignored — that sentence earns credit rather than losing it.
In water the copper exists as the aqua complex; ammonia then displaces water to give the tetraamminecopper(II) ion. Ammonia lies higher in the spectrochemical series than water (Napkins after Wedding on the guest list), so it produces a larger Δ. A larger gap means the absorbed photon carries more energy and therefore has a shorter wavelength; the band shifts, and the complementary colour transmitted shifts with it — from pale blue to an intense blue-violet. The intensity also rises because the ammine complex forms in high concentration. This colour change is exactly why ammonia is a confirmatory test for Cu2+.
Bonding in Metal Carbonyls: Synergic Bonding
Carbon monoxide is a strange guest. It is electrically neutral, so there is no ionic attraction pulling it towards the metal, and yet it forms some of the most stable complexes in the whole of inorganic chemistry — and it does so with metals in the zero oxidation state, where a purely electrostatic model predicts nothing should happen at all. Carbonyls in which every ligand is CO are called homoleptic carbonyls.
| Carbonyl | Metal oxidation state | EAN | Hybridisation | Shape |
|---|---|---|---|---|
| [Ni(CO)4] | 0 | 28 − 0 + 8 = 36 | sp3 | Tetrahedral |
| [Fe(CO)5] | 0 | 26 − 0 + 10 = 36 | dsp3 | Trigonal bipyramidal |
| [Cr(CO)6] | 0 | 24 − 0 + 12 = 36 | d2sp3 | Octahedral |
| [Co2(CO)8] | 0 | — (dinuclear, with bridging CO groups) | — | Two metal centres joined by CO bridges |
Each half strengthens the other: more σ donation makes the metal more electron-rich, which drives more back-donation, which drains the metal and encourages more σ donation. That mutual reinforcement is the synergic effect.
The consequences are testable, and this is exactly the detail examiners reward. Because electron density is being pushed into an orbital that is antibonding with respect to C–O, the carbon–oxygen bond gets weaker and longer, and its infrared stretching frequency falls below that of free CO. At the same time the metal–carbon bond gets stronger and shorter, acquiring partial double-bond character. Two bonds, moving in opposite directions, from one mechanism.
In the dinner-party picture, CO is the guest who hands the host a gift and then immediately accepts a gift back — and the exchange makes them both more committed to the evening. Every other ligand you have met only gives.
Model answer. Nickel is in the zero oxidation state, so its configuration 3d84s2 rearranges to 3d104s0. The vacant 4s and three 4p orbitals hybridise to give four sp3 orbitals directed to the corners of a tetrahedron, and each accepts a lone pair from the carbon atom of a CO molecule. Since all electrons are paired, the complex is diamagnetic (μ = 0 BM), and the EAN is 28 − 0 + 8 = 36.
The M–C bond is not purely σ. Filled nickel d orbitals also back-donate electron density into the empty π* antibonding orbitals of CO. Populating an antibonding orbital reduces the C–O bond order below three, so the C–O bond lengthens and weakens, while the Ni–C bond gains π character and shortens. This mutual strengthening is called synergic bonding.
Marking note: the four marks are usually (i) sp3/tetrahedral, (ii) σ donation from C, (iii) back-donation into π*, (iv) consequence for the two bond lengths. Write all four as separate sentences.
Diagram: orbital-overlap picture of synergic σ donation and π back-donation in M–C≡O — to be added by illustrator.
Stability of Coordination Compounds
When a metal ion sits in a solution containing ligands, complex formation is an equilibrium, not a one-way street. The position of that equilibrium is measured by the stability constant (or formation constant), K. The bigger K is, the more the complex is favoured — the more firmly the guests stay at the table.
Complexes normally form one ligand at a time, so there is a stepwise constant for each addition, and an overall constant β that is the product of them all. If you are comfortable with equilibrium expressions from the Solutions chapter, this is the same algebra applied to a different reaction.
Overall: M + 4L ⇌ ML4, with β4 = K1 × K2 × K3 × K4, so log β4 = log K1 + log K2 + log K3 + log K4. Stepwise constants normally decrease down the series, because each incoming ligand faces more crowding and less residual charge on the metal.
log β4 = 4.0 + 3.3 + 2.7 + 2.0 = 12.0, so β4 = 1012.0 = 1.0 × 1012.
Two things to notice. First, the constants fall steadily (4.0 → 3.3 → 2.7 → 2.0), which is the normal pattern. Second, an overall constant of 1012 means that at equilibrium the free metal ion concentration is vanishingly small — which is precisely why a complexed metal ion fails the ordinary qualitative tests for that metal.
Three factors decide how large the stability constant will be:
- Charge on the metal ion. Higher charge → stronger electrostatic attraction → more stable. Fe3+ complexes are generally more stable than the corresponding Fe2+ complexes.
- Size of the metal ion. Smaller ion → higher charge density → more stable, for the same charge.
- Nature of the ligand. Stronger donors and, above all, chelating ligands give much larger constants — the chelate effect again.
Importance and Applications of Coordination Compounds
This section is pure recall, and it is where a well-prepared student picks up marks quickly. Organise it under five headings and it stops being a list of trivia.
1. Qualitative and quantitative analysis
- Ni2+ gives a bright red precipitate with dimethylglyoxime (DMG) in an ammoniacal medium — a chelate, and the standard confirmatory test for nickel.
- Cu2+ gives the deep blue tetraamminecopper(II) ion with excess ammonia.
- Fe3+ gives a blood-red thiocyanato complex with SCN−.
- Ag+: silver chloride dissolves in aqueous ammonia by forming the diamminesilver(I) ion, which is how AgCl is separated from AgBr and AgI in salt analysis.
- EDTA titrations determine the hardness of water: Ca2+ and Mg2+ form EDTA chelates with different stability constants, so they can be estimated selectively. The underlying acid–base ideas are the ones you met with acids, bases and salts — a ligand is simply a Lewis base and the metal ion a Lewis acid.
2. Extraction and purification of metals
Cyanide process for silver and gold. The metal is leached out of low-grade ore as a soluble complex, then displaced by a more reactive metal:
4 Au + 8 CN− + 2 H2O + O2 → 4 [Au(CN)2]− + 4 OH−
2 [Au(CN)2]− + Zn → [Zn(CN)4]2− + 2 Au
Mond process for purifying nickel. Impure nickel is warmed in carbon monoxide to form volatile tetracarbonylnickel(0); the vapour is separated from the non-volatile impurities and then decomposed at a higher temperature to give very pure nickel, releasing the CO for reuse:
Ni + 4 CO (330–350 K) → [Ni(CO)4] (450–470 K) → Ni + 4 CO
3. Biological systems — life runs on complexes
| Molecule | Central metal | What it does |
|---|---|---|
| Haemoglobin | Fe | Carries oxygen in blood; the iron sits at the centre of a porphyrin ring |
| Chlorophyll | Mg | The green pigment of plants; harvests light for photosynthesis |
| Vitamin B12 (cyanocobalamin) | Co | Essential for red blood cell formation; deficiency causes pernicious anaemia |
| Carboxypeptidase-A | Zn | A digestive enzyme — a metalloenzyme that cleaves peptide bonds |
A lovely detail worth quoting in an answer: haemoglobin and chlorophyll are built on the same kind of ring and differ mainly in the metal at the centre. Change the host, change the job entirely — oxygen transport versus light harvesting.
4. Medicine
- cis-[Pt(NH3)2Cl2] — cisplatin — is used to treat tumours. The trans isomer is ineffective, one of chemistry’s sharpest demonstrations that geometry matters.
- Chelation therapy. EDTA is administered to treat lead poisoning: it chelates the lead ion into a stable, water-soluble complex that the body excretes. Related chelating drugs are used for excess copper and excess iron.
- Certain gold and platinum coordination compounds are used in other therapeutic contexts. (If you quote specific drug names in an exam, stick to the ones in your textbook.)
5. Catalysis, photography and electroplating
- Wilkinson’s catalyst, chlorotris(triphenylphosphine)rhodium(I), homogeneously catalyses the hydrogenation of alkenes.
- Black-and-white photography. Unexposed silver bromide is dissolved out of the developed film by sodium thiosulphate (“hypo”) as the soluble bis(thiosulphato)argentate(I) complex — the fixing step.
- Electroplating. Silver and gold are plated from cyanido complexes rather than from simple salts, because the low free-metal-ion concentration makes deposition slow and even, giving a smooth, adherent coating.
(i) Analysis: nickel(II) is confirmed by the bright red chelate it forms with dimethylglyoxime in ammoniacal solution.
(ii) Metallurgy: in the Mond process nickel is purified through volatile [Ni(CO)4], which is formed at 330–350 K and decomposed back to pure nickel at 450–470 K.
(iii) Biology: haemoglobin is an iron coordination compound that transports oxygen; chlorophyll is the corresponding magnesium complex in plants.
(iv) Medicine: cis-[Pt(NH3)2Cl2], cisplatin, is used in cancer chemotherapy; EDTA is used in chelation therapy for lead poisoning.
(v) Catalysis: Wilkinson’s catalyst, [(Ph3P)3RhCl], catalyses alkene hydrogenation.
Marking note: one mark per part, and the mark is for the named complex. “Used in medicine” earns nothing; “cisplatin, cis-[Pt(NH3)2Cl2]” earns the mark.
Coordination Compounds Formula, Ligand and Nomenclature Quick List
Everything worth revising the night before, in one place. If you can reproduce these three tables from memory, you can answer almost any objective or short-answer question in this unit.
| What you need | The rule or value |
|---|---|
| Spin-only magnetic moment | μ = √[n(n+2)] BM → n = 0,1,2,3,4,5 gives 0.00, 1.73, 2.83, 3.87, 4.90, 5.92 BM |
| Effective atomic number | EAN = Z − (oxidation number) + 2 × (number of donor atoms); target 36, 54 or 86 |
| Octahedral splitting | eg at +0.6Δo (2 orbitals), t2g at −0.4Δo (3 orbitals) |
| Tetrahedral splitting | t2 at +0.4Δt (3 orbitals), e at −0.6Δt (2 orbitals) |
| Relation between the gaps | Δt = (4/9)Δo ≈ 0.44Δo → tetrahedral complexes are effectively always high spin |
| CFSE | (number of t2g electrons) × (−0.4Δo) + (number of eg electrons) × (+0.6Δo) |
| High spin vs low spin | Δo > P → low spin; Δo < P → high spin; only matters for d4–d7 |
| Hybridisation → shape | d2sp3/sp3d2 octahedral · dsp2 square planar · sp3 tetrahedral · dsp3 trigonal bipyramidal · sp linear |
| Colour condition | A partly filled d shell (d1–d9) is required for a d–d transition; d0 and d10 are colourless |
| Spectrochemical order | I− < Br− < SCN− < Cl− < S2− < F− < OH− < C2O42− < H2O < NCS− < EDTA4− < NH3 < en < CN− < CO |
| Complex you must know cold | Metal, dn | Hybridisation & shape | n unpaired | μ / BM |
|---|---|---|---|---|
| [Co(NH3)6]3+ | Co(III), d6 | d2sp3, octahedral (inner) | 0 | 0.00 |
| [CoF6]3− | Co(III), d6 | sp3d2, octahedral (outer) | 4 | 4.90 |
| [Fe(CN)6]3− | Fe(III), d5 | d2sp3, octahedral (inner) | 1 | 1.73 |
| [FeF6]3− | Fe(III), d5 | sp3d2, octahedral (outer) | 5 | 5.92 |
| [Fe(CN)6]4− | Fe(II), d6 | d2sp3, octahedral (inner) | 0 | 0.00 |
| [Ni(CN)4]2− | Ni(II), d8 | dsp2, square planar | 0 | 0.00 |
| [NiCl4]2− | Ni(II), d8 | sp3, tetrahedral | 2 | 2.83 |
| [Cr(NH3)6]3+ | Cr(III), d3 | d2sp3, octahedral | 3 | 3.87 |
| [Mn(H2O)6]2+ | Mn(II), d5 | sp3d2, octahedral | 5 | 5.92 |
| [Ni(CO)4] | Ni(0), d10 | sp3, tetrahedral | 0 | 0.00 |
Practice Worksheet
Ten original questions in the CBSE style — a mixture of one-markers, the standard four-mark bonding question and the applications recall that comes up every year. Work each one on paper first, then open the answer. If you want more of the same flavour, the official Class 12 Chemistry sample paper with its marking scheme shows you exactly how these marks are split. Treat this as your NCERT-intext-style revision drill: attempt, check, correct, repeat.
Q1. A cobalt compound of composition CoCl3·4NH3 gives one mole of AgCl per mole of compound with excess AgNO3. Write its coordination formula, its IUPAC name, the coordination number of cobalt and the number of ions it gives in solution.
Formula: [Co(NH3)4Cl2]Cl.
Oxidation number: x + 4(0) + 2(−1) = +1 → x = +3.
Name: tetraamminedichloridocobalt(III) chloride.
Coordination number: 4 + 2 = 6.
Ions in solution: 2 — one [Co(NH3)4Cl2]+ and one Cl−.
Q2. Give the oxidation number of the central metal in (a) K2[Zn(OH)4], (b) [Cr(NH3)3(H2O)3]Cl3, (c) [Fe(H2O)5NO]SO4, (d) Na3[Co(NO2)6].
(b) Sphere charge +3, all ligands neutral: x = +3.
(c) Sphere charge +2, NO taken as NO+: x + 5(0) + 1 = +2 → +1. (The brown-ring complex — iron in the unusual +1 state, d7, three unpaired electrons, μ = 3.87 BM.)
(d) Sphere charge −3: x + 6(−1) = −3 → +3.
Q3. Write the IUPAC names of (a) [Pt(NH3)2Cl2], (b) K3[Cr(C2O4)3], (c) [Co(NH3)4(H2O)Cl]Cl2, (d) [Ni(CO)4].
(b) Cr: x + 3(−2) = −3 → +3. Anionic sphere → chromate. → potassium trioxalatochromate(III) (equivalently potassium tris(oxalato)chromate(III)).
(c) Co: x + 4(0) + 0 + (−1) = +2 → +3. Alphabetical order ammine, aqua, chlorido. → tetraammineaquachloridocobalt(III) chloride.
(d) Ni is 0; CO is carbonyl. → tetracarbonylnickel(0).
Q4. Write the formulas for (a) potassium hexacyanidoferrate(II), (b) tris(ethane-1,2-diamine)chromium(III) chloride, (c) hexaamminenickel(II) nitrate, (d) sodium tetrahydroxidozincate(II).
(b) Cr(III) with three neutral bidentate en → charge +3, needs three Cl−. → [Cr(en)3]Cl3. Coordination number 6.
(c) Ni(II) with six neutral NH3 → charge +2, needs two NO3−. → [Ni(NH3)6](NO3)2.
(d) Zn(II) with four OH− → 2 − 4 = −2, needs two Na+. → Na2[Zn(OH)4].
Q5. Calculate the effective atomic number of the metal in (a) [Cr(CO)6], (b) [Fe(CN)6]3−, (c) [Co(NH3)6]Cl3, (d) [PtCl6]2−. Which of them obey the noble-gas rule? (Z: Cr 24, Fe 26, Co 27, Pt 78.)
(a) 24 − 0 + 12 = 36 — krypton. Obeys.
(b) Fe is +3: 26 − 3 + 12 = 35. Does not obey — one electron short of krypton, and the ion is indeed a good one-electron oxidising agent, readily reduced to [Fe(CN)6]4−, which has EAN 36.
(c) Co is +3: 27 − 3 + 12 = 36. Obeys.
(d) Pt: x + 6(−1) = −2 → +4. EAN = 78 − 4 + 12 = 86 — radon. Obeys.
Q6. Using valence bond theory, predict the hybridisation, geometry, number of unpaired electrons and spin-only magnetic moment of [Mn(CN)6]3− and [MnF6]3−. (Z of Mn = 25.)
[Mn(CN)6]3−: CN− is a strong-field ligand, so the four d electrons pair up into t2g (t2g4), freeing two inner 3d orbitals. Hybridisation d2sp3, geometry octahedral, inner orbital, low spin. n = 2, μ = √[2 × 4] = √8 = 2.83 BM.
[MnF6]3−: F− is weak-field, so the arrangement stays t2g3eg1. No inner 3d orbital is free, so 4d orbitals are used: sp3d2, octahedral, outer orbital, high spin. n = 4, μ = √[4 × 6] = √24 = 4.90 BM.
Q7. (i) How many isomers in total does [Co(en)2Cl2]+ have, and of what kinds? (ii) What kind of isomerism relates [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2·H2O, and how would you distinguish them?
(ii) Hydrate (solvate) isomerism — water has moved from inside the coordination sphere to the crystal lattice. Distinguish them with excess AgNO3: the first gives 3 mol of AgCl per mole, the second gives 2 mol. Their colours also differ (violet and blue-green respectively).
Q8. Calculate the crystal field stabilisation energy, in units of Δo, for an octahedral d5 ion in (a) a weak field and (b) a strong field. Which case applies to [Fe(CN)6]3−, and what is its magnetic moment?
(b) Strong field (low spin), t2g5: CFSE = 5(−0.4Δo) = −2.0Δo (crystal-field term only; the extra pairing energy is not included).
CN− is a strong-field ligand, so [Fe(CN)6]3− is low spin, t2g5, with n = 1. μ = √[1 × 3] = 1.73 BM.
Q9. Explain why [Cu(NH3)4]2+ is deep blue while [Cu(CN)4]3− is colourless, even though both contain copper and both are four-coordinate.
A d9 ion has a partly filled d shell, so a d–d transition is possible: it absorbs part of the visible spectrum and transmits the complementary colour, giving the intense blue. A d10 ion has every d orbital full, so there is no vacancy to promote an electron into, no d–d transition, and therefore no colour. The lesson: colour depends on the d-electron count, not on the element.
Q10. (i) Describe the bonding in [Ni(CO)4] and state its shape and magnetic behaviour. (ii) Explain why the C–O bond in the complex is weaker than in free CO. (iii) Name the industrial process that uses this complex and give the two temperatures involved.
(ii) Besides the σ donation from carbon, filled nickel d orbitals back-donate electron density into the vacant π* antibonding orbitals of CO. Filling an antibonding orbital lowers the C–O bond order below three, so that bond becomes longer and weaker, while the Ni–C bond gains π character and becomes shorter and stronger. This mutual reinforcement is the synergic effect.
(iii) The Mond process for purifying nickel. The carbonyl is formed at 330–350 K and decomposed back to pure nickel at 450–470 K.
That is the whole chapter. If a section still feels slippery, do not reread it — redo it. Cover the worked example, attempt it cold, then compare. Understanding that survives a blank page is the only kind that shows up in a board exam.
And keep the bar low enough to clear. You do not need to master coordination chemistry tonight. You need one more correct answer today than you managed yesterday — one more oxidation number found without hesitating, one more name written without checking the order. Stack thirty of those small wins and the chapter is simply done.