Class 12 · Chemistry · Chapter 5
Coordination Compounds
The chapter starts with Werner's experiments on cobalt ammine chlorides, sets up the vocabulary (ligand, coordination number, coordination sphere), teaches IUPAC naming, then the six kinds of isomerism, and finally explains bonding, magnetism and colour through valence bond theory and crystal field theory before closing with metal carbonyls and applications. It carries 7 marks and turns up every year as naming, isomerism and a why-is-this-complex-paramagnetic question.
Before you startWhat the board asks from this chapter
Unit 5, Coordination Compounds, carries 7 marks in the CBSE unit-wise weightage for the 70-mark theory paper. In a typical paper that is one or two 1-mark items (MCQ or assertion–reason), one 2- or 3-mark question on naming, isomerism or VBT/CFT, and the chapter often supplies a part of a 5-mark question or a 4-mark case study together with the d- and f-block chapter.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Oxidation number of the metal, number of ions a complex gives in solution, which ligand is ambidentate or hexadentate, which complex is most stable or has the highest magnetic moment, order of wavelength absorbed, relation between Δt and Δo. |
| Assertion–Reason | 1 | Pairs on why [Ni(CN)4]2− is diamagnetic, why tetrahedral complexes are high spin, optical activity of [Co(en)3]3+, anhydrous CuSO4 being white, and linkage isomerism of NO2−. |
| Very short answer | 2 | Write the IUPAC name or the formula from the name; give one example each of two isomerism types; define chelate effect or ambidentate ligand; explain why a named complex is paramagnetic or colourless in two lines. |
| Short answer | 3 | Three explain-why parts drawn from VBT and CFT (inner/outer orbital, strong/weak field, colour), or a name-plus-isomerism question on three complexes, or hybridisation, shape and magnetic moment of a given complex with the orbital diagram. |
| Long answer | 5 | Usually shared with the d- and f-block chapter: hybridisation, geometry and magnetic character of two or three complexes, drawing geometrical and optical isomers, crystal field splitting with diagram and the d4 case, or Werner's postulates with the CoCl3·xNH3 evidence. |
| Case-based | 4 | A passage on Werner's cobalt ammine chlorides and AgNO3, or on crystal field splitting and colour, followed by four sub-questions, one of which is a spin-only magnetic moment or a count of ions. |
Hot topics, in order
- IUPAC nomenclature: name from formula and formula from name, including bis/tris, -ate endings and nitrito-N/nitrito-O
- Valence bond theory: hybridisation, shape and magnetic character of [Ni(CN)4]2−, [NiCl4]2−, [Ni(CO)4], [Fe(CN)6]3−, [CoF6]3−, [Co(NH3)6]3+
- Isomerism: geometrical (cis/trans, fac/mer), optical ([Co(en)3]3+), linkage, ionisation, coordination and hydrate isomers with tests
- Crystal field theory: octahedral and tetrahedral splitting diagrams, Δo versus pairing energy, high and low spin d4, spectrochemical series
- Colour of complexes: d–d transitions, why [Ti(H2O)6]3+ is violet and [Sc(H2O)6]3+ or anhydrous CuSO4 is not coloured
- Werner's theory and the CoCl3·xNH3 silver nitrate evidence; double salts versus complexes
- Ligand types: unidentate, didentate, polydentate, ambidentate, chelate effect and stability
- Applications: EDTA for hardness and lead poisoning, cisplatin, DMG test for Ni2+, cyanide extraction of gold and silver, haemoglobin and chlorophyll
The whole chapter, in orderNotes
01 5.1 Werner's Theory of Coordination Compounds PYQ 2013
Alfred Werner (Nobel Prize, 1913) worked out the structure of coordination compounds from a very simple experiment. Cobalt(III) chloride combines with ammonia to give a series of compounds of formula CoCl3·xNH3. When excess AgNO3 is added to their solutions, the amount of AgCl precipitated is different for each: CoCl3·6NH3 (yellow) gives 3 mol AgCl, CoCl3·5NH3 (purple) gives 2 mol, CoCl3·4NH3 (green, and a violet form) gives 1 mol, and CoCl3·3NH3 gives none. Molar conductance falls in the same order. So only some of the chloride is free in solution; the rest is held tightly to the cobalt along with the ammonia.
Werner's postulates explain this. (i) In coordination compounds a metal shows two types of linkage, primary and secondary. (ii) The primary valences are normally ionisable and are satisfied by negative ions. (iii) The secondary valences are non-ionisable, are satisfied by neutral molecules or negative ions, and their number, the coordination number, is fixed for a given metal. (iv) The groups bound by secondary linkages have a definite spatial arrangement, which gives the complex its geometry. For cobalt the secondary valence is six, so the four compounds are [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2, [Co(NH3)4Cl2]Cl and [Co(NH3)3Cl3]. Werner also proposed that coordination number 6 gives an octahedral arrangement and coordination number 4 gives tetrahedral or square planar.
The board likes the difference between a double salt and a complex. Both are formed from two or more stable compounds in stoichiometric ratio. A double salt such as carnallite KCl·MgCl2·6H2O, Mohr's salt FeSO4·(NH4)2SO4·6H2O or potash alum K2SO4·Al2(SO4)3·24H2O dissociates completely into its simple ions in water. A complex such as K4[Fe(CN)6] does not give Fe2+ and CN−; the complex ion [Fe(CN)6]4− keeps its identity in solution.
02 5.2 Definitions of Some Important Terms PYQ 2015 · 2011
A coordination entity is a central metal atom or ion bonded to a fixed number of ions or molecules, for example [CoCl3(NH3)3], [Ni(CO)4], [PtCl2(NH3)2], [Fe(CN)6]4−. The central atom or ion is the one to which the ligands are bound; it is a Lewis acid. Ligands are the ions or molecules bound to it; each ligand donates a lone pair, so it is a Lewis base.
Ligands are classified by the number of donor atoms they use. A unidentate ligand binds through one atom: Cl−, H2O, NH3. A didentate ligand binds through two: ethane-1,2-diamine (en, H2NCH2CH2NH2) and oxalate C2O42−. A polydentate ligand binds through several: EDTA4− is hexadentate, binding through two nitrogen and four oxygen atoms. When a di- or polydentate ligand uses two or more donor atoms on the same metal it forms a ring, called a chelate, and the ligand is a chelating ligand. The number of ligating groups is the denticity. Chelate complexes are more stable than those with the equivalent number of unidentate ligands; this extra stability is the chelate effect. An ambidentate ligand has two different donor atoms but uses only one at a time: NO2− binds through N (nitrito-N) or through O (nitrito-O); SCN− binds through S (thiocyanato-S) or N (thiocyanato-N).
Coordination number is the number of ligand donor atoms directly bonded to the metal: 6 in [PtCl6]2−, 4 in [Ni(NH3)4]2+, 6 in [Fe(C2O4)3]3− and [Co(en)3]3+ because each didentate ligand counts twice. Only σ bonds are counted. The metal with its ligands inside the square bracket is the coordination sphere; the ions outside it are counter ions. The spatial arrangement of the ligand atoms around the metal is the coordination polyhedron, commonly octahedral, square planar or tetrahedral. The oxidation number of the central atom is the charge it would carry if all ligands were removed with their electron pairs; it is written in Roman numerals in parentheses, as in [Cu(CN)4]3−, Cu(I). A homoleptic complex has only one kind of ligand, [Co(NH3)6]3+; a heteroleptic complex has more than one kind, [Co(NH3)4Cl2]+.
03 5.3 Nomenclature of Coordination Compounds PYQ 2017 · 2016 · 2015 · 2013
Writing the formula: the central atom is listed first, then the ligands in alphabetical order regardless of charge (polydentate and abbreviated ligands are also placed alphabetically, by the first letter of the abbreviation). The whole coordination entity goes in square brackets, polyatomic ligands in parentheses, no spaces between ligand and metal. The charge on a complex ion is written outside the bracket as a superscript with the number before the sign, [Co(CN)6]3−; for a salt the cation is written first and the counter ion balances the charge.
Naming: (1) the cation is named first, then the anion, whichever is the complex. (2) Ligands are named in alphabetical order before the name of the metal, ignoring numerical prefixes. (3) Anionic ligands end in -o: chlorido, bromido, cyanido, hydroxido, oxalato, sulphato, carbonato, nitrito-N, nitrito-O, thiocyanato-S. Neutral ligands keep their name (methylamine, pyridine, ethane-1,2-diamine) except aqua for H2O, ammine for NH3, carbonyl for CO and nitrosyl for NO. (4) Numbers of ligands are shown by di, tri, tetra, penta, hexa; if the ligand name itself contains a numerical prefix or is complicated, use bis, tris, tetrakis with the ligand name in parentheses: bis(ethane-1,2-diamine), tris(oxalato) is also accepted as trioxalato. (5) The oxidation state of the metal follows its name in Roman numerals in parentheses. (6) If the complex is an anion the metal name ends in -ate: ferrate, cobaltate, nickelate, cuprate, chromate, platinate, zincate, argentate, aurate. In a cationic or neutral complex the metal name is unchanged.
Worked examples the board uses: [Cr(NH3)3(H2O)3]Cl3 triamminetriaquachromium(III) chloride; [Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethane-1,2-diamine)cobalt(III) sulphate; [Ag(NH3)2][Ag(CN)2] diamminesilver(I) dicyanidoargentate(I); K3[Fe(CN)6] potassium hexacyanidoferrate(III); [Ni(CO)4] tetracarbonylnickel(0); [Pt(NH3)2Cl(NO2)] diamminechloridonitrito-N-platinum(II); [Co(NH3)4Cl(NO2)]Cl tetraamminechloridonitrito-N-cobalt(III) chloride; [Pt(NH3)2Cl(NH2CH3)]Cl diamminechlorido(methylamine)platinum(II) chloride; K2[PdCl4] potassium tetrachloridopalladate(II).
04 5.4 Isomerism I: Stereoisomerism (geometrical and optical) PYQ 2017 · 2016 · 2015 · 2011
Isomers have the same formula but different arrangement of atoms. Stereoisomers have the same bonds but different spatial arrangement; structural isomers have different bonds. Stereoisomerism is of two kinds.
Geometrical isomerism arises in heteroleptic complexes because ligands can be next to each other (cis) or opposite (trans). In square planar [Ma2b2] there are cis and trans forms, the classic being [Pt(NH3)2Cl2]. Square planar [Mabcd] gives three isomers, for example [Pt(NH3)(Br)(Cl)(py)], because each ligand can be trans to any one of the other three. In octahedral [Ma4b2] there are cis and trans forms, [Co(NH3)4Cl2]+; the same holds for [M(aa)2b2] with didentate aa, [CoCl2(en)2]+. Octahedral [Ma3b3] gives the facial (fac) isomer, three identical ligands on one triangular face, and the meridional (mer) isomer, three identical ligands around a meridian; [Co(NH3)3(NO2)3] is the NCERT example. Tetrahedral complexes show no geometrical isomerism because all four positions are adjacent to one another.
Optical isomerism: two isomers that are non-superimposable mirror images, called dextro (d) and laevo (l), rotate plane polarised light in opposite directions. It is common in octahedral complexes with didentate ligands. [Co(en)3]3+ and [Cr(C2O4)3]3− are optically active. In [PtCl2(en)2]2+ and [CoCl2(en)2]+ only the cis isomer is optically active; the trans isomer has a plane of symmetry and is inactive. So [CoCl2(en)2]+ has three isomers in all: trans, and the d and l forms of cis. This three-isomer count is the most repeated drawing question from the chapter.
05 5.4 Isomerism II: Structural isomerism (linkage, coordination, ionisation, solvate) PYQ 2017 · 2015 · 2014 · 2013 · 2010
Linkage isomerism arises with an ambidentate ligand. The NO2− ligand in [Co(NH3)5(NO2)]Cl2 can bind through nitrogen, giving the yellow nitrito-N complex, or through oxygen, giving the red nitrito-O complex [Co(NH3)5(ONO)]Cl2. SCN− behaves the same way through S or N.
Coordination isomerism arises when both cation and anion are complexes and the ligands are interchanged between the two metals: [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
Ionisation isomerism arises when a counter ion is itself a potential ligand and can swap places with a ligand inside the sphere: [Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4. They give different ions in solution, and that is the test the board asks for: [Co(NH3)5Br]SO4 gives a white precipitate of BaSO4 with BaCl2 but nothing with AgNO3; [Co(NH3)5(SO4)]Br gives a pale yellow precipitate of AgBr with AgNO3 but nothing with BaCl2. For [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl the same logic works with BaCl2 and AgNO3.
Solvate (hydrate) isomerism is ionisation isomerism where the ligand that swaps is the solvent, usually water. CrCl3·6H2O exists as violet [Cr(H2O)6]Cl3, grey-green [Cr(H2O)5Cl]Cl2·H2O and dark green [Cr(H2O)4Cl2]Cl·2H2O. They are distinguished by the moles of AgCl precipitated per mole (3, 2 and 1) and by the number of water molecules lost over concentrated H2SO4 (0, 1 and 2 respectively). A common 3-mark question gives one of these observations and asks for the structural formula and IUPAC name.
06 5.5.1–5.5.3 Valence Bond Theory, magnetic properties and its limitations PYQ 2017 · 2016 · 2013
According to VBT the metal atom or ion, under the influence of ligands, uses its (n−1)d, ns, np or ns, np, nd orbitals to form a set of equivalent hybrid orbitals, which overlap with the filled orbitals of the ligands. The hybridisation fixes the geometry: coordination number 4 gives sp3 (tetrahedral) or dsp2 (square planar); 5 gives sp3d (trigonal bipyramidal); 6 gives d2sp3 (octahedral, inner orbital, uses the inner (n−1)d orbitals) or sp3d2 (octahedral, outer orbital, uses the outer nd orbitals). Strong field ligands such as CN−, CO and NH3 (for Co3+) force pairing of the metal d electrons so that inner d orbitals become vacant; weak field ligands such as F−, Cl− and H2O do not.
The examples to reproduce with orbital diagrams: [Co(NH3)6]3+, Co3+ is 3d6; NH3 pairs the electrons into three 3d orbitals, two 3d, one 4s and three 4p form d2sp3: inner orbital, diamagnetic. [CoF6]3−: no pairing, sp3d2, outer orbital, four unpaired electrons, paramagnetic. [Ni(CN)4]2−, Ni2+ is 3d8; CN− pairs the electrons into four 3d orbitals, leaving one 3d empty for dsp2: square planar, diamagnetic. [NiCl4]2−: no pairing, sp3, tetrahedral, two unpaired electrons, paramagnetic. [Ni(CO)4]: Ni(0) is 3d84s2; CO pushes the 4s electrons into 3d giving 3d10, sp3, tetrahedral, diamagnetic. [Fe(CN)6]4−, Fe2+ d6, d2sp3, diamagnetic; [Fe(CN)6]3−, Fe3+ d5, d2sp3, one unpaired electron, weakly paramagnetic; [FeF6]3− and [Fe(H2O)6]3+, sp3d2, five unpaired electrons, strongly paramagnetic. [Ni(NH3)6]2+ is outer orbital (sp3d2) because d8 cannot free two inner d orbitals even with pairing.
Magnetic moment is measured and compared with μ = √(n(n+2)) BM. For d1, d2, d3 the number of unpaired electrons is the same in the free ion and the octahedral complex. For d4 to d6 the result depends on the ligand: [Mn(CN)6]3− has two unpaired electrons while [MnCl6]3− has four; [Fe(CN)6]3− has one while [FeF6]3− has five; [CoF6]3− has four while [Co(C2O4)3]3− is diamagnetic. VBT explains this as inner versus outer orbital complexes.
Limitations of VBT: it involves a number of assumptions; it gives no quantitative interpretation of magnetic data; it does not explain the colour of complexes; it gives no quantitative interpretation of thermodynamic or kinetic stability; it cannot predict whether a 4-coordinate complex will be tetrahedral or square planar; and it does not distinguish weak and strong field ligands.
07 5.5.4 Crystal Field Theory PYQ 2015 · 2013 · 2011
CFT treats the metal–ligand bond as purely ionic: the ligands are point charges (anions) or point dipoles (polar molecules). In a free ion the five d orbitals are degenerate. When ligands approach, the electrons in orbitals pointing at the ligands are repelled more than those in orbitals pointing between them, so the degeneracy is lifted. This is crystal field splitting.
Octahedral field: six ligands lie on the axes. The dx2−y2 and dz2 orbitals point at the ligands and are raised in energy; they form the eg set. The dxy, dyz and dzx orbitals point between the ligands and are lowered; they form the t2g set. The gap is Δo, the crystal field splitting energy. The average energy is preserved, so t2g lies 0.4Δo below and eg 0.6Δo above the barycentre. Δo depends on the field strength of the ligand and on the charge on the metal ion. Ligands arranged in order of increasing field strength give the spectrochemical series: I− < Br− < S2− < SCN− < Cl− < F− < OH− < C2O42− < H2O < NCS− < edta4− < NH3 < en < CN− < CO. It is an experimental series based on absorption spectra.
Filling: d1, d2, d3 go into t2g one each, by Hund's rule. For d4 the fourth electron has a choice: if Δo < P (pairing energy), it goes into eg, giving t2g3eg1, a high spin complex with a weak field ligand; if Δo > P, it pairs in t2g, giving t2g4eg0, a low spin complex with a strong field ligand. The same choice arises for d5, d6 and d7. So [Mn(H2O)6]2+ (d5, weak field) is t2g3eg2 with five unpaired electrons, while [Mn(CN)6]4− is t2g5eg0 with one. [Co(NH3)6]3+ (d6) is t2g6, diamagnetic; [CoF6]3− is t2g4eg2 with four unpaired electrons.
Tetrahedral field: four ligands lie between the axes, so the splitting is reversed, the e set (dx2−y2, dz2) lies lower and t2 higher, and it is smaller: Δt = (4/9)Δo. Because Δt is usually less than the pairing energy, tetrahedral complexes are almost always high spin; low spin tetrahedral complexes are rarely observed.
08 5.5.5–5.5.6 Colour in coordination compounds and limitations of CFT PYQ 2017
A complex is coloured because it absorbs light of one colour from the visible region and transmits the rest; the colour we see is the complementary colour of the light absorbed. In CFT the absorbed energy promotes an electron from t2g to eg, a d–d transition, and the energy needed is Δo, which for most complexes falls in the visible range. The NCERT example is [Ti(H2O)6]3+, a d1 ion: it absorbs light of about 500 nm in the blue-green region to move its single electron from t2g to eg, and the transmitted light gives the complex its violet colour. Some values from the NCERT table: [CoCl(NH3)5]2+ absorbs 535 nm (yellow) and looks violet; [Co(NH3)5(H2O)]3+ absorbs 500 nm (blue-green) and looks red; [Co(NH3)6]3+ absorbs 475 nm (blue) and looks yellow-orange; [Cu(H2O)4]2+ absorbs 600 nm (red) and looks blue.
Three consequences are examined. First, if there is no ligand there is no splitting and no colour: anhydrous CuSO4 is white, CuSO4·5H2O is blue. Second, if the d subshell is empty or full there is no d–d transition: [Sc(H2O)6]3+ (d0) and Zn2+ complexes (d10) are colourless. Third, the ligand changes the colour because it changes Δo: [Ni(H2O)6]2+ is green, and adding ethane-1,2-diamine step by step gives [Ni(H2O)4(en)]2+ pale blue, [Ni(H2O)2(en)2]2+ blue-purple and [Ni(en)3]2+ violet. A stronger ligand means larger Δo, higher energy absorbed, shorter wavelength absorbed. This is why the wavelength of absorption follows [Ni(NO2)6]4− < [Ni(NH3)6]2+ < [Ni(H2O)6]2+. The colour of gemstones is the same effect in a lattice: ruby is Al2O3 with about 0.5 to 1 per cent Cr3+ replacing Al3+ in octahedral sites; emerald is Cr3+ in beryl, Be3Al2Si6O18.
Limitations of CFT. (i) It treats ligands as point charges, so anionic ligands should give the largest splitting; in fact they lie at the low end of the spectrochemical series. (ii) It ignores the covalent character of the metal–ligand bond, which is real and explains why CO and CN− are such strong field ligands.
09 5.6 Bonding in Metal Carbonyls PYQ 2015 · 2010 · 2009
Homoleptic carbonyls, complexes containing only carbonyl ligands, are formed by most transition metals. They have simple, well defined shapes: [Ni(CO)4] is tetrahedral, [Fe(CO)5] trigonal bipyramidal, [Cr(CO)6] octahedral. The binuclear carbonyls are [Mn2(CO)10], made of two square pyramidal Mn(CO)5 units joined by a Mn–Mn bond, and [Co2(CO)8], which has a Co–Co bond bridged by two CO groups in addition to the six terminal CO groups. The metal is in oxidation state zero in all of these.
The metal–carbon bond has both σ and π character. The σ bond is formed by donation of the lone pair on the carbon atom of CO into a vacant orbital of the metal. The π bond is formed by donation of a pair of electrons from a filled d orbital of the metal into the vacant antibonding π* orbital of carbon monoxide. The metal-to-ligand π bonding creates a synergic effect: electron density pushed back into the ligand strengthens the metal–CO bond, and the extra negative charge on the metal from σ donation is relieved. This is why CO, a poor Lewis base towards ordinary acids, binds transition metals so strongly and sits at the top of the spectrochemical series.
The board asks two things here: draw and describe the synergic bonding (a diagram with the σ arrow from C to M and the π arrow from M d orbital to the π* of CO), and state the shapes of [Ni(CO)4], [Fe(CO)5] and [Cr(CO)6]. The oxidation state of Ni in [Ni(CO)4] is zero and its hybridisation is sp3, with the 4s electrons pushed into the 3d subshell so the complex is diamagnetic. Metal carbonyls are also the basis of the Mond process for refining nickel: impure nickel with CO at 330–350 K gives volatile [Ni(CO)4], which decomposes at 450–470 K to pure nickel.
10 5.7 Importance and Applications of Coordination Compounds PYQ 2013 · 2011
The board asks for one example under each heading, so learn them as a list. In qualitative and quantitative analysis: metal ions are detected and estimated through coloured complexes; Ni2+ gives a red precipitate with dimethylglyoxime (DMG); Cu2+ gives deep blue [Cu(NH3)4]2+; the hardness of water is estimated by titrating Ca2+ and Mg2+ against EDTA, and the two can be estimated separately because their EDTA complexes have different stability constants.
In metallurgy: silver and gold are extracted by leaching with cyanide in the presence of air, 4Au + 8CN− + 2H2O + O2 → 4[Au(CN)2]− + 4OH−, and the metal is recovered with zinc, 2[Au(CN)2]− + Zn → 2Au + [Zn(CN)4]2−. Nickel is purified by the Mond process through [Ni(CO)4]. In catalysis: Wilkinson's catalyst, [(Ph3P)3RhCl], is used for hydrogenation of alkenes. In biology: chlorophyll (the pigment of photosynthesis) is a magnesium complex, haemoglobin (the oxygen carrier of blood) is an iron complex, and vitamin B12 (cyanocobalamine, the anti-pernicious anaemia factor) is a cobalt complex; enzymes such as carboxypeptidase A and carbonic anhydrase are metal complexes too.
In medicine: cisplatin, cis-[Pt(NH3)2Cl2], is used in cancer chemotherapy; EDTA is used in the treatment of lead poisoning by chelation; excess copper and iron are removed by the chelating ligands D-penicillamine and desferrioxamine B respectively. In electroplating: articles are coated smoothly with silver and gold from baths of the cyanide complexes [Ag(CN)2]− and [Au(CN)2]− rather than from simple salts. In photography: after the film is developed, hypo (sodium thiosulphate) dissolves the unreacted AgBr as the soluble complex [Ag(S2O3)2]3−, which fixes the image.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Oxidation number of the central metal | x + (sum of ligand charges) = charge on the complex ion | Neutral ligands count 0; for a neutral salt the charge on the complex ion balances the counter ions. |
| Spin-only magnetic moment | μ = √(n(n + 2)) BM | n = 1, 2, 3, 4, 5 → 1.73, 2.83, 3.87, 4.90, 5.92 BM; 1 BM = 9.27 × 10−24 A m2. |
| Octahedral splitting | t2g (dxy, dyz, dzx) at −0.4Δo; eg (dx2−y2, dz2) at +0.6Δo | Barycentre preserved: 3 × 0.4 = 2 × 0.6. |
| Tetrahedral splitting | Δt = (4/9)Δo; e below t2 | Small splitting, so tetrahedral complexes are high spin. |
| High spin or low spin (d<sup>4</sup>–d<sup>7</sup>) | Δo < P → high spin; Δo > P → low spin | d4: t2g3eg1 (weak field) or t2g4eg0 (strong field). |
| Hybridisation and geometry (VBT) | sp3 tetrahedral; dsp2 square planar; sp3d trigonal bipyramidal; d2sp3 and sp3d2 octahedral | d2sp3 = inner orbital (low spin), sp3d2 = outer orbital (high spin). |
| Energy of light absorbed | Δo = hν = hc/λ | Stronger ligand → larger Δo → shorter wavelength absorbed; observed colour is complementary. |
| Isomer counts to remember | [Ma2b2] square planar: 2; [Mabcd] square planar: 3; [Ma4b2] octahedral: 2; [Ma3b3]: 2 (fac, mer); [M(aa)2b2]: 3 (trans, d-cis, l-cis); [M(aa)3]: 2 (d, l) | Tetrahedral complexes: no geometrical isomers. |
| Werner evidence | CoCl3·6NH3 → 3 AgCl; ·5NH3 → 2; ·4NH3 → 1; ·3NH3 → 0 | Moles of AgCl per mole of complex = number of Cl− outside the sphere; conductance falls in the same order. |
| Cyanide extraction of gold | 4Au + 8CN− + 2H2O + O2 → 4[Au(CN)2]− + 4OH−; 2[Au(CN)2]− + Zn → 2Au + [Zn(CN)4]2− | Same scheme for silver. |
Practise with a pencilDiagrams that earn marks
Splitting of d orbitals in an octahedral crystal field
On the left draw five short horizontal lines at one level labelled 'free metal ion, degenerate d orbitals'. Draw a dashed horizontal line to the right at the same level (barycentre, 'average energy of d orbitals in spherical field'). Above it draw two lines labelled eg (dx2−y2, dz2) with '+0.6Δo', below it three lines labelled t2g (dxy, dyz, dzx) with '−0.4Δo'. Mark the gap between them with a double-headed arrow labelled Δo. Write 'energy' up the y-axis.
Splitting of d orbitals in a tetrahedral crystal field
Same layout as the octahedral diagram but reversed: two lines (e set, dx2−y2 and dz2) below the barycentre and three lines (t2 set, dxy, dyz, dzx) above it. Mark the gap Δt visibly smaller than in the octahedral case and write Δt = (4/9)Δo beside it. Note the labels drop the g subscript because a tetrahedron has no centre of symmetry.
Geometrical isomers: cis and trans [Co(NH<sub>3</sub>)<sub>4</sub>Cl<sub>2</sub>]<sup>+</sup>, fac and mer [Co(NH<sub>3</sub>)<sub>3</sub>(NO<sub>2</sub>)<sub>3</sub>]
Draw an octahedron as Co at the centre with four ligands in a square plane (two solid, two dashed wedges) and one above and one below. For cis put the two Cl adjacent (one in the plane, one on top, or two adjacent in the plane); for trans put them at the top and bottom. For fac put the three NO2 groups on three mutually adjacent positions forming a triangular face; for mer put them in a line through the metal (top, bottom and one in the plane). Label every ligand and write the charge on the bracket.
Optical isomers of [Co(en)<sub>3</sub>]<sup>3+</sup> and cis-[CoCl<sub>2</sub>(en)<sub>2</sub>]<sup>+</sup>
Draw the octahedron with each en shown as a curved line joining two adjacent positions (label each end N). Draw a vertical dashed line as the mirror and draw the mirror image beside it. Label them d and l and write 'non-superimposable mirror images'. For cis-[CoCl2(en)2]+ put the two Cl adjacent and the two en on the remaining pairs, then mirror it; add the trans form separately with the note 'plane of symmetry, optically inactive'.
Synergic bonding in a metal carbonyl
Draw M on the left and C≡O on the right. Show a filled lobe on carbon (lone pair) overlapping an empty metal orbital with an arrow from C to M labelled 'σ donation, OC → M'. Below it show a filled metal d orbital (four-lobed) overlapping the empty π* orbital of CO (lobes on C and O with opposite signs) with an arrow from M to CO labelled 'π back-donation, M → CO'. Write 'synergic effect: each bond strengthens the other' beneath.
TextbookNCERT questions the board keeps asking
5.1 Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanidonickelate(II) (iii) Tris(ethane-1,2-diamine)chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate (vi) Iron(III) hexacyanidoferrate(II)3 m
(i) [Co(NH3)4(H2O)2]Cl3 (ii) K2[Ni(CN)4] (iii) [Cr(en)3]Cl3 (iv) [Pt(NH3)BrCl(NO2)]− (v) [PtCl2(en)2](NO3)2 (vi) Fe4[Fe(CN)6]3
Check each by charge: in (iv) Pt(II) with Br−, Cl− and NO2− gives +2 − 3 = −1, so the entity is an anion, which is why the name ends in platinate. In (vi) each [Fe(CN)6]4− needs 4 positive charges; four Fe3+ (12+) balance three anions (12−).
5.2 FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why.2 m
FeSO4 and (NH4)2SO4 in 1:1 ratio form the double salt Mohr's salt, FeSO4·(NH4)2SO4·6H2O. A double salt dissociates completely in water into its simple ions, Fe2+, NH4+ and SO42−, so Fe2+ is present and gives its test.
CuSO4 and NH3 in 1:4 ratio form the coordination compound [Cu(NH3)4]SO4. The complex ion [Cu(NH3)4]2+ retains its identity in solution and does not release free Cu2+ ions, so the test of Cu2+ is not given.
5.5 Specify the oxidation numbers of the metals in the following coordination entities: (i) [Co(H2O)(CN)(en)2]2+ (ii) [CoBr2(en)2]+ (iii) [PtCl4]2− (iv) K3[Fe(CN)6] (v) [Cr(NH3)3Cl3]2 m
(i) x + 0 + (−1) + 0 = +2, so x = +3: Co(III). (ii) x + 2(−1) + 0 = +1, so x = +3: Co(III). (iii) x + 4(−1) = −2, so x = +2: Pt(II). (iv) Three K+ balance [Fe(CN)6]3−; x + 6(−1) = −3, so x = +3: Fe(III). (v) x + 0 + 3(−1) = 0, so x = +3: Cr(III).
5.7 Using IUPAC norms write the systematic names of the following: (i) [Co(NH3)6]Cl3 (ii) [Pt(NH3)2Cl(NH2CH3)]Cl (iii) [Ti(H2O)6]3+ (iv) [Co(NH3)4Cl(NO2)]Cl (v) [Mn(H2O)6]2+ (vi) [NiCl4]2− (vii) [Ni(NH3)6]Cl2 (viii) [Co(en)3]3+ (ix) [Ni(CO)4]3 m
(i) Hexaamminecobalt(III) chloride (ii) Diamminechlorido(methylamine)platinum(II) chloride (iii) Hexaaquatitanium(III) ion (iv) Tetraamminechloridonitrito-N-cobalt(III) chloride (v) Hexaaquamanganese(II) ion (vi) Tetrachloridonickelate(II) ion (vii) Hexaamminenickel(II) chloride (viii) Tris(ethane-1,2-diamine)cobalt(III) ion (ix) Tetracarbonylnickel(0)
Points to notice: in (vi) the complex is an anion, hence nickelate; in (ix) the metal is in oxidation state zero because CO is neutral; in (viii) tris is used because the ligand name already contains 'di'.
5.9 How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]2 m
(i) None. [Cr(C2O4)3]3− is a homoleptic complex of three identical didentate ligands, so there is only one spatial arrangement; it shows optical isomerism (d and l forms) but not geometrical isomerism.
(ii) Two. [Co(NH3)3Cl3] is of the type [Ma3b3] and exists as the facial (fac) isomer, with the three Cl on one triangular face, and the meridional (mer) isomer, with the three Cl in a plane through the metal.
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?3 m
The complex is square planar of the type [Mabcd], so three geometrical isomers exist, one for each ligand that can be placed trans to NH3: (1) NH3 trans to Br (so Cl trans to py); (2) NH3 trans to Cl (so Br trans to py); (3) NH3 trans to py (so Br trans to Cl). Draw each as a square with Pt at the centre and the four ligands at the corners.
None of them is optically active. A square planar complex has the molecular plane as a plane of symmetry, so each isomer is superimposable on its mirror image.
5.13 Aqueous copper sulphate solution (blue in colour) gives (i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.2 m
In aqueous solution copper sulphate exists as the blue aqua complex [Cu(H2O)4]2+.
(i) With KF the water ligands are replaced by fluoride ions to give [CuF4]2−, which is a green precipitate: [Cu(H2O)4]2+ + 4F− → [CuF4]2− + 4H2O.
(ii) With KCl the water ligands are replaced by chloride ions to give [CuCl4]2−, which is soluble and bright green: [Cu(H2O)4]2+ + 4Cl− → [CuCl4]2− + 4H2O.
The colour changes because F− and Cl− produce a different crystal field splitting from H2O, so a different wavelength is absorbed.
5.14 What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?2 m
Cyanide first reduces Cu2+ to Cu+ and the excess CN− then forms the complex: 2CuSO4 + 10KCN → 2K3[Cu(CN)4] + 2K2SO4 + (CN)2. The coordination entity is [Cu(CN)4]3−, tetracyanidocuprate(I).
This complex is extremely stable (CN− is a strong field ligand and the complex has a very high stability constant), so the concentration of free copper ions in solution is negligible. When H2S is passed there are not enough free copper ions to exceed the solubility product of copper sulphide, and no precipitate forms.
5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6]4− (ii) [FeF6]3− (iii) [Co(C2O4)3]3− (iv) [CoF6]3−5 m
(i) [Fe(CN)6]4−: Fe2+ is 3d6. CN− is a strong field ligand and pairs the six electrons into three 3d orbitals. The two vacant 3d, one 4s and three 4p orbitals hybridise as d2sp3. Octahedral, inner orbital, no unpaired electron, diamagnetic.
(ii) [FeF6]3−: Fe3+ is 3d5. F− is a weak field ligand and causes no pairing, so the five 3d orbitals each hold one electron. Hybridisation uses 4s, three 4p and two 4d orbitals: sp3d2. Octahedral, outer orbital, five unpaired electrons, strongly paramagnetic.
(iii) [Co(C2O4)3]3−: Co3+ is 3d6. Oxalate acts as a strong enough field towards Co3+ to pair the electrons into three 3d orbitals; d2sp3, octahedral, inner orbital, diamagnetic.
(iv) [CoF6]3−: Co3+ is 3d6. F− causes no pairing, so the configuration keeps four unpaired electrons; sp3d2, octahedral, outer orbital, paramagnetic with μ = √(4 × 6) = 4.90 BM.
5.19 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why.2 m
[Cr(NH3)6]3+: Cr3+ is 3d3. The three electrons occupy three 3d orbitals singly; the remaining two 3d orbitals, the 4s and three 4p orbitals form d2sp3 hybrids. The three unpaired electrons make the complex paramagnetic (μ = √15 = 3.87 BM).
[Ni(CN)4]2−: Ni2+ is 3d8. CN− is a strong field ligand and pairs the eight electrons into four 3d orbitals. The vacant 3d, 4s and two 4p orbitals form dsp2 hybrids, giving a square planar complex with no unpaired electron, so it is diamagnetic.
5.20 A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.2 m
In [Ni(H2O)6]2+, Ni2+ is 3d8 and H2O is a weak field ligand. Two unpaired electrons remain and d–d transitions between the t2g and eg levels are possible; the complex absorbs red light and appears green.
In [Ni(CN)4]2−, CN− is a strong field ligand that pairs all eight d electrons (dsp2, square planar). With no unpaired electron there is no d–d transition in the visible region, so no visible light is absorbed and the solution is colourless.
5.21 [Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?2 m
Both contain Fe2+ (3d6) in an octahedral field, but the ligands differ in field strength. CN− is a strong field ligand and produces a large Δo; H2O is a weak field ligand and produces a small Δo. The energy, and hence the wavelength, of the light absorbed in the d–d transition is therefore different for the two complexes (Δo = hc/λ), so the complementary colour transmitted is different: [Fe(CN)6]4− is pale yellow while [Fe(H2O)6]2+ is pale green.
5.10 The hexaaquamanganese(II) ion contains five unpaired electrons, while the hexacyanido ion contains only one unpaired electron. Explain using crystal field theory.2 m
Mn2+ is 3d5. In [Mn(H2O)6]2+, H2O is a weak field ligand, so Δo < P. The electrons occupy all five orbitals singly, t2g3eg2, a high spin complex with five unpaired electrons.
In [Mn(CN)6]4−, CN− is a strong field ligand, so Δo > P. It is energetically cheaper to pair electrons in t2g than to promote them to eg: t2g5eg0, a low spin complex with only one unpaired electron.
5.23 Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes: (i) K3[Co(C2O4)3] (ii) cis-[CrCl2(en)2]Cl (iii) (NH4)2[CoF4] (iv) [Mn(H2O)6]SO43 m
(i) Co: x + 3(−2) = −3, x = +3; Co3+ is d6, and with oxalate it is low spin t2g6eg0; coordination number 6 (three didentate ligands).
(ii) Cr: x + 2(−1) + 0 = +1, x = +3; Cr3+ is d3, t2g3; coordination number 6.
(iii) Co: x + 4(−1) = −2, x = +2; Co2+ is d7, high spin e4t23 in the tetrahedral field; coordination number 4.
(iv) Mn: x + 0 = +2; Mn2+ is d5, high spin t2g3eg2; coordination number 6.
5.32 What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6]4−, [Ni(NH3)6]2+, [Ni(H2O)6]2+?2 m
The metal ion is Ni2+ in all three, so the splitting depends only on the ligand. Field strength from the spectrochemical series: H2O < NH3 < NO2−, so Δo increases in the order [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4−.
Since Δo = hc/λ, a larger splitting means a shorter wavelength absorbed. Order of wavelength of absorption: [Ni(NO2)6]4− < [Ni(NH3)6]2+ < [Ni(H2O)6]2+.
Previous years and frequently askedBoard questions with model answers
MCQ The number of ions produced from the complex [Co(NH3)6]Cl2 in aqueous solution is:- 6
- 4
- 3
- 2
Frequently asked1 m
MCQ The oxidation number of cobalt in K[Co(CO)4] is:- +1
- +3
- −1
- −3
Frequently asked1 m
MCQ The crystal field splitting energy in a tetrahedral complex (Δt) is related to that in an octahedral complex (Δo) of the same metal and ligands as:- Δt = Δo
- Δt = (4/9)Δo
- Δt = (9/4)Δo
- Δt = ½Δo
Frequently asked1 m
MCQ Which of the following is an ambidentate ligand?- NH3
- NO2−
- en
- C2O42−
Frequently asked1 m
MCQ Amongst the following, the most stable complex is:- [Fe(H2O)6]3+
- [Fe(NH3)6]3+
- [Fe(C2O4)3]3−
- [FeCl6]3−
Frequently asked1 m
MCQ Amongst the following ions, the one with the highest magnetic moment value is:- [Cr(H2O)6]3+
- [Fe(H2O)6]2+
- [Zn(H2O)6]2+
- [Ni(H2O)6]2+
Frequently asked1 m
MCQ Which of the following complexes shows geometrical isomerism?- [Ni(CO)4]
- [Pt(NH3)2Cl2]
- [Co(NH3)6]Cl3
- [Ni(NH3)4]2+
Frequently asked1 m
MCQ The correct order of the wavelength of light absorbed by the following complexes is:- [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4−
- [Ni(NO2)6]4− < [Ni(NH3)6]2+ < [Ni(H2O)6]2+
- [Ni(NH3)6]2+ < [Ni(H2O)6]2+ < [Ni(NO2)6]4−
- All three absorb at the same wavelength
Frequently asked1 m
Assertion–Reason Assertion (A): [Ni(CN)4]2− is diamagnetic and square planar. Reason (R): CN− is a strong field ligand that pairs the 3d electrons of Ni2+, freeing one 3d orbital for dsp2 hybridisation.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): cis-[Pt(NH3)2Cl2] and trans-[Pt(NH3)2Cl2] are geometrical isomers. Reason (R): Geometrical isomerism is commonly shown by tetrahedral complexes.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): [FeF6]3− is diamagnetic. Reason (R): F− is a weak field ligand.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Anhydrous CuSO4 is white whereas CuSO4·5H2O is blue. Reason (R): In the absence of ligands crystal field splitting does not occur, so no d–d transition is possible.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer Give one example each of (i) linkage isomerism, (ii) coordination isomerism and (iii) ionisation isomerism.Frequently asked2 m
(i) Linkage: [Co(NH3)5(NO2)]Cl2 (nitrito-N, yellow) and [Co(NH3)5(ONO)]Cl2 (nitrito-O, red).
(ii) Coordination: [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
(iii) Ionisation: [Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4.
Write both members of each pair; a single formula does not show the isomerism.
Very short answer Using IUPAC norms write the formulas of (i) Pentaamminenitrito-O-cobalt(III) ion and (ii) Potassium trioxalatochromate(III). Write the IUPAC name of [Co(NH3)5(CO3)]Cl.Frequently asked2 m
(i) [Co(NH3)5(ONO)]2+: Co(III) with five neutral NH3 and one NO2− bound through oxygen gives charge +3 − 1 = +2.
(ii) K3[Cr(C2O4)3]: Cr(III) with three oxalate (2−) ligands gives a 3− anion, balanced by three K+.
[Co(NH3)5(CO3)]Cl: pentaamminecarbonatocobalt(III) chloride. Check: x + 0 + (−2) = +1 (one Cl− outside), so x = +3.
Very short answer What is meant by the chelate effect? Give an example.CBSE 20152 m
When a didentate or polydentate ligand binds a metal ion through two or more of its donor atoms it forms a ring (a chelate). Complexes containing such rings are more stable than complexes with the same number of unidentate ligands of similar donor atoms; this extra stability is called the chelate effect.
Example: [Ni(en)3]2+, with three five-membered chelate rings, is far more stable than [Ni(NH3)6]2+ although both have six nitrogen donors. Similarly [Fe(C2O4)3]3− is more stable than [FeCl6]3−.
Very short answer Why are low spin tetrahedral complexes not formed?CBSE 20172 m
In a tetrahedral field only four ligands surround the metal and none of them points directly at a d orbital, so the crystal field splitting is small: Δt = (4/9)Δo. This splitting is almost always less than the pairing energy P. It is therefore energetically cheaper to place electrons in the higher t2 set than to pair them in the lower e set, and the complex stays high spin. Low spin tetrahedral complexes are hence rarely observed.
Very short answer [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?Frequently asked2 m
In [NiCl4]2− nickel is Ni2+, 3d8. Cl− is a weak field ligand and causes no pairing, so two electrons remain unpaired in the 3d subshell while the 4s and three 4p orbitals form sp3 hybrids; the complex is paramagnetic (μ = √8 = 2.83 BM).
In [Ni(CO)4] nickel is Ni(0), 3d84s2. CO is a strong field ligand; it forces the two 4s electrons into the 3d subshell giving 3d10 with all electrons paired. The empty 4s and 4p orbitals form sp3 hybrids and the complex is diamagnetic.
Short answer (a) Write the hybridisation and magnetic character of the complex [Fe(CN)6]4− on the basis of valence bond theory (atomic number of Fe = 26). (b) Draw the geometrical isomers of [Pt(NH3)2Cl2].Frequently asked3 m
(a) Fe (Z = 26) is [Ar]3d64s2; Fe2+ is 3d6. CN− is a strong field ligand and pairs the six electrons into three 3d orbitals. The two vacant 3d, one 4s and three 4p orbitals give d2sp3 hybridisation. The complex is octahedral, an inner orbital complex, and with no unpaired electron it is diamagnetic. Draw the orbital box diagram showing the six paired electrons and the six hybrid orbitals filled by the CN− lone pairs. (2 marks)
(b) [Pt(NH3)2Cl2] is square planar. Draw Pt at the centre of a square: the cis isomer has the two Cl on adjacent corners and the two NH3 on the other two adjacent corners; the trans isomer has the two Cl on opposite corners and the two NH3 on the other pair of opposite corners. Label them cis (cisplatin) and trans. (1 mark)
Short answer Explain the following giving reasons: (i) [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. (ii) [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. (iii) [Ti(H2O)6]3+ is coloured whereas [Sc(H2O)6]3+ is colourless.Frequently asked3 m
(i) Fe3+ is 3d5. H2O is a weak field ligand: no pairing, sp3d2, five unpaired electrons, μ = 5.92 BM, strongly paramagnetic. CN− is a strong field ligand: the five electrons pair into three 3d orbitals leaving one unpaired, d2sp3, μ = 1.73 BM, weakly paramagnetic.
(ii) Co3+ is 3d6; NH3 pairs the electrons into three 3d orbitals, leaving two inner 3d orbitals free for d2sp3 hybridisation, so the complex is inner orbital. Ni2+ is 3d8; even after pairing, four 3d orbitals are occupied and only one is free, which is not enough for d2sp3. Nickel must use the outer 4d orbitals, sp3d2, so the complex is outer orbital.
(iii) Ti3+ is 3d1. The single electron in t2g can absorb visible light in the blue-green region (about 500 nm) and jump to eg; this d–d transition makes the complex violet. Sc3+ is 3d0; with no d electron there is no d–d transition and the complex is colourless.
Short answer What is crystal field splitting energy? Write the electronic configuration of a d4 ion in an octahedral field when (i) Δo > P and (ii) Δo < P. Give one example of each type.Frequently asked3 m
When ligands approach a metal ion the five degenerate d orbitals split into a lower t2g set (dxy, dyz, dzx) and a higher eg set (dx2−y2, dz2) in an octahedral field. The energy difference between the two sets is the crystal field splitting energy, Δo. (1 mark)
(i) Δo > P (strong field ligand): the fourth electron pairs in t2g; configuration t2g4eg0, low spin, two unpaired electrons. Example: [Mn(CN)6]3−. (1 mark)
(ii) Δo < P (weak field ligand): the fourth electron enters eg; configuration t2g3eg1, high spin, four unpaired electrons. Example: [Mn(H2O)6]3+ or [MnCl6]3−. (1 mark)
Short answer (a) Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ionisation isomers. (b) A solution of CrCl3·6H2O gives one mole of AgCl per mole of the compound on treatment with excess AgNO3. Write its structural formula and IUPAC name.Frequently asked3 m
(a) The two compounds give different ions in solution. [Co(NH3)5Cl]SO4 ionises to give SO42−, so it gives a white precipitate of BaSO4 with BaCl2 solution but no precipitate with AgNO3. [Co(NH3)5(SO4)]Cl ionises to give Cl−, so it gives a white precipitate of AgCl with AgNO3 but none with BaCl2. The same formula with different ions in solution is the definition of ionisation isomerism. (2 marks)
(b) One mole of AgCl per mole means only one Cl− is outside the coordination sphere; the other two chlorides and four water molecules complete the coordination number of six, with two waters of crystallisation. Formula: [Cr(H2O)4Cl2]Cl·2H2O. Name: tetraaquadichloridochromium(III) chloride dihydrate. (1 mark)
Short answer Write the IUPAC name and indicate the type(s) of isomerism shown by each of the following: (i) K[Cr(H2O)2(C2O4)2] (ii) [Co(en)3]Cl3 (iii) [Co(NH3)5(NO2)](NO3)2Frequently asked3 m
(i) Potassium diaquadioxalatochromate(III). Cr: x + 0 + 2(−2) = −1, x = +3. It shows geometrical isomerism (cis and trans arrangement of the two water molecules) and the cis form also shows optical isomerism.
(ii) Tris(ethane-1,2-diamine)cobalt(III) chloride. Three identical didentate ligands give a chiral octahedron: optical isomerism (d and l forms).
(iii) Pentaamminenitrito-N-cobalt(III) nitrate. It shows linkage isomerism (NO2− can bind through O as nitrito-O) and ionisation isomerism (NO2− and NO3− can exchange positions, giving [Co(NH3)5(NO3)](NO2)(NO3)).
Long answer (a) On the basis of valence bond theory give the hybridisation, shape and magnetic moment of (i) [Co(NH3)6]3+ (ii) [CoF6]3− (iii) [NiCl4]2−. (Atomic numbers: Co = 27, Ni = 28.) (b) Draw the optical isomers of [Co(en)3]3+.Frequently asked5 m
(a) (i) Co (Z = 27) is 3d74s2; Co3+ is 3d6. NH3 is a strong field ligand for Co3+: the six electrons pair into three 3d orbitals. Two 3d, one 4s and three 4p orbitals form d2sp3 hybrids. Shape octahedral, inner orbital complex, no unpaired electron, μ = 0, diamagnetic. (1½ marks)
(ii) Co3+ 3d6 with the weak field ligand F−: no pairing, four unpaired electrons remain in 3d. The 4s, three 4p and two 4d orbitals form sp3d2 hybrids. Shape octahedral, outer orbital complex, μ = √(4 × 6) = √24 = 4.90 BM, paramagnetic. (1½ marks)
(iii) Ni (Z = 28) is 3d84s2; Ni2+ is 3d8. Cl− is weak field: no pairing, two unpaired electrons. The 4s and three 4p orbitals form sp3 hybrids. Shape tetrahedral, μ = √(2 × 4) = √8 = 2.83 BM, paramagnetic. (1 mark)
For each part draw the orbital box diagram of the free ion, then of the ion in the complex with the ligand lone pairs (shown as xx) in the hybrid orbitals.
(b) [Co(en)3]3+: draw the octahedron with Co at the centre and each en as an arc joining two adjacent N positions (three arcs). Draw a vertical mirror line and the mirror-image octahedron on the other side in which the arcs wind the opposite way. Label the two structures d and l, and state that they are non-superimposable mirror images, hence optically active. (1 mark)
Long answer (a) Draw the crystal field splitting diagram for d orbitals in an octahedral field and explain how the magnitude of Δo decides the configuration of a d4 ion. (b) What is the spectrochemical series? (c) Explain why [Ni(H2O)6]2+ is green but [Ni(CN)4]2− is colourless.Frequently asked5 m
(a) Draw the five degenerate d orbitals of the free ion on the left, the barycentre as a dashed line, eg (dx2−y2, dz2) raised by 0.6Δo and t2g (dxy, dyz, dzx) lowered by 0.4Δo, with the gap marked Δo. The eg orbitals point at the six ligands on the axes and are repelled more; t2g point between them. (1 mark for the diagram)
The first three electrons occupy t2g singly. For the fourth electron there are two options: enter eg, costing Δo, or pair in t2g, costing the pairing energy P. If Δo < P (weak field ligand) the configuration is t2g3eg1, high spin; if Δo > P (strong field ligand) it is t2g4eg0, low spin. (1 mark)
(b) The spectrochemical series is the arrangement of ligands in order of increasing crystal field splitting they produce, determined experimentally from absorption spectra: I− < Br− < S2− < SCN− < Cl− < F− < OH− < C2O42− < H2O < NCS− < edta4− < NH3 < en < CN− < CO. Ligands to the left of H2O are weak field, to the right strong field. (1 mark)
(c) In [Ni(H2O)6]2+, Ni2+ (3d8) with weak field water has two unpaired electrons and t2g → eg d–d transitions occur by absorbing red light, so the solution looks green. In [Ni(CN)4]2−, the strong field CN− pairs all eight electrons (dsp2, square planar); there is no unpaired electron and no d–d transition in the visible region, so it is colourless. (2 marks)
Long answer (a) State Werner's postulates of coordination compounds. How did he explain that CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3 and CoCl3·3NH3 precipitate 3, 2, 1 and 0 moles of AgCl respectively with excess AgNO3? (b) Discuss the nature of bonding in metal carbonyls.Frequently asked5 m
(a) Werner's postulates: (i) In coordination compounds metals show two types of linkages, primary and secondary. (ii) Primary valences are normally ionisable and are satisfied by negative ions. (iii) Secondary valences are non-ionisable, are satisfied by neutral molecules or negative ions, and the secondary valence (coordination number) is fixed for a metal. (iv) The ions or groups bound by secondary linkages have characteristic spatial arrangements corresponding to different coordination numbers. (2 marks)
Cobalt(III) has a secondary valence of six. Only chloride outside the coordination sphere is ionisable and precipitates with AgNO3: [Co(NH3)6]Cl3 has 3 Cl− outside (3 mol AgCl); [Co(NH3)5Cl]Cl2 has 2 outside (2 mol); [Co(NH3)4Cl2]Cl has 1 outside (1 mol); [Co(NH3)3Cl3] has all three chlorides inside the sphere and gives no precipitate. (1 mark)
(b) In metal carbonyls the metal–carbon bond has both σ and π character. The σ bond forms by donation of the lone pair on carbon into a vacant orbital of the metal (OC → M). The π bond forms by back-donation of a pair of electrons from a filled d orbital of the metal into the vacant antibonding π* orbital of CO (M → CO). The two bonds reinforce each other: σ donation increases electron density on the metal, which promotes π back-donation, which in turn strengthens the σ bond. This is synergic bonding. Draw the two overlaps with arrows in the correct directions. Examples: [Ni(CO)4] tetrahedral, [Fe(CO)5] trigonal bipyramidal, [Cr(CO)6] octahedral. (2 marks)
Case-based Read the passage and answer the questions that follow. Werner prepared a series of compounds of cobalt(III) chloride with ammonia, CoCl3·6NH3 (yellow), CoCl3·5NH3 (purple), CoCl3·4NH3 (green) and CoCl3·3NH3. When excess AgNO3 was added to their solutions the moles of AgCl precipitated per mole of compound were 3, 2, 1 and 0. Their molar conductances decreased in the same order. Werner explained this by proposing that cobalt has a primary valence of three and a fixed secondary valence of six, and that only the chloride satisfying the primary valence is free to ionise.
(i) Write the formulas of the four compounds as coordination compounds.
(ii) Write the IUPAC name of the compound that gives two moles of AgCl.
(iii) How many ions does one formula unit of CoCl3·4NH3 give in solution? Name the type of stereoisomerism it can show.
(iv) Why does CoCl3·3NH3 not conduct electricity in solution?Frequently asked4 m
(i) [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2, [Co(NH3)4Cl2]Cl, [Co(NH3)3Cl3]. (1 mark)
(ii) [Co(NH3)5Cl]Cl2: pentaamminechloridocobalt(III) chloride. (1 mark)
(iii) [Co(NH3)4Cl2]Cl gives two ions, [Co(NH3)4Cl2]+ and Cl−. Being [Ma4b2] octahedral, it shows geometrical (cis–trans) isomerism; the green compound is the trans form and the violet one the cis form. (1 mark)
(iv) In [Co(NH3)3Cl3] all three chlorides satisfy the secondary valence and lie inside the coordination sphere; the compound is a neutral molecule and gives no ions in solution, so it is a non-electrolyte. (1 mark)
Case-based Read the passage and answer the questions that follow. According to crystal field theory the five degenerate d orbitals of a metal ion split in an octahedral field into a lower t2g set and a higher eg set separated by Δo. Ligands that produce a large Δo are strong field ligands and those producing a small Δo are weak field ligands; the experimental order is the spectrochemical series. A complex is coloured when an electron absorbs visible light of energy Δo and moves from t2g to eg. [Ti(H2O)6]3+ absorbs light of about 500 nm in the blue-green region and appears violet.
(i) Write the electronic configuration of the d electrons in [Ti(H2O)6]3+ in terms of t2g and eg, and state why it is coloured.
(ii) Which of [Co(NH3)6]3+ and [CoF6]3− absorbs light of shorter wavelength? Give a reason.
(iii) Calculate the spin-only magnetic moment of [Fe(H2O)6]2+ (Z of Fe = 26).
(iv) Why is [Zn(H2O)6]2+ colourless?Frequently asked4 m
(i) Ti3+ is 3d1: t2g1eg0. The single electron absorbs blue-green light of about 500 nm to jump from t2g to eg; the complementary colour, violet, is transmitted. (1 mark)
(ii) [Co(NH3)6]3+. NH3 is a stronger field ligand than F−, so Δo is larger and, since Δo = hc/λ, the wavelength absorbed is shorter. (1 mark)
(iii) Fe2+ is 3d6; H2O is weak field, so the configuration is t2g4eg2 with n = 4 unpaired electrons. μ = √(n(n+2)) = √(4 × 6) = √24 = 4.90 BM. (1 mark)
(iv) Zn2+ is 3d10. All d orbitals are completely filled, so no d–d transition is possible and no visible light is absorbed. (1 mark)
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Naming ligands in the order they appear in the formula instead of alphabetically. In [Co(NH3)4Cl(NO2)]Cl the name starts with tetraammine, then chlorido, then nitrito-N; the examiner deducts for any other order.
- Forgetting the -ate ending for an anionic complex: writing potassium hexacyanidoiron(III) instead of hexacyanidoferrate(III). The same mistake appears with cuprate, nickelate, argentate and chromate.
- Saying a complex is coloured because it has unpaired electrons. Colour comes from a d–d transition of energy Δo in the visible region; a d0 or d10 ion is colourless because no such transition exists, and a filled low spin ion such as [Ni(CN)4]2− is colourless for the same reason.
- Claiming geometrical isomerism for a tetrahedral complex, or optical isomerism for the trans isomer of [Co(en)2Cl2]+. Tetrahedral positions are all adjacent; the trans isomer has a plane of symmetry. Only cis is optically active, so the total isomer count is three.
- Mixing up inner and outer orbital: d2sp3 uses the inner (n−1)d orbitals and is inner orbital, low spin; sp3d2 uses the outer nd orbitals and is outer orbital, high spin. Writing the hybridisation without naming the orbitals loses the reasoning mark.
- Reversing the arrows in synergic bonding: σ donation goes from the CO lone pair to the metal, π back-donation goes from a filled metal d orbital to the π* orbital of CO.
- Treating NO2− or SCN− as didentate. They are ambidentate: two possible donor atoms, but only one is used at a time, so they are unidentate in any given complex and give linkage isomers.
- Computing the magnetic moment with the wrong number of unpaired electrons because the ligand field strength was ignored: [Fe(CN)6]3− has one unpaired electron (1.73 BM), not five (5.92 BM); [CoF6]3− has four, not zero.
How to prepareStrategy for this chapter
Spend the first sitting on vocabulary and naming, because a third of the marks from this chapter come from questions a well-drilled student answers in two minutes: ligand types with examples, coordination number, and the six naming rules. Practise both directions, name from formula and formula from name, on the NCERT intext and exercise lists until the -ate endings, bis/tris, nitrito-N versus nitrito-O and the alphabetical order come without thought. Then learn the isomerism section as pairs: for each type memorise one example pair, and for stereoisomerism practise drawing the three isomers of [Co(en)2Cl2]+, fac and mer of [Ma3b3], and the mirror images of [Co(en)3]3+ with a ruler and labels.
The second sitting is bonding. Write out the VBT orbital box diagrams for the eight standard complexes ([Co(NH3)6]3+, [CoF6]3−, [Ni(CN)4]2−, [NiCl4]2−, [Ni(CO)4], [Fe(CN)6]4−, [Fe(CN)6]3−, [FeF6]3−) with hybridisation, shape and μ, and draw the octahedral and tetrahedral splitting diagrams from memory. Learn the spectrochemical series as a sentence and the d4 high spin versus low spin argument with the Δo versus P comparison. Then make one card of why-questions on colour: Ti3+ versus Sc3+, anhydrous versus hydrated CuSO4, [Ni(H2O)6]2+ versus [Ni(CN)4]2−.
Finish with Werner's table, the metal carbonyl shapes and synergic bonding diagram, and one line per application. In the exam, give the free-ion configuration, the ligand field strength, the hybridisation and the unpaired-electron count as four separate steps; that is how the marking scheme is written.