Inorganic Chemistry-2 BCH3T06

Syllabus topics (official RTMNU)

(A) MO theory

• LCAO approximation, wave equation for molecular orbitals.
• Difference between bonding and antibonding MO in terms of energy and electron density distribution curves, order of energy levels in MO.
• Molecular Orbital diagrams for homonuclear diatomic molecules of elements(with Z= 1 to 9) Concepts of nonbonding MO in HF molecule.
• Coulson’s MO diagram of CO and NO diatomic molecule.

(B) Noble gases

• Chemical properties of the noble gases, Chemistry of Xenon and Krypton, Preparation, properties and structures of xenon fluorides (XeF2, XeF4 and XeF6), structure and bonding in XeOF2, XeOF4 and XeO2F2.

(A) Molecular orbital (MO) theory

LCAO approximation and wave equation for MOs

• In a molecule, electrons occupy molecular orbitals spread over all the nuclei.
• MOs are formed by Linear Combination of Atomic Orbitals: ψ_MO = c_Aψ_A ± c_Bψ_B.
• Addition gives the bonding MO: ψ_b = ψ_A + ψ_B; subtraction gives the antibonding MO: ψ_a = ψ_A − ψ_B.
• Electron density: ψ_b² = ψ_A² + ψ_B² + 2ψ_Aψ_B (extra density between nuclei); ψ_a² = ψ_A² + ψ_B² − 2ψ_Aψ_B (less density, a node between nuclei).
• Conditions for combination: AOs of comparable energy, same symmetry about the bond axis, and good overlap.
• n atomic orbitals give n molecular orbitals.

Bonding vs antibonding MO

Electron density curve (density on y-axis, internuclear axis on x-axis): for the bonding MO the curve stays high between the two nuclei; for the antibonding MO it drops to zero at the midpoint.

Order of energy levels

• Up to N₂ (s–p mixing is strong): σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z.
• For O₂ and F₂: σ2p_z lies below π2p_x = π2p_y.
• Bond order = ½(N_b − N_a). Unpaired electrons → paramagnetic.

Homonuclear diatomic molecules (Z = 1 to 9)

O₂ configuration (valence): σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹. MO theory explains paramagnetism of O₂, which VBT cannot.

Non-bonding MO in HF

• H 1s combines only with F 2p_z (same symmetry along the bond axis, similar energy) to give σ and σ* MOs.
• F 2s (too low in energy) and F 2p_x, 2p_y (wrong symmetry) do not combine: they stay as non-bonding MOs, holding F’s three lone pairs.
• 8 valence electrons: 2 in σ, 6 in non-bonding MOs; bond order 1. σ MO lies closer to F, so the bond is polar (H^δ+–F^δ−).

Coulson’s MO diagram of CO and NO

• Heteronuclear: O is more electronegative, so its AOs are drawn lower in energy than those of C or N. The diagram is unsymmetrical.
• CO: 10 valence electrons, isoelectronic with N₂. Bond order = 3; diamagnetic. The highest filled MO (a σ orbital) is mainly on carbon, so CO donates its lone pair through carbon to metals (metal carbonyls).
• NO: 11 valence electrons; the extra electron goes into a π* MO. Bond order = (8 − 3)/2 = 2.5; paramagnetic (1 unpaired electron). Losing it gives NO⁺, bond order 3, shorter bond.

(B) Noble gases

Chemical properties

• Monatomic with stable ns²np⁶ configuration, very high ionisation energy and nearly zero electron affinity, so they are very unreactive.
• 1962: Bartlett made O₂⁺[PtF₆]⁻; noting that the IE of Xe (1170 kJ mol⁻¹) is close to that of O₂ (1175 kJ mol⁻¹), he made Xe⁺[PtF₆]⁻, the first noble gas compound.
• Only Kr, Xe and Rn form true compounds, mainly with the most electronegative elements F and O. Xe and Kr also form clathrates (trapped in ice or quinol cages).

Xenon fluorides: preparation

• Xe + F₂ (2 : 1 excess Xe), 400 °C, nickel vessel → XeF₂
• Xe + 2F₂ (1 : 5), 400 °C, about 6 atm → XeF₄
• Xe + 3F₂ (1 : 20), about 300 °C, 50–60 atm → XeF₆

Properties

• Colourless crystalline solids; strong fluorinating and oxidising agents.
• Hydrolysis: 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂; 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂; XeF₆ + 3H₂O → XeO₃ + 6HF.
• Partial hydrolysis of XeF₆: XeF₆ + H₂O → XeOF₄ + 2HF; XeF₆ + 2H₂O → XeO₂F₂ + 4HF.
• Fluoride donors/acceptors: XeF₂ + PF₅ → [XeF]⁺[PF₆]⁻; XeF₆ + MF → M⁺[XeF₇]⁻ (M = Na, K, Rb, Cs).

Structures (VSEPR)

In the oxyfluorides the Xe=O π bonds are formed by d-orbitals of Xe (pπ–dπ) and do not change the shape.

Krypton

• KrF₂ is the main krypton compound: made by electric discharge through Kr and F₂ at very low temperature. Linear, sp³d, like XeF₂.
• Unstable at room temperature; a very powerful fluorinating agent. KrF₂ + SbF₅ → [KrF]⁺[SbF₆]⁻.

Likely exam questions

Long answer

1. Explain the LCAO approximation. Draw the MO diagram of O₂ and N₂ and explain their bond order and magnetic nature.
2. How are XeF₂, XeF₄ and XeF₆ prepared? Discuss their structures on the basis of hybridisation.

Short answer

1. Distinguish between bonding and antibonding molecular orbitals.
2. Explain the concept of non-bonding MO with reference to HF.
3. Give the structures of XeOF₂, XeOF₄ and XeO₂F₂.

Syllabus topics (official RTMNU)

(A) Chemistry of elements of first transition series

• Characteristic properties of the elements of first transition series with reference to their: Electronic configuration, Atomic and ionic radii, Ionization potential, Variable oxidation states, Magnetic properties , Colour, Complex formation tendency and catalytic activity.

(B) Chemistry of elements of second and third transition series

• Electronic configuration of 4d and 5d transition series.
• Comparative treatment with their 3d-analogous (Group Cr-Mo-W, Co-Rh-Ir,) in respect of oxidation states and magnetic behaviour.

(A) Elements of the first transition series (3d series)

Transition elements are d-block elements that have partly filled d-orbitals in the atom or in a common ion. First series: Sc (21) to Zn (30).

Electronic configuration

• General: [Ar] 3d^1–10 4s^1–2.
• Exceptions: Cr [Ar] 3d⁵ 4s¹ and Cu [Ar] 3d¹⁰ 4s¹, due to extra stability of half-filled and fully filled d-subshells.
• Zn (3d¹⁰ in atom and ion) is not a true transition element; Cu is (Cu²⁺ is 3d⁹).

Atomic and ionic radii

• Radii decrease from Sc to about Cr because nuclear charge increases while added electrons enter the inner 3d shell, which shields the 4s electrons poorly.
• From Cr to Ni they stay nearly constant (increased shielding balances nuclear charge); they rise slightly at Cu and Zn (filled 3d repels).
• For the same charge, ionic radii (M²⁺, M³⁺) decrease across the series.

Ionisation potential

• IE increases gradually and irregularly from Sc to Zn (smaller increase than in s- and p-blocks) because of shielding by 3d electrons.
• Second IE of Cr and Cu is high (breaking a stable 3d⁵ or 3d¹⁰). Third IE of Mn is high (3d⁵ of Mn²⁺), and of Zn is very high (3d¹⁰).

Variable oxidation states

• Energies of 3d and 4s are close, so both can be lost: many oxidation states.
• Number of states increases to Mn (+2 to +7) then decreases. Sc only +3; Zn only +2.
• +2 is common across the series; highest states are found in oxides and oxyanions (MnO₄⁻, Cr₂O₇²⁻). Lower states are reducing (Cr²⁺, Fe²⁺), higher states oxidising (MnO₄⁻, Cr₂O₇²⁻).

Magnetic properties

• Unpaired d-electrons make ions paramagnetic; d⁰ and d¹⁰ ions are diamagnetic.
• Spin-only formula: μ = √[n(n + 2)] BM (n = unpaired electrons).
• Examples: Ti³⁺ (d¹) 1.73 BM; Cu²⁺ (d⁹) 1.73 BM; Ni²⁺ (d⁸) 2.83 BM; Cr³⁺ (d³) 3.87 BM; Fe²⁺ (d⁶) 4.90 BM; Mn²⁺ and Fe³⁺ (d⁵) 5.92 BM (maximum); Zn²⁺ 0.
• Fe, Co, Ni are ferromagnetic.

Colour

• Due to d–d transitions: in a complex the d-orbitals split; an electron absorbs visible light to jump to the higher level, and the complementary colour is seen.
• Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) are colourless.
• Examples: Ti³⁺ violet, V³⁺ green, Cr³⁺ green, Mn²⁺ pale pink, Fe²⁺ pale green, Fe³⁺ yellow, Co²⁺ pink, Ni²⁺ green, Cu²⁺ blue.
• Charge transfer gives intense colour in MnO₄⁻ (purple) and Cr₂O₇²⁻ (orange) even though they are d⁰.

Complex formation tendency

Strong, because the ions are small with high charge (high polarising power) and have vacant d-orbitals of suitable energy to accept lone pairs. Examples: [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺, [Ni(CO)₄], [Co(NH₃)₆]³⁺.

Catalytic activity

• Due to variable oxidation states (form unstable intermediates) and large surface area for adsorption.
• Examples: Fe in Haber process (NH₃); V₂O₅ in contact process (SO₃); Ni in hydrogenation of oils; MnO₂ in decomposition of KClO₃; TiCl₄ in Ziegler–Natta polymerisation.

(B) Elements of the second and third transition series

Electronic configuration

• 4d series (Y to Cd): [Kr] 4d^1–10 5s^0–2. Many irregularities: Nb 4d⁴5s¹, Mo 4d⁵5s¹, Ru 4d⁷5s¹, Rh 4d⁸5s¹, Pd 4d¹⁰5s⁰, Ag 4d¹⁰5s¹.
• 5d series (La, Hf to Hg): [Xe] 4f¹⁴ 5d^1–10 6s^1–2 (La has no 4f). Exceptions: Pt 5d⁹6s¹, Au 5d¹⁰6s¹. W is 5d⁴6s².

General comparison with the 3d series

• 4d and 5d atoms are larger than 3d, but 4d and 5d members of a group have almost the same size because of lanthanide contraction (Zr 160 pm, Hf 159 pm). So Zr/Hf, Nb/Ta, Mo/W are very alike.
• Higher oxidation states become more stable down the group; lower states (+2, +3) less common for heavier elements.
• Heavier metals form stronger metal–metal bonds, have higher melting points and enthalpies of atomisation.

Group Cr–Mo–W

Group Co–Rh–Ir

Magnetic behaviour

• 4d and 5d orbitals are larger, so ligand-field splitting (Δ) is greater (about 30–50 % more than 3d). Hence their complexes are almost always low spin. Example: [CoF₆]³⁻ is high spin and paramagnetic, but Rh(III) and Ir(III) complexes are low spin and diamagnetic.
• Spin–orbit coupling is strong in heavier atoms, so observed moments are lower than spin-only values and change with temperature; the simple μ = √[n(n + 2)] formula does not work well.
• Metal–metal bonding and antiferromagnetic coupling further reduce the moments.

Likely exam questions

Long answer

1. Discuss the characteristic properties of first transition series elements with respect to oxidation states, magnetic properties and colour.
2. Compare the 4d and 5d elements with their 3d analogues (Cr–Mo–W and Co–Rh–Ir) with respect to oxidation states and magnetic behaviour.

Short answer

1. Why are transition metal ions coloured? Why is Zn²⁺ colourless?
2. Calculate the spin-only magnetic moment of Fe³⁺ and Cu²⁺ ions.
3. Why do transition elements act as good catalysts? Give two examples.

Syllabus topics (official RTMNU)

(A) Chemistry of Lanthanides

• Position in periodic table , electronic configuration, Oxidation states, Atomic and ionic radii, Lanthanide contraction and its consequences, Complex forming tendency.
• Occurrence and separation of lanthanides (ion exchange and solvent extraction).

(B) Chemistry of Actinides

• Position in periodic table, electronic configuration, Oxidation states, Atomic and ionic radii, actinide contraction.

(A) Chemistry of lanthanides

Position in the periodic table

• The 14 elements after lanthanum, Ce (58) to Lu (71), in which the differentiating electron enters the 4f subshell. They are f-block, inner transition elements.
• They belong to group 3, period 6, and all would fit in the single place of La. To keep the table compact they are placed in a separate row at the bottom.
• Reason for one position: they have very similar properties because the outer shells (5d, 6s) are nearly the same.

Electronic configuration

• General: [Xe] 4f^1–14 5d^0–1 6s².
• Examples: La [Xe]5d¹6s²; Ce [Xe]4f¹5d¹6s²; Pr [Xe]4f³6s²; Eu [Xe]4f⁷6s²; Gd [Xe]4f⁷5d¹6s² (keeps half-filled 4f); Yb [Xe]4f¹⁴6s²; Lu [Xe]4f¹⁴5d¹6s².
• Ln³⁺ ions: [Xe]4fⁿ, from La³⁺ (4f⁰) to Lu³⁺ (4f¹⁴).

Oxidation states

• +3 is the common and most stable state for all (loss of 2 s and one d or f electron).
• +4 and +2 occur when they give an empty, half-filled or filled 4f: Ce⁴⁺ (4f⁰) and Tb⁴⁺ (4f⁷); Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴); also Sm²⁺.
• Ce⁴⁺ is a good oxidising agent (ceric ammonium nitrate, ceric sulphate); Eu²⁺ and Sm²⁺ are reducing.

Atomic and ionic radii and lanthanide contraction

• Lanthanide contraction: steady decrease in atomic and ionic radii from La to Lu. Ln³⁺ radius falls from about 103 pm (La³⁺) to about 86 pm (Lu³⁺).
• Cause: with each step nuclear charge rises by +1 and an electron is added to 4f. 4f orbitals are diffuse and shield poorly, so the effective nuclear charge increases and pulls the shells inward.
• Atomic radii show small irregularities: Eu and Yb are larger (they use only 2 electrons in metallic bonding).

Consequences of lanthanide contraction

1. Similar size of 4d and 5d elements: Zr (160 pm) and Hf (159 pm), Nb/Ta, Mo/W have almost identical radii and chemistry; Zr and Hf are very hard to separate.
2. Decrease in basicity: La(OH)₃ is most basic and Lu(OH)₃ least basic, because smaller ions polarise OH⁻ more (more covalent).
3. Difficulty in separating lanthanides, as neighbours differ very little in size.
4. Position of yttrium: Y³⁺ (about 90 pm) is close to Ho³⁺, so Y occurs with heavy lanthanides.
5. Higher density, ionisation energy and nobility of 5d metals (Os, Ir, Pt, Au) compared with 4d.
6. Complex-forming tendency and hydrolysis increase from La to Lu.

Complex forming tendency

• Weak compared with d-block, because Ln³⁺ ions are large (low charge density) and 4f orbitals are buried inside, so they take no part in bonding (no crystal field stabilisation to speak of).
• Bonding is mainly ionic; they prefer hard O- and N-donor ligands and chelating agents: EDTA, citrate, oxalate, β-diketones (acetylacetone).
• High coordination numbers (8, 9 common). Tendency increases from La³⁺ to Lu³⁺ as size decreases.

Occurrence

• Monazite sand: phosphates of Ce, La, Nd, Pr with thorium; large deposits on the Kerala and Tamil Nadu coasts in India.
• Bastnaesite: LnCO₃F (fluorocarbonate). Xenotime: YPO₄ with heavy lanthanides.
• Pm does not occur naturally in useful amounts (radioactive).

Separation of lanthanides

(i) Ion exchange method (best modern method):

1. A solution of Ln³⁺ ions is passed through a column of cation-exchange resin (sulphonated polystyrene, H⁺ form): 3RH + Ln³⁺ → R₃Ln + 3H⁺.
2. The column is eluted with a complexing agent such as buffered ammonium citrate–citric acid (or EDTA) at controlled pH.
3. Smaller Ln³⁺ ions (Lu³⁺) have larger hydrated radii, are held less strongly by the resin and form stronger citrate complexes, so they come out first; La³⁺ comes out last.
4. Fractions are collected and each Ln is precipitated as oxalate and ignited to the oxide.

(ii) Solvent extraction:

• Ln nitrates in nitric acid are shaken with tri-n-butyl phosphate (TBP) in kerosene; distribution coefficients differ slightly from one Ln to the next (heavier ones extract better).
• Repeated, continuous counter-current extraction gives good separation. Example: Gd(NO₃)₃ separated from La(NO₃)₃.

(B) Chemistry of actinides

Position in the periodic table

• The 14 elements after actinium, Th (90) to Lr (103), where the differentiating electron enters 5f.
• Group 3, period 7; placed as a separate second row at the bottom, below the lanthanides.
• All are radioactive. Elements after uranium (Np onwards) are transuranic and man-made.

Electronic configuration

• General: [Rn] 5f^1–14 6d^0–1 7s².
• Examples: Th [Rn]6d²7s² (no 5f electron); Pa [Rn]5f²6d¹7s²; U [Rn]5f³6d¹7s²; Pu [Rn]5f⁶7s²; Cm [Rn]5f⁷6d¹7s².
• 5f and 6d energies are very close in early actinides, so configurations are irregular.

Oxidation states

• +3 is common, but early actinides show many higher states because 5f, 6d and 7s electrons are close in energy.
• Th +4; Pa +5; U +3 to +6 (+6 most stable, as UO₂²⁺ uranyl); Np and Pu up to +7. From Am onwards +3 dominates; No shows +2 (5f¹⁴).

Atomic and ionic radii; actinide contraction

• Radii of An³⁺ and An⁴⁺ ions decrease steadily from Th to Lr: actinide contraction.
• Cause: 5f electrons shield even more poorly than 4f, so effective nuclear charge rises with atomic number.
• The contraction from element to element is greater than the lanthanide contraction.

Likely exam questions

Long answer

1. What is lanthanide contraction? Explain its causes and consequences.
2. Describe the separation of lanthanides by the ion exchange method and by solvent extraction.

Short answer

1. Give the electronic configuration and oxidation states of lanthanides. Why do Ce and Eu show +4 and +2 states?
2. Why is the complex forming tendency of lanthanides weak?
3. Compare lanthanides and actinides (any four points).

Syllabus topics (official RTMNU)

(A) Non-aqueous solvents

• Classification of solvents based on proton donor-acceptor properties and ionizing nature.
• Aqueous and non-aqueous solvents.
• Properties of liquid ammonia and liq.
• SO2 as solvents, solubilities of various compounds in liquid ammonia and liquid SO2.
• Following reactions in liq.
• NH3 and liq.
• SO2: (i) acid base (ii) redox (iii) precipitation (metathesis).

(B) Hard and Soft Acids and Bases

• Classification of Acids and Bases as Hard and Soft.
• Pearson’s HSAB Concept and its applications.
• Acid-base strength versus hardness and softness.
• Theoretical basis for hardness.
• Symbiosis.
• Limitations of HSAB concept.

(A) Non-aqueous solvents

Classification of solvents

(i) Based on proton donor–acceptor properties

• Protic solvents (contain ionisable H):
  • Protogenic (acidic, proton donors): HF, H₂SO₄, CH₃COOH.
  • Protophilic (basic, proton acceptors): liquid NH₃, pyridine, hydrazine.
  • Amphiprotic (both): H₂O, alcohols.
• Aprotic solvents (no proton donation): benzene, CCl₄ (non-polar); liquid SO₂, BrF₃, N₂O₄ (polar, can self-ionise without protons).

(ii) Based on ionising nature

• Ionising solvents: polar, high dielectric constant, dissolve ionic compounds and let them ionise: H₂O (78.5), HF (about 84), liquid NH₃ (about 22).
• Non-ionising solvents: low dielectric constant: benzene (2.3), CCl₄ (2.2), CS₂.

Aqueous vs non-aqueous solvents

• Water: high dielectric constant, strong H-bonding, dissolves ionic compounds, liquid over a wide range (0–100 °C), self-ionises: 2H₂O ⇌ H₃O⁺ + OH⁻.
• Non-aqueous solvents allow reactions not possible in water (e.g. with alkali metals, water-sensitive compounds), and can reverse the direction of reactions because solubilities are different.

Liquid ammonia as a solvent

• m.p. −77.7 °C, b.p. −33.4 °C; dielectric constant about 22; strong H-bonding (less than water).
• Self-ionisation: 2NH₃ ⇌ NH₄⁺ + NH₂⁻. Acids = substances giving NH₄⁺ (ammonium salts); bases = substances giving NH₂⁻ (amides).
• Dissolves alkali and alkaline earth metals giving blue, conducting, paramagnetic solutions of solvated (ammoniated) electrons: Na + (x + y)NH₃ → [Na(NH₃)_x]⁺ + [e(NH₃)_y]⁻. Concentrated solutions are bronze coloured and diamagnetic.

Solubility in liquid NH₃: lower dielectric constant than water, so ionic compounds are generally less soluble. Ammonium salts, nitrates, nitrites, cyanides, thiocyanates and iodides dissolve well; fluorides, oxides, hydroxides, sulphates, carbonates are almost insoluble. Halide solubility increases F⁻ < Cl⁻ < Br⁻ < I⁻. Many organic compounds (amines, alcohols, esters) dissolve better than in water. AgCl and AgI dissolve (ammine complexes) but BaCl₂ does not.

Reactions in liquid NH₃

1. Acid–base (neutralisation): NH₄Cl + NaNH₂ → NaCl + 2NH₃ (acid + base → salt + solvent). Amphoteric behaviour: Zn(NH₂)₂ + 2NaNH₂ → Na₂[Zn(NH₂)₄]; Zn(NH₂)₂ + 2NH₄Cl → [Zn(NH₃)₄]Cl₂.
2. Redox: metal–ammonia solutions are powerful reducing agents. K₂[Ni(CN)₄] + 2K → K₄[Ni(CN)₄] (Ni(II) → Ni(0)); 2Na + O₂ → Na₂O₂; 2Na + 2NH₃ → 2NaNH₂ + H₂ (slow; fast with Fe catalyst). Also used in Birch reduction of aromatic rings.
3. Precipitation (metathesis): because solubilities differ from water, the reaction is reversed: Ba(NO₃)₂ + 2AgCl → BaCl₂↓ + 2AgNO₃ (in water, AgCl precipitates instead).

Liquid sulphur dioxide as a solvent

• m.p. −75.5 °C, b.p. −10 °C; dielectric constant about 15; aprotic.
• Proposed self-ionisation: 2SO₂ ⇌ SO²⁺ + SO₃²⁻ (not proven experimentally). Acids = SO²⁺ (thionyl) donors such as SOCl₂; bases = SO₃²⁻ donors such as Cs₂SO₃, K₂SO₃.
• Solubility: covalent substances dissolve well (organic compounds, PCl₃, BCl₃, CS₂, Br₂). Among ionic salts, iodides and thiocyanates are soluble; chlorides less; fluorides, oxides, sulphides, sulphates are insoluble. Halide solubility: I⁻ > Br⁻ > Cl⁻ > F⁻.

Reactions in liquid SO₂

1. Acid–base: SOCl₂ + Cs₂SO₃ → 2CsCl + 2SO₂. Amphoteric behaviour: Al₂(SO₃)₃ (formed from AlCl₃ and a soluble sulphite) dissolves in excess sulphite as [Al(SO₃)₃]³⁻, and adding the acid SOCl₂ gives Al₂(SO₃)₃ back.
2. Redox: SbCl₅ + 2KI → SbCl₃ + 2KCl + I₂ (iodide oxidised to iodine).
3. Precipitation (metathesis): chlorides are less soluble than iodides, so: SOCl₂ + 2KI → 2KCl↓ + SOI₂ (SOI₂ is unstable and decomposes, giving I₂).

(B) Hard and soft acids and bases (HSAB)

Classification (Pearson, 1963)

Pearson’s HSAB principle

Applications

• Occurrence of minerals: hard metal ions occur as oxides, carbonates, silicates (Al₂O₃, CaCO₃, MgCO₃); soft ones as sulphides (HgS, Ag₂S, Cu₂S, PbS).
• Stability of complexes: [HgI₄]²⁻ is much more stable than [HgF₄]²⁻; [AlF₆]³⁻ is stable but Al³⁺ iodo complexes are not.
• Predicting the direction of reactions: LiI + CsF → LiF + CsI; HgF₂ + BeI₂ → BeF₂ + HgI₂.
• Ambidentate ligands: SCN⁻ binds soft Hg²⁺/Pt²⁺ through S and harder metal ions through N.
• Solubility: AgF (hard base) is soluble in water; AgI (soft–soft) is insoluble.
• Toxicity: soft Hg²⁺, Cd²⁺, Pb²⁺ bind strongly to soft –SH groups of enzymes.

Acid–base strength vs hardness and softness

• Strength and hardness are different properties. OH⁻ and F⁻ are both hard, but OH⁻ is the much stronger base.
• A very strong acid–base pair can react even when mismatched: H⁺ (hard) + H⁻ (soft) → H₂ goes to completion because both are very strong.
• Pearson’s equation: log K = S_AS_B + σ_Aσ_B (S = strength, σ = softness factor).

Theoretical basis of hardness and softness

• Ionic–covalent theory: hard–hard bonding is mainly ionic (electrostatic); soft–soft bonding is mainly covalent.
• π-bonding theory: soft acids have filled d-orbitals and soft bases have empty d- or π*-orbitals, so π back-bonding strengthens soft–soft bonds (Pt–PR₃, Ni–CO).
• Electronegativity and polarisability: hard species have high electronegativity and low polarisability.
• Klopman / frontier orbitals: hard interactions are charge-controlled (large HOMO–LUMO gap); soft ones are frontier-orbital controlled (small gap). Absolute hardness η = (I − A)/2.

Symbiosis

Hard ligands attached to a centre make it harder; soft ligands make it softer (Jørgensen). So hard ligands tend to gather together and soft ones together.

• BF₃ + F⁻ → BF₄⁻ is favoured; BH₃ + H⁻ → BH₄⁻ is favoured.
• [Co(NH₃)₅]³⁺ prefers F⁻ over I⁻; [Co(CN)₅]²⁻ prefers I⁻.

Limitations of HSAB

• Mainly qualitative; no single exact scale of hardness.
• Ignores the intrinsic strength of acids and bases.
• Many borderline species and exceptions; same element changes with oxidation state (Cu⁺ soft, Cu²⁺ borderline).
• Solvent effects are not considered.

Likely exam questions

Long answer

1. Discuss liquid ammonia as a non-aqueous solvent with reference to acid–base, redox and precipitation reactions.
2. What is Pearson’s HSAB principle? Classify acids and bases as hard and soft and give its applications.

Short answer

1. Classify solvents on the basis of proton donor–acceptor properties.
2. Explain acid–base reactions in liquid SO₂ with examples.
3. What is symbiosis? Give two limitations of the HSAB concept.