Boranes and Wade’s Rules: Predicting Cluster Shapes

September 6, 2026

| Chemical Science Exams, Inorganic Chemistry

Inorganic Chemistry · Main Group Boranes and Wade’s Rules: Predicting Cluster Shapes Boron clusters look chaotic until you count skeletal electron pairs. Then each structure follows from a single number. BSc & MSc · Inorganic Chemistry · Method The short answer: Count the skeletal electron pairs. For n boron vertices, n+1 pairs gives a closo structure, n+2 gives nido, n+3 gives arachno. Each type is derived from the closo polyhedron by removing vertices, so the shapes are related rather than independent. Why boranes need special treatment Boron has three valence electrons but four valence orbitals, so it cannot form enough…

Michaelis–Menten Enzyme Kinetics: Derivation and Interpretation

September 6, 2026

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| Chemical Science Exams, Physical Chemistry

Physical Chemistry · Kinetics Michaelis–Menten Enzyme Kinetics: Derivation and Interpretation A steady-state derivation applied to a biological catalyst, giving two constants whose meanings are constantly confused. BSc & MSc · Physical Chemistry · Concept The short answer: Apply the steady-state approximation to the enzyme–substrate complex and the rate becomes v = Vmax[S]/(KM + [S]). KM is the substrate concentration at half maximal rate and indicates how tightly the substrate binds; Vmax reflects how fast the enzyme turns over once saturated. The mechanism E + S ⇌ ES (k1 forward, k−1 reverse) ES → E + P (k2) The enzyme…

Stability Constants and the Chelate Effect

September 5, 2026

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| Chemical Science Exams, Inorganic Chemistry

Inorganic Chemistry · Coordination Stability Constants and the Chelate Effect Why a ligand that bites twice binds far more tightly than two ligands that bite once — and why the answer is entropy, not bond strength. BSc & MSc · Inorganic Chemistry · Concept The short answer: Complex formation proceeds stepwise, each step with its own constant, and the overall constant is their product. Chelating ligands give far larger overall constants than comparable monodentate ligands. The dominant reason is entropic: one chelate molecule replaces several monodentate ones, increasing the number of free particles. Stepwise and overall constants Ligands add one…

The Phase Rule and Reading Phase Diagrams

September 5, 2026

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| Chemical Science Exams, Physical Chemistry

Physical Chemistry · Equilibria The Phase Rule and Reading Phase Diagrams One short equation that tells you how many variables you are free to change, and a diagram that shows the consequence. BSc & MSc · Physical Chemistry · Concept The short answer: The phase rule states F = C − P + 2, where F is the degrees of freedom, C the number of components and P the number of phases in equilibrium. Applied to a one-component diagram it explains why an area has two degrees of freedom, a line one, and the triple point none. The rule F…

Electron Transfer: Inner Sphere versus Outer Sphere

September 5, 2026

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| Chemical Science Exams, Inorganic Chemistry

Inorganic Chemistry · Mechanism Electron Transfer: Inner Sphere versus Outer Sphere Two mechanisms distinguished by one question — does a ligand bridge the two metals during transfer, or not? BSc & MSc · Inorganic Chemistry · Concept The short answer: In outer sphere transfer the coordination shells stay intact and the electron tunnels between them. In inner sphere transfer a bridging ligand connects the two metals and the electron passes through it. The classic evidence is ligand transfer: if the bridging ligand ends up on the other metal, the mechanism was inner sphere. The two mechanisms Outer sphere Inner sphere…

Hückel Molecular Orbital Theory for Conjugated Systems

September 5, 2026

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| Chemical Science Exams, Physical Chemistry

Physical Chemistry · Quantum Hückel Molecular Orbital Theory for Conjugated Systems A drastically simplified model that nonetheless predicts aromaticity, reactivity and spectra correctly for planar conjugated molecules. BSc & MSc · Physical Chemistry · Concept The short answer: Treat only the π electrons, assume every carbon contributes one p orbital, and set all Coulomb integrals equal to α and all resonance integrals between neighbours equal to β. Solving the resulting determinant gives orbital energies of the form α + mβ, from which delocalisation energy and aromaticity follow. The assumptions Hückel theory works because it throws almost everything away and keeps…

Aldol Condensation and Enolate Chemistry

September 4, 2026

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| Chemical Science Exams, Organic Chemistry

Organic Chemistry · Carbonyl Aldol Condensation and Enolate Chemistry Almost every carbon–carbon bond formed at a carbonyl runs through an enolate. Learn the enolate and a dozen named reactions collapse into one idea. BSc & MSc · Organic Chemistry · Concept The short answer: A hydrogen alpha to a carbonyl is acidic because the resulting anion is resonance stabilised as an enolate. That enolate is a nucleophile, and it attacks another carbonyl to give a beta-hydroxy carbonyl. Heating then eliminates water to give the conjugated enone, which is the condensation step. Why the alpha hydrogen is acidic A hydrogen on…

Supramolecular Chemistry: Crown Ethers, Cryptands and Host–Guest Binding

September 4, 2026

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| Chemical Science Exams, Inorganic Chemistry

Inorganic Chemistry · Entrance Exams Supramolecular Chemistry: Crown Ethers, Cryptands and Host–Guest Binding Chemistry beyond the covalent bond — where selectivity comes from the size of a hole and the number of contacts, not from making or breaking bonds. BSc & MSc · Inorganic Chemistry · Concept The short answer: Supramolecular chemistry studies assemblies held together by non-covalent forces. A crown ether binds an alkali metal ion whose radius matches its cavity, which is how selectivity is achieved without any covalent chemistry. Cryptands enclose the ion in three dimensions and bind far more strongly still. The extra stability of a…

Term Symbols and Russell–Saunders Coupling, Step by Step

September 4, 2026

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| Chemical Science Exams, Inorganic Chemistry

Inorganic Chemistry · Spectra Term Symbols and Russell–Saunders Coupling, Step by Step A mechanical procedure that looks abstract. Follow the five steps in order and any ground-state term symbol takes about a minute. BSc & MSc · Inorganic Chemistry · Method The short answer: Combine the individual orbital angular momenta into L and the spins into S, then couple them into J. The term symbol is written 2S+1LJ. Hund’s rules then pick the ground state: maximum multiplicity first, then maximum L, then J by whether the shell is less or more than half filled. What a term symbol encodes 2S+1LJ…

Asymmetric Synthesis: How Enantioselectivity Is Actually Achieved

September 4, 2026

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| Chemical Science Exams, Organic Chemistry

Stereochemistry · Entrance Exams Asymmetric Synthesis: How Enantioselectivity Is Actually Achieved Two enantiomers have identical energies, so no achiral reagent can ever prefer one. Every method in this area is a way of breaking that symmetry. BSc & MSc · Organic Chemistry · Concept The short answer: Enantiomeric products come from enantiomeric transition states, which have identical energies, so an achiral system must give a racemate. Introducing something chiral makes the two competing transition states diastereomeric instead, and diastereomers differ in energy. Whether the chirality comes from an auxiliary, a reagent or a catalyst, that is the single mechanism behind…

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