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SQA Advanced Higher Chemistry

Advanced Higher · SQAChemistry44 catatan dalam 9 folder, 218 KB

44 catatan belajar, dibuat oleh cookie, dalam 4 bagian yang mengikuti topik kursus.

Notes for SQA Advanced Higher Chemistry, in folders for the four areas of the course specification (inorganic chemistry, physical chemistry, organic chemistry and instrumental analysis, and researching chemistry) in its order, with one note for each topic and the practical techniques in the researching chemistry folder. Follows version 4.0 of the course specification for exams from 2027.

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  • Inorganic chemistry
    • Electromagnetic radiation and atomic spectra6 KB
    • Atomic orbitals and quantum numbers4 KB
    • Electronic configurations and ionisation energies5 KB
    • Shapes of molecules and ions5 KB
    • Transition metal atoms, ions and oxidation states5 KB
    • Ligands and complex ions5 KB
    • Colour and d orbital splitting4 KB
    • Catalysis by transition metals4 KB
  • Physical chemistry
    • The equilibrium constant4 KB
    • Water, pH and the ionic product3 KB
    • Acids, bases and salts5 KB
    • Buffer solutions4 KB
    • Indicators4 KB
    • Enthalpy and entropy4 KB
    • Free energy and feasibility4 KB
    • Rate equations and orders of reaction4 KB
    • Reaction mechanisms and the rate-determining step4 KB
  • Organic chemistry and instrumental analysis
    • Molecular orbitals
      • Molecular orbitals, sigma bonds and pi bonds5 KB
      • Hybridisation, conjugation and colour6 KB
    • Synthesis
      • Bond fission, curly arrows and reaction types5 KB
      • Structural formulae and nomenclature6 KB
      • Haloalkanes6 KB
      • Alcohols and ethers6 KB
      • Alkenes6 KB
      • Carboxylic acids and esters5 KB
      • Amines3 KB
      • Benzene and electrophilic substitution5 KB
      • Planning synthetic routes5 KB
    • Stereochemistry
      • Isomerism6 KB
    • Experimental determination of structure
      • Elemental microanalysis and mass spectrometry5 KB
      • Infrared spectroscopy5 KB
      • Proton NMR spectroscopy6 KB
      • High-resolution NMR and interpreting spectra5 KB
    • Pharmaceutical chemistry
      • Drugs, receptors and enzymes5 KB
  • Researching chemistry
    • Apparatus and experimental skills6 KB
    • Stoichiometric calculations: moles, solutions and gases6 KB
    • Excess reactants, yield and empirical formulae5 KB
    • Gravimetric analysis4 KB
    • Standard solutions and primary standards5 KB
    • Volumetric analysis and titrations7 KB
    • Colorimetry4 KB
    • Distillation, reflux and vacuum filtration4 KB
    • Recrystallisation and solvent extraction6 KB
    • Melting point and thin-layer chromatography6 KB

Catatan pertama

Inorganic chemistry / Electromagnetic radiation and atomic spectra

Waves and photons

Electromagnetic radiation can be described as a wave, with a wavelength and a frequency, and it can also be described as a stream of particles called photons. This is its dual nature. The wave description accounts for how radiation travels and spreads out; the particle description is needed whenever radiation is absorbed or emitted by matter, because energy is then exchanged in whole photons and never in fractions of one.

The wavelength λ\lambda is the distance between two adjacent crests and is measured in metres, although wavelengths of visible light are normally quoted in nanometres (1 nm=1⋅10−9 m\pu{1 nm} = \pu{1e-9 m}). The frequency ff is the number of waves passing a point each second, in hertz (Hz=s−1\pu{Hz} = \pu{s-1}). All electromagnetic radiation travels at the same speed cc in a vacuum, about 3.00⋅108 m s−1\pu{3.00e8 m s-1}, so wavelength and frequency are linked by

c=fλc = f\lambda

A longer wavelength therefore means a lower frequency. The electromagnetic spectrum is the full range of radiation arranged in order of wavelength. From the longest wavelength to the shortest it runs through radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. Visible light covers roughly 400 nm (violet) to 700 nm (red).

Photon energy

A photon carries a quantised amount of energy that is proportional to the frequency of the radiation:

E=hforE=hcλE = hf \qquad\text{or}\qquad E = \frac{hc}{\lambda}

Here EE is the energy of one photon in joules, hh is Planck's constant (6.63⋅10−34 J s\pu{6.63e-34 J s}), ff is the frequency in Hz\pu{Hz} and λ\lambda is the wavelength in metres. Because EE is proportional to ff, the photons in high frequency radiation transfer more energy than the photons in low frequency radiation. An ultraviolet photon is more energetic than a photon of red light, which is why ultraviolet radiation is able to break bonds and damage skin.

Chemists usually want the energy for a mole of photons, found by multiplying by the Avogadro constant LL (6.02⋅1023 mol−1\pu{6.02e23 mol-1}):

E=LhforE=LhcλE = Lhf \qquad\text{or}\qquad E = \frac{Lhc}{\lambda}

This is normally converted to kJ mol−1\pu{kJ mol-1}, so that it can be compared with bond enthalpies and ionisation energies.

Worked example

Light of wavelength 520 nm is absorbed by a sample. Find the frequency, the energy of one photon and the energy of one mole of photons.

λ=520 nm=5.20⋅10−7 mf=cλ=3.00×1085.20×10−7=5.77⋅1014 HzEphoton=hf=6.63×10−34×5.77×1014=3.83⋅10−19 JEmole=L×Ephoton=6.02×1023×3.83×10−19=2.31⋅105 J mol−1 \begin{aligned} \lambda &= \pu{520 nm} = \pu{5.20e-7 m} \\ f &= \frac{c}{\lambda} = \frac{3.00\times10^{8}}{5.20\times10^{-7}} = \pu{5.77e14 Hz} \\ E_{\text{photon}} &= hf = 6.63\times10^{-34}\times 5.77\times10^{14} = \pu{3.83e-19 J} \\ E_{\text{mole}} &= L \times E_{\text{photon}} = 6.02\times10^{23}\times 3.83\times10^{-19} = \pu{2.31e5 J mol-1} \end{aligned}

The energy per mole is about 231 kJ mol−1\pu{231 kJ mol-1} (the small difference from 230 comes from rounding the intermediate values).

Energy levels and emission spectra

The electrons in an atom can only have certain fixed energies, so the atom has a set of discrete (quantised) energy levels. When energy is supplied, for example by heating or an electric discharge, an electron can be promoted from a lower energy level to a higher one. This excited state is unstable, and when the electron falls back to a lower level the atom emits a photon whose energy equals the difference between the two levels:

ΔE=Ehigher−Elower=hf\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

Since only certain energy differences are possible, only certain frequencies are emitted. The light from excited atoms therefore gives a line spectrum, a series of sharp lines at particular wavelengths on a dark background, and not a continuous band of colours. This is direct evidence that electrons in atoms occupy discrete energy levels. If energy levels were continuous, a continuous spectrum would be seen.

For example, if an electron in a hydrogen atom drops between two levels that are 3.03⋅10−19 J\pu{3.03e-19 J} apart, the photon emitted has wavelength

λ=hcE=6.63×10−34×3.00×1083.03×10−19=6.56⋅10−7 m=656 nm\lambda = \frac{hc}{E} = \frac{6.63\times10^{-34}\times 3.00\times10^{8}}{3.03\times10^{-19}} = \pu{6.56e-7 m} = \pu{656 nm}

which is a red line in the visible region.

Atomic absorption and emission spectroscopy

Every element has its own set of energy levels, and so its own characteristic emission and absorption spectra. The lines act like a fingerprint, and they can be used both to identify an element in a sample and to measure how much of it is present.

Emission spectroscopy

In emission spectroscopy a high temperature, such as that of a flame or a plasma, is used to excite the electrons in the atoms of the sample. As the electrons drop back to lower energy levels they emit photons. The instrument measures the intensity of the light emitted at different wavelengths to give an emission spectrum.

Absorption spectroscopy

In absorption spectroscopy the sample is first atomised, and electromagnetic radiation is directed at it. Photons whose energy exactly matches the gap between two energy levels are absorbed as electrons are promoted. The intensity of the light that comes through is measured across a range of wavelengths, and the absorption spectrum shows dark lines (missing wavelengths) on a bright background. The absorption lines of an element fall at the same wavelengths as its emission lines, because the same pairs of energy levels are involved.

Finding concentration

The wavelengths present show which elements are in the sample. The intensity of the light emitted or absorbed at a characteristic wavelength is related to the concentration of that element, so a sample can be compared against solutions of known concentration to measure how much of the element it contains.

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