Chemistry: The Ozone Story

    a Electronegativity:

      Test yourself
    • Electronegativity is a measure of the ability of an atom in a molecule to attract electrons in a chemical bond to itself
    • Going across periods, electronegativity increases, and going down a group, electronegativity decreases (excluding the noble gases). Therefore, fluorine is the most electronegative element with 4.0
    • In a covalent bond between two atoms, the electrons are usually found in orbitals between them. In a non-polar bond (e.g. in O2), they are found directly in the middle of the two nuclei
    • If one bonding atom has a higher electronegativity than the others, the electrons will be closer to that atom
    • The atom farthest from the bonding electrons will have a partial positive charge (δ+) and the atoms closest will have a partial negative charge (δ-)
    • Bond polarity in a covalent bond can be predicted by using the relative electronegativity of an atom. In a C-F bond, fluorine has a higher electronegativity (4.0) than carbon (2.6) so it attracts the shared electrons more strongly and the polarity of the bond is: C δ+ — F δ-, where carbon has a partial positive charge (δ+) and fluorine has a partial negative charge (δ-)

    Polar bonds:

    • The greater the electronegativity difference, the more polar the bond. Anything with a difference over around 0.4 is considered polar, and forms a permanent dipole
    • Sometimes an arrow is used to show that a bond is polar. It points towards the partially negative atom, from the positive atom

    b, d, k(i) Intermolecular bonds & dipoles:

      Test yourself
    • A dipole is something a molecule has if it contains a positive end and a negative end because of a difference in charge, forming polar bonds
    • A polar molecule is said to have a dipole

    Instantaneous dipoles:

    • These form when electrons in the bond are not evenly distributed, making one atom gain a slight negative charge and the other a slight positive charge. This happens in every molecule although it is most noticeable in non-polar substances

    Instantaneous dipole-induced dipole bonds:

    • If a molecule is near to another with an instantaneous dipole, it is induced and becomes polarised
    • This effect forms instantaneous dipole-induced dipole bonds, the weakest type of intermolecular bond

    Permanent dipoles:

    • These occur when the two atoms have different electronegativities, and the charges do not balance (e.g. CO2 is non-polar due to the 180° bond angle cancelling the charges, but HCl forms a permanent dipole)
    • Permanent dipole-permanent dipole bonds (e.g. in HCl) are relatively strong and substances containing them are more likely to be liquids or solids rather than gases. These are formed between polar molecules. The intermolecular bonds can be shown with dashed lines

    Intermolecular dipoles in alkanes:

    • If the carbon chain is longer, there is more molecular surface and therefore stronger instantaneous dipole-induced dipole bonds. This increases the energy required to separate them
    • Branched-chain alkanes have smaller contact areas, so the intermolecular forces are weaker

    Determining if a liquid contains permanent dipoles:

    • Put an electrostatically charged rod next to a jet of a liquid (the rod can be positively or negatively charged). If the jet is deflected, it shows that the liquid is polar
    • This is because the partially positive sides will be attracted to the negatively charged rod, or vice-versa
    • If a liquid is deflected, it likely contains permanent dipole-permanent dipole bonds

    Comparing intermolecular bonds experimentally:

    • Evaporation is endothermic, so the surrounding temperature decreases when something evaporates. If a substance evaporates more easily, the surrounding temperature will decrease faster. With a thermometer and timer, different liquids can be compared
    • If a liquid evaporates very fast, the intermolecular bonds are weaker
    • Therefore, liquids with hydrogen bonding (e.g. water) will cause more of a temperature decrease

    Intermolecular bonds & boiling points:

    • In a liquid or a solid, there are intermolecular bonds causing molecules to be attracted to each another
    • The low boiling points of noble gases show that the bonds between the atoms are very weak, and therefore can be separated easily
    • In the halogens (e.g. F2, Cl2, Br2), fluorine has the lowest boiling point due to the weakest instantaneous dipole-induced dipole bonds (it has fewer electrons)
    • Larger molecules have larger electron clouds, so the instantaneous dipole-induced dipole bonds are stronger. Similarly, atoms with a larger surface area also have stronger instantaneous dipole-induced dipole bonds due to a bigger exposed electron cloud

    c Hydrogen bonding:

      Test yourself
    • Hydrogen bonding is the strongest type of intermolecular bond
    • For a hydrogen bond to form:
      - A hydrogen must be covalently bonded to fluorine, nitrogen or oxygen
         - This is because they all are very electronegative, attracting the bonding electrons
      - There must be a F/N/O on another molecule, so that the hydrogen has a lone pair to hydrogen bond to
    • In water, the oxygen atom has 2 lone pairs of electrons and there are twice as many hydrogen atoms as oxygen atoms, which bond to these lone pairs. So each molecule will form up to 2 hydrogen bonds
    • The structure is shown below. When drawing it, make sure to draw the O-H/N-H/F-H bond in line with the dotted hydrogen bond from the H

    Water ice:

    • In water ice, the structure is open with four groups around each oxygen atom, creating a lattice. This creates a lot of open space, making ice less dense than water, explaining why ice floats

    Properties of liquids with lots of hydrogen bonding:

    • They:
      - Have a high viscosity (don't flow easily)
      - Have high melting and boiling points
      - Are soluble in water, because they form hydrogen bonds with the water molecules
    • Water doesn't follow the first point because there is not much hydrogen bonding in liquid state
    • Usually, organic molecules which form hydrogen bonds contain -OH (alcohol) or -NH (amine) groups. This is because the H must be covalently bonded to an F/N/O atom

    e, h Kinetics:

      Test yourself
    • Activation enthalpy is related to the minimum energy a pair of particles need to react when they collide
    • This energy is used to break/stretch old bonds and form new ones. This intermediate stage is known as the transition state
    • Catalysts provide an alternate pathway with a lower activation enthalpy
    • An example of an enthalpy profile diagram (for an exothermic reaction):

      Here, the Δ H is negative. This is always the case for exothermic reactions
    • In the exam, write the molecular formulae of the reactants and products above the lines
    • A particle with lower activation enthalpy requires less energy (e.g. through heat) to react so it is less stable
    • If two particles collide, they only react (successful collision) if they are above the activation enthalpy and are in the correct positions

    Homogeneous catalysis:

    • Homogenous catalysts are in the same physical state as the reactants in the reaction
    • An example of homogeneous catalysis would be the destruction of ozone in the stratosphere by chlorine and bromine atoms (the radicals act as catalysts because they are not destroyed in the reaction overall; they are recreated)

    f, g Rate of reaction:

      Test yourself
    • As the concentration of a reactant increases, the rate of the reaction also increases. This is because there are more particles in a given volume which will result in more successful collisions
    • Increasing pressure gives the same effect, as it decreases the distance between the particles, allowing for more collisions per second
    • Increasing temperature will also increase the rate of a reaction because it makes the particles move faster (since it increases kinetic energy), increasing the number of collisions per second. This extra energy also brings more particles above the activation enthalpy
    • Increasing surface area (reducing particle size), and increasing radiation intensity (for reactions requiring radiation) also increase the rates of reactions

    Maxwell-Boltzmann distributions:

    • Below, a Maxwell-Boltzmann distribution is shown. These plot the distribution of kinetic energy in a gas; most particles have little energy, well below the activation energy. Only a small proportion to the right of the activation energy line have a chance of having a successful collision and reacting

    The effect of catalysis:

    • A homogeneous catalyst reduces the activation enthalpy by providing a different, more efficient pathway, containing an intermediate compound. Therefore, with a catalyst, more particles have sufficient kinetic energy and therefore the reaction completes faster


      The enthalpy profile diagram will have two bumps due to the intermediate compound being formed:

    The effect of temperature:

    • This graph shows what happens to the distribution as the temperature is increased. More molecules now have higher energies. Note that the area under the curve is still the same

    Measuring rate of reaction:

    • To measure the rate, the loss of reactant or formation of product needs to be measured from the:
      - Volume of gas produced (e.g. with a gas syringe), or loss of mass as a gas is produced
      - pH change
      - Temperature change
      - Colour change [with a colorimeter (from the Year 2 DM module)]
      - Titration results from samples taken at set time intervals
    • Once you have this data, plot a graph of the measurement (y) against time (x) with a curved trendline
      - Draw a tangent and measure the gradient at a specific time
      - This is the rate of reaction at that time
      - Measure the gradient at time 0 to find the initial rate

    i Gas concentrations:

      Test yourself
    • When the concentration of a gas is small it can be easier to express it in parts per million () rather than as a percentage
    • For example, carbon dioxide has a concentration of about 399 , which means that out of one million particles in a random sample of air, 399 of them will be CO2 molecules

    Converting ppm to percentage concentration:

    • Divide the amount in by 10 000 and add a % sign
    • Alternatively, divide by 1 000 000 and convert this decimal to a percentage by multiplying by 100
      (you only need to pick one method, they'll both give the same answer)

    Converting percentage concentration to ppm:

    • Multiply the percentage by 10 000
    • Alternatively, convert the percentage to a decimal by dividing by 100. Then multiply this by 1 000 000
      (you only need to pick one method, they'll both give the same answer)

    j Haloalkanes:

      Test yourself
    • A haloalkane is an alkane with at least one hydrogen atom replaced with a halogen atom
    • Naming haloalkanes follows the same process as naming alcohols (see m in DF) but the name of the halogen atom is added as the prefix to the name of the parent alkane
    • So CH3CH2CH2Cl would be called 1-chloropropane
    • The prefixes of bromo- and chloro- are listed in alphabetical order, and the numbers used to show where bromine and chlorine are located are the lowest possible
    • Use di, tri, or tetra if there is more than one of a specific halogen
    • For example, a carbon bonded to one hydrogen and three chlorine atoms is called trichloromethane
    • The carbon-halogen bond is polar; the electronegative halogen pulls the electrons away from the carbon. This makes the haloalkane carbon-deficient and susceptible to attack by a nucleophile

    Amines:

    • Amines are organic compounds based on ammonia but with alkyl groups
    • They have the general formula R-NH2 (Year 2 students - this is referring to primary amines only)
    • For example:
      methylamine: CH3NH2
      ethylamine: CH3CH2NH2
      propylamine: CH3CH2CH2NH2


       Ethylamine

    k(ii), m Organic Reactions:

      Test yourself
    • The reaction mechanism between hydroxide ions (OH-) and 1-bromobutane by nucleophilic substitution is:
    • Here, the nucleophile (OH-) attacks the carbon atom in the C-Br bonds
    • The (OH-) donates two electrons to form a new dative covalent bond
    • The C-Br bond breaks heterolytically and the bromine atoms receive two electrons, producing a bromide ion (the leaving group)
    • The reaction must be done under reflux
    • The general equation for this reaction is R-X + NaOH → ROH + NaX (where R is an alkyl group (e.g. CH3) and X is a halogen)

    Water as a nucleophile:

    • A hydrolysis reaction involves a bond being broken in a molecule, using water
    • Water can also act as a nucleophile because it has lone pairs on the oxygen atom
    • This reaction is slower than with OH- ions
    • The reactants must be heated under reflux
    • An unstable intermediate has now formed. This produces an alcohol and a H+ ion
    • This is a hydrolysis reaction; R-X + H2O → R-OH + H+ + X-

    Ammonia as a nucleophile:

    • Ammonia acts like water - the lone pair on the nitrogen attacks the haloalkane
    • The haloalkane is heated with concentrated ammonia solution in a sealed tube
    • An amine with an NH2 group is formed:


       (hover over the image to zoom in, pinch-zoom on mobile)
      - The Cδ+ attracts a lone pair from the ammonia and the C-Br bond breaks
      - An ammonia molecule removes a hydrogen from the NH3 group
      - This leaves an amine (e.g. C2H7N, ethylamine) and NH4+

    l Substitution & nucleophiles:

      Test yourself
    • A substitution reaction is a reaction where an atom or group in a molecule is replaced by another atom or group
    • A nucleophile is a molecule or negatively charged ion with a lone pair of electrons that can be donated to a positively charged atom to form a covalent bond
    • Nucleophiles are also known as electron-pair donors

    n Haloalkane reactivity:

      Test yourself
    • As you go down group 7, the C-X bond (where X is a halogen atom) becomes less polar
    • From this, you would assume that C-F is the most reactive
    • However, experimental evidence shows that this is not the case; C-I is actually the most reactive
    • The actual explanation is that the C-I bond is much stronger (has a higher bond enthalpy) than the C-F bond

    Experimental evidence:

    • Add chloro, bromo and iodo alkanes into separate test tubes. Add some silver nitrate solution and ethanol (as a solvent). The silver halide formed is insoluble (forming a precipitate)
    • This reaction is Ag+(aq) + X-(aq) → AgX(s)
    • With the iodoalkane, the precipitate forms fastest, showing that it is most reactive
    • This is an experiment which provides the experimental evidence mentioned above

    o, p(iii) Radicals:

      Test yourself
    • A radical is a molecule with unpaired electrons. They try to fill their outer shells by taking an electron from another atom/molecule

    Heterolytic fission:

    • Heterolytic fission is when both shared electrons in a covalent bond go to just one of the atoms when the bond breaks
    • This produces two ions; one +ve and one -ve
    • Polar molecules typically undergo heterolytic fission
    • A diagram is shown below. Double-headed arrows are used because two electrons move

    Homolytic fission:

    • Homolytic fission is when one shared electron goes to each atom in the bond
    • The atoms have no overall electronic charge, because they both get back their electron that they were sharing in the bond
    • This forms two radicals
    • Single-headed arrows are used because only one electron is transferred to each atom in the bond

    p → p(ii) The 3 stages of a radical chain reaction:

      Test yourself
    • Initiation: radicals are formed, often by photodissociation (see t). For example, Cl2 + → Cl● + Cl●
    • Propagation: this is a chain reaction; radicals are used in reactions and are produced again. For this reason, they are catalysing it. For example, with chlorine:
      Cl● + H2 → HCl + H●
      H● + Cl2 → HCl + Cl●
    • Termination: when two radicals collide with each other, the chain reaction stops because the radicals have been removed

    An example reaction between an alkane and a halogen:

    • Initiation:
      - Cl2 + → Cl● + Cl●
    • Propagation:
      - Cl● + CH4 → HCl + CH3●
      - CH3● + Cl2 → CH3Cl + Cl●
    • Termination:
      - Cl● + Cl● → Cl2
      - or CH3● + Cl● → CH3Cl
      - or CH3● + CH3● → C2H6

    q Depletion of ozone due to haloalkanes:

      Test yourself
    • Haloalkanes such as chloromethane (CH3Cl) and bromomethane (CH3Br) reach the stratosphere naturally
    • Solar radiation causes these molecules to split up giving chlorine and bromine radicals. Chlorine radicals are used in the below equations, but it works the same with bromine radicals
    • There are two propagation reactions:
      - Cl● + O3 → O2 + ClO●
      - ClO● + O● → O2 + Cl●
    • These can be combined into an overall reaction by combining both and cancelling species which appear on both sides. The steps are:
      - Add the left-hand sides: Cl● + O3 + ClO● + O●
      - Add the right-hand sides: O2 + ClO● + O2 + Cl●
         - This simplifies to Cl● + ClO● + 2O2
      - Now add the → back in: Cl● + ClO● + O3 + O● → Cl● + ClO● + 2O2
      - Both sides contain Cl● and ClO●, so they cancel
      - This leaves us with O3 + O● → 3O2. This is the overall reaction
    • Other radicals such as nitrogen oxides can also destroy ozone (NO2 + → NO● + O●)

    Bond enthalpies:

    • C-F bonds are the least likely to break, and C-I the most likely
    • This is because C-F has the highest bond enthalpy and C-I the lowest
    • However, most C-I bonds react before reaching the stratosphere, so they have little effect on the ozone layer
    • Bromine atoms do reach the ozone layer and are more destructive than chlorine radicals, but the stratospheric bromine radical concentration is much lower than chlorine

    r The effects of ozone:

      Test yourself
    • The ozone layer absorbs most high-energy UV radiation, preventing it from reaching the surface
    • The UV that gets through can break chemical bonds such as those in DNA molecules, damaging genes and increasing the risk of skin cancer. It also can damage proteins in the skin, making people look older
    • In the troposphere (the lowest level of the atmosphere) however, ozone is a pollutant and can cause photochemical smog, leading to irritation and respiratory problems. Ozone is also toxic to humans
      - It is produced with sunlight and nitrogen dioxide/hydrocarbons
      - Power stations and vehicles contribute to this
    • However, UV radiation is one of the main ways humans produce vitamin D

    Ozone formation and destruction:

    • Ozone is created when UV radiation dissociates O2 molecules creating two oxygen radicals. One of these radicals can react with another oxygen atom to create an O3 molecule
    • UV can also break apart ozone by splitting it into an O2 molecule and an O● radical
    • Therefore an equilibrium is set up; O2 + O ⇌ O3

    s Ultraviolet radiation:

      Test yourself
    • The electromagnetic spectrum (in order of increasing frequency/energy):
      - Radio waves
      - Microwaves
      - Infrared
      - Visible light
      - Ultraviolet
      - X-rays
      - Gamma rays
    • The Sun mostly emits the green ones on the list above
    • The Earth absorbs radiation from the sum and re-emits it as infrared, at a lower frequency (because the Earth is cooler than the Sun)

    t Energy levels:

      Test yourself
    • Electrons in any molecule have fixed energy levels, quantised energy levels
    • Electrons may be excited to a higher energy level when they absorb radiation
    • With even more energy, the bonding electrons will no longer be able to hold the bond and radicals will be formed. This process is called photodissociation
    • Ionisation is also possible if there is enough energy for an electron to leave
    • The addition of energy from photons is often shortened to in symbol equations

    u Energy, wavelength and frequency:

      Test yourself
    • Electromagnetic radiation travels at 3.00 × 108 in a vacuum (the speed is about the same in air on Earth too)
    • speed of light c () = wavelength (m) × frequency ( or )
    • the energy of a photon E (J) = Planck constant h × frequency ( or )