Subject: Chemistry
Level: AS
Exam Boards: AQA, EDEXCEL


Atomic Structure

MODULE 1

1.1.1

a)

  Proton Neutron Electron
Relative Charge +1 0 -1
Relative Mass 1 1 1/2000

b) Most of the mass within an atom is within the centre of the atom, because all the protons and neutrons are found in the nucleus in the centre of the atom. The positive charge within the atom is within the nucleus, from the protons, while the negative charge is found within the electrons, which surround the atom in orbitals, which make up shells.

c) Protons give the element its atomic number, i.e. what element it is and also, in neutral atoms also indicates the number of electrons in shells. The mass number is the sum of all the protons and neutrons, and indicates how heavy the nucleus is, it also gives what isotope of an element an atom is.

d) 1.You can deduce the number of protons in an atom from its atomic number, and the number of neutrons from its mass number, by calculating the mass number – the atomic number (i.e. the total mass – the number of protons). In a neutral atom, you can deduce the number of electrons by the atomic number, because the negative charge must equal the positive charge. 2.The number of protons and neutrons can similarly be deduced from an ion by looking at its atomic number and mass number, and the number of electrons can be deduced from looking at the proton number and the ionic charge. The charge is the difference between the number of protons and electrons, if there is a positive ionic charge then there are more protons and if there is a negative ionic charge then there are more electrons.

e) Isotopes are atoms of an element with different numbers of neutrons and therefore different masses.

f) 12C is used as the standard measurement for relative masses, they are measured against the mass of 1/12 of 12C.

g) Relative isotopic mass is the mass of an atom of an isotope compared with one twelfth of the mass of carbon twelve.   Relative atomic mass (Ar) is the weighted mean mass of an atom of an element compared with one twelfth of the mass of an atom of carbon-12.

h) You can calculate the relative atomic mass of an element given the relative abundances of its isotopes:

E.g. A sample contains 75% 35Cl and 25% 37Cl – calculate the relative atomic mass of chlorine:

(0.75 x 35) + (0.25 x 37) = 35.5

i) Relative molecular mass (Mr) is the weighted mean mass of a molecule compared with one twelfth of the mass of an atom of 12C. Relative Formula Mass is the weighted mean mass of a formula unit compared with one twelfth of the mass of an atom of 12C. You may be asked to calculate the value of a RFM or RMM by adding the values of the RAM of individual atoms within a compound.

1.1.2

a)

  1. Amount of Substance is the quantity whose unit is a mole. It is a means of counting atoms.
  2. A mole is the amount of any substance containing as many particles as there are carbon atoms in exactly 12g of the carbon-12 isotope.
  3. Avogadro’s Constant (NA) is the number of atoms per mole of the carbon-12 isotope (6.02 x 1023).

b) Molar mass, M, is the mass per mole of a substance. The units of molar mass are g mol-1

c)

  1. Empirical Formula is the simplest whole-number ratio of atoms of each element present in a compound.
  2. Molecular formula is the number of atoms of each element present in a molecule. d) An empirical formula is the lowest ratio that a formula can be put into, it can be calculated using experimental data.
  3. Find moles of each element (mass/RAM) – if you are given a percentage concentration then use each of these as the mass.
  4. Ratio moles (divide by the smallest number)
  5. Record formula

e) An equation is constructed from the reactants and the products. The ratio of each reactant and product in the balanced equation is the ratio of moles required for the reaction.

f) The concentration of a solution is the amount of a solute, in mol, dissolved per 1dm3 (1000 cm3) of solution. A standard solution is a solution of known concentration. Standard solutions are normally used in titrations to determine unknown information about another substance. A species is any type of particle that takes part in a chemical reaction.

g) Stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction. From calculations, you can determine the ratio of moles of reactants needed for each to react completely.

h) A concentrated solution has a large amount of solute per dm3 – concentrated solutions usually have a concentration of 10M or more. A dilute solution has a small amount of solute per dm3 – these are usually used in the lab, and have a concentration of 1-2M.

1.1.3

a) An acid is a substance which is a proton donor. All acids contain hydrogen, which when dissolved in water (aqueous solution) releases H+ ions.

b) Hydrochloric acid – HCl Sulphuric acid – H2SO4 Nitric Acid – HNO3

c) A base is a species that is a proton acceptor. The common bases are metal oxides, metal hydroxides and ammonia (NH3).

d) An alkali is a soluble base, which releases hydroxide (OH-) ions when in aqueous solution.

e) The most common alkalis are sodium hydroxide (NaOH), potassium hydroxide (KOH) and aqueous ammonia (NH3).

f) A salt is any chemical compound formed from an acid when an H+ ion from the acid has been replaced by a metal or another positive ion, such as the ammonium ion, NH4+.

g) When an acid reacts with a carbonate, it forms a salt, water, and carbon dioxide:

e.g.

CaCO3 + 2HCl → CaCl2 + H2O + CO2
When an alkali or base reacts with an acid, a neutralisation reaction occurs, to form a salt and water:

e.g. aOH + HCl → NaCl + H2

h) A base will readily accept an H+ ion from an acid, to form a new compound, e.g. OH- forms H2O and NH3 forms NH4+.

i) Hydrated refers to a crystalline compound containing water molecules. Anhydrous refers to a substance that contains no water molecules. Water of Crystallisation refers to water molecules that form an essential part of the crystalline structure of a compound.

j) Using percentage composition or mass composition, you can find the formula of a salt by finding the ratio of water molecules to molecules of the anhydrous salt. This, along with stoichiometry, allows the formula to be found. This can also be obtained using experimental data.

k) Perform acid-base titrations and carry out structured titrations – this is a practical element, but can be described in an exam as:

  • Use a burette with either acid or alkali in the top, and drop it down into a conical containing the other solution.
  • Do a rough run using an indicator such as methyl orange, and then use this value to repeat the titration more exactly.
  • If a salt needs to be obtained, repeat the titre without the indicator, filter and evaporate to give pure crystals.

1.1.4

a) An oxidation number is a measure of the number of electrons that an atom uses to bond with atoms of other elements. They are also known as oxidation states. Oxidation numbers are derived from a set of rules:

  • Elements have an oxidation number of 0 (e.g. Na = 0, Cl2 = 0)
  • Neutral molecules have an overall oxidation number of 0 (e.g. NaCl = 0, where Na = +1 and Cl = -1)
  • Charged species have an overall oxidation number the same as the charge (e.g. [SO4]2- = -2)
  • Metal ions have an oxidation number equal to their group number (1-3) because they have a charge that is equal to their group number (e.g. Na+ = +1, Mg2+ = +2)
  • Combined hydrogen has an oxidation number of +1, except in metal hydrides where it has an oxidation number of -1.
  • Combined oxygen has an oxidation number of -2, except in peroxides, where it is -1 or when combined with fluorine, which is more electronegative than oxygen.
  • Combined halogens usually have an oxidation number of -1.

b) OIL RIG Oxidation is the loss of electrons or an increase in oxidation number of a species. Reduction is the gain of electrons or a decrease in oxidation number of a species.

c) Roman numerals can be used to indicate the magnitude of the oxidation state of an element when the name may be ambiguous, e.g. nitrate(III) and nitrate(V), which are both nitrate groups, but in nitrate(III) the nitrogen has an oxidation number of +3, whereas in nitrate(V) the nitrogen has an oxidation number of +5.

d) You can write formulae using oxidation numbers – if the overall compound is neutral, then the overall sum of the oxidation numbers must be zero. If the compound has an overall charge then the sum of the oxidation numbers must match that charge. E.g. SO42- has a charge of 2- and so its oxidation state must be -2. Oxygen has an oxidation number of -2, which means that in this case, sulphur must be +6, in order to balance the compound with a 2- charge.

e)

  1. Metals generally form ions by losing electrons, which increases their oxidation number. They form positive ions. Most metals have 1-3 electrons in their outer shell, and so to gain stability, they need to lose electrons.
  2. Non-metals generally gain electrons to form ions, which decreases their oxidation number. They form negative ions. Non-metals have 4-8 electrons in their outer shell and so in order to gain stability, it is easier to gain electrons. The noble gases, which have 8 outer shell electrons, are inert and do not easily form ions.

f) The reaction of a metal with dilute acid is a redox reaction. The generic equation is metal + acid → salt + hydrogen. E.g. Mg + 2HCl → MgCl2 + H2 E.g. Ca + H2SO4 → CaSO4 + H2

g) INTERPRET AND MAKE PREDICTIONS FORM REDOX EQUATIONS IN TERMS OF OXIDATION NUMBERS AND ELECTRON LOSS OR GAI.

MODULE 2

1.2.1

a) The first ionisation energy of an element is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of unipositive gaseous ions. Successive ionisation energies are a measure of the energy required to remove each electron in turn. For example, the second ionisation energy of an element is the energy required to remove one electron from each ion in one mole of gaseous 1+ ions to form one mole of 2+ gaseous ions.

b) Ionisation energies are affected by nuclear charge, electron shielding and distance of the outermost electron from the nucleus:

  • Nuclear charge – the more protons there are in the nucleus, the greater the positive charge, which means that the electrons are more tightly held. As you go across the period, the electrons are held in the same space but held by a stronger nuclear charge, so the ionisation energies increase.
  • Electron shielding – ionisation energies decrease as you go down a group as there are more shells of electrons between the nucleus and the outer shell electrons, which reduces the effective nuclear charge on the electrons and makes it easier to remove them.
  • Distance of outermost electrons from the radius – as you go down a group, the atomic radius increases due to more shells of electrons, which means there is a greater distance between the nucleus and the outermost electrons, which causes the effective nuclear charge to be weaker.

c) PREDICT SUCCESSIVE IONISATION ENERGIES

d) A shell is a group of atomic orbitals with the same principal quantum number, n. This is also known as a main energy level. Principal Quantum Number, n, is a number representing the relative overall energy of each orbital which increases with distance from the nucleus. The sets of orbitals with the same n value are referred to as electron shells or energy levels. The first energy level can hold only two electrons, in an s orbital. The second can hold 8, in an s orbital and a p orbital. The 3rd and 4th can both hold 18, in s, p and d orbitals.

e) **An orbital is a region that can hold up to two electrons with opposite spins. **

f) An s orbital is a roughly spherical shaped orbital, which holds 2 electrons. A p orbital is propeller shaped and can also hold 2 electrons.

g) An s subshell is made up from one s-orbital and can hold one pair of electrons. A p subshell is made up from three p-orbitals, perpendicular to each other, and can hold 6 electrons in total, two in each p orbital. A d subshell is made up from 5 d-orbitals, each of which can hold two electrons, so the d-subshell can hold 10 electrons in total.

h) An s orbital has less energy than a p orbital, which has less energy than a d orbital. The first energy level is the lowest energy, and then the second energy level etc. The orbitals fill in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p…

  1. DEDUCE THE ELECTRON CONFIGURATIONS OF ATOMS AND IONS
  2. CLASSIFY THE ELEMENTS INTO S P AND D BLOCKS

1.2.2

a) Ionic bonding is the electrostatic attraction between oppositely charged ions.

b) DOT AND CROSS DIAGRAMS IONIC

c) You can predict the charge of ions from their position on the ionic table, due to the electron movement that will make them the most stable. Groups 1, 2 and 3 lose 1, 2 and 3 electrons respectively to form their ions, therefore usually have a charge of 1+, 2+ or 3+. Group 7 usually gains one electron, group 6 gains 2 and group 5 gains 3, therefore these groups usually have a charge of -1, -2, and -3 respectively.

d) Some common radical groups are ions: Nitrate: NO3- Carbonate: CO32- Sulphate: SO42- Ammonium: NH4+ e) A covalent bond is a shared pair of electrons

f) DOT AND CROSS COVALENT

g) The shape of a simple molecule is determined by the repulsion between electron pairs surrounding the central atom to give maximum stability.

h) Lone pairs of electrons repel more than bonded pairs of electrons, and so the angle created by lone pairs is greater than the angle created by bond pairs.

i)** EXPLAIN SHAPES OF AND BOND ANGLES IN MOLECULES AND IONS SURROUNDING A CENTRAL ATOM **

j)** PREDICT THE SHAPES OF AND BOND ANGLES IN MOECULES ANALAGOUS TO THOSE STUDIED**

k) Electronegativity is the ability of an atom to attract the bonding electrons in a covalent bond.

l) When covalently bonded atoms have different electronegativity, the electrons in the bond will sit closer to the more electronegative species, which causes a permanent dipole within the bond, which causes a polar bond, as the less electronegative end will be + and the more electronegative end will be -.

m) There are three types of intermolecular forces – Van der Waals, dipole-dipole and hydrogen bonds.

  • Van der Waals (e.g. noble gases) – Electrons in a molecule are constantly in flux, and at different times there may be more electrons in one end of the molecule than the other, which creates an instantaneous dipole on the molecule. When this comes up against another molecule, it can create a temporary induced dipole, where the negative end of that molecule is electrostatically attracted to the positive end of the other molecule. This attraction is what causes Van der Waals.
  • Permanent Dipoles (e.g. Hydrogen chloride) – Some molecules have a polar bond, due to the electronegativity of the species in the molecule, and if all the polar bonds act in the same direction, the molecule is polar, i.e. it has a + and - end. This can be attracted to another molecule that has opposite orientation.

n) Hydrogen bonding is an attraction between an electron deficient hydrogen atom, and a lone pair of electrons on a highly electronegative atom on a different molecules. There are only a few species in which hydrogen bonding can occur: nitrogen, oxygen and fluorine. Some examples of molecules which have hydrogen bonding are H2O and NH3

o) Water has some anomalous properties for its simple molecular structure, which can be explained by hydrogen bonding:

  • Water is denser than ice – this is because ice is an open lattice with hydrogen bonds holding the water molecules apart. When ice melts, the hydrogen bonds collapse, allowing the water molecules to move closer together, which increases the density of water.
  • Water has a relatively high freezing and boiling point – The hydrogen bonds are much stronger intermolecular forces than van der Waals, which are found in most simple molecules. As water has hydrogen bonds, more energy is required to overcome these than if there were just Van der Waals.

p) **Metallic bonding is the attraction of positive metal ions to delocalised electrons. **

MODULE 3

a) The periodic table is arranged in order of increasing atomic number, which denotes the number of protons in the nucleus of an atom of a given element. It is arranged in groups and periods, with each period having repeating trends in physical and chemical properties and each group having similar physical and chemical properties.

b) Periodicity is the regular periodic variation of properties of element s with atomic number and position in the periodic table.

c) In each group, elements have the same number of outer shell electrons, as well as having the same kind of orbitals in the outer shell, which results in them having similar properties. The repeating pattern of similarity is caused by the repeating pattern of similar electronic configuration.

d) DESCRIBE AND EXPLAIN THE VARIATION OF THE FIRST IONISATION ENERGIES OF ELEMENTS

e) In periods two and three, as you go across the period, the number of protons in the nucleus of the atom increases. As the atoms are neutral, this means that the number of electrons surrounding the nucleus also increases.



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