UNIT 1: Principles & Applications of Science I

Transcription

1 Level 3 Applied Science UNIT 1: Principles & Applications of Science I CHEMISTRY SECTION Name:.. Teacher:.. Level 3 Applied Science Unit 1 (Chemistry) 1

2 Contents 1. The Periodic Table, Atoms, & Ions Page Vid # Done revised 1.1 Introduction to the atom Elemental symbols in the Periodic Table Introduction to the Periodic Table of elements Making ions Trends in ionic radius Ionisation energy Electron affinity Compounds, Bonding & Intermolecular Forces 2.1 Introduction to bonding in compounds Ionic bonding Covalent bonding Metallic bonding Electronegativity Introduction to intermolecular forces Permanent dipole-dipole forces Temporary dipole-induced dipole forces Hydrogen bonding Trends in melting and boiling points across a period Trends in melting and boiling points down a group Orbital Theory 3.1 Sub-shells and orbitals Electron in box diagrams and electron configurations Blocks in the periodic table Ionisation energies re-visited Balanced Equations & Chemical Reactions 4.1 balancing equations Reactions of period 2 and 3 elements with oxygen Reaction of metals with oxygen, water, and dilute acids Redox Oxidation numbers for transition metals and oxyanions Reactivity series Displacement reactions Uses and applications of the substances in this unit Quantitative Chemistry 5.1 Moles and masses Moles and solutions Moles and equations Percentage yield Questions 51 Checklist 54 Videos can be found at hccappliedscience.weebly.com under unit 1 and chemistry Level 3 Applied Science Unit 1 (Chemistry) 2

3 1.1 Introduction to the atom 1. The Periodic Table, Atoms & Ions The atom consists of a central nucleus where p (positively charged) and neutrons (neutral) are found. In s around the nucleus are electrons (negatively charged). Relatively, protons and neutrons have a s mass. So if we say the mass of a proton is 1 then the mass of a neutron is also. Relative to protons and neutrons, the mass of an electron is so small that it is negligible You must be able to recall the relative mass and charge for each of the sub-atomic particles: Particle Relative mass Relative charge Position in atom Proton 1 Neutron Electron Most of the mass is found in the n of the atom. Note that we use dots (or crosses) to show electrons in shells. Up to t electrons are found in the first shell outside the nucleus and e electrons in the next shell. For a neutral atom (one that has no charge and therefore is not an ion), the number of positive protons e the number of negative electrons. The atom shown above has 7 protons and 7 electrons and is nitrogen. We can tell how many protons, neutrons and electrons are in an atom by looking at the elemental symbols in the periodic table (see next section). Complete questions 1-2 Level 3 Applied Science Unit 1 (Chemistry) 3

4 1.2 Elemental symbols in the Periodic Table You can work out how many protons and neutrons are in the nucleus of a particular element when given its elemental symbol along with its atomic number and mass number: Mass number Atomic number 14 7 N Elemental symbol Atomic number: the number of protons in the nucleus of an atom Mass number: the number of protons and neutrons in the nucleus of an atom For a neutral atom (one that has no charge and therefore is not an ion), the number of positive protons e the number of negative electrons. Therefore the atomic number also tells you how many e are present in a neutral atom. The number of neutrons can be found by s the atomic number from the mass number. Nitrogen has - = neutrons. Note that in the periodic table the average of the mass numbers, the relative atomic mass, is given, not the mass number. Relative atomic mass Atomic number 35.5 Cl 17 Elemental symbol For example, in nature two i of chlorine are found; 35 Cl and 37 Cl. Both i have an atomic number of 17, telling us that chlorine has 17 protons and also 17 electrons when a neutral atom. The only difference between any i of the same element is the number of n and therefore the m numbers. The two isotopes of chlorine react the same despite having a different number of neutrons and different mass numbers because they still have the same number of e. Chemical reactivity depends on e movement. You might think to yourself that the mean of 35 and 37 is ( ) / 2 =. But the relative atomic mass (R.A.M.) is not worked out in this simple way it considers the relative contribution of each isotope. The symbol in the Periodic Table shows an average of Because the average is closer to 35 this must mean there are m 35 Cl atoms present in a sample of chlorine found in nature and less 37 Cl. So to work out the number of neutrons in an atom, you do not look at the relative atomic mass, as this is just the average mass, but instead you look at the mass number given for that p isotope. Relative atomic mass: the average mass of an atom of an element compared to 1/12 th of the mass of an atom of 12 C Level 3 Applied Science Unit 1 (Chemistry) 4

5 Group Group Group Group Group Group Group Group Look at the definition for R.A.M. above if you take an atom of pure 12 C, only 1/12 th of that atom is used as the s for weighing atomic masses. Atomic masses are weighed r to 1/12 th of 12 C. The relative atomic masses do not have any units because they are only relative numbers. Examples for calculating the number of sub-atomic particles when given pure isotopes: Pure Isotope Atomic no. (protons) Mass no. (protons + neutrons) No. of electrons No. of neutrons R.A.M in periodic table 16 O 32 S 10 B 24 Mg Complete questions Introduction to the Periodic Table of elements + Research how you calculate relative atomic mass, with at least one example calculation shown Identify the groups and periods. Remember that period 1 contains the elements H and He and is not the row from Li to Ne (a common mistake)! Also identify how you can find the metals and non-metals. Period Period Period Period Level 3 Applied Science Unit 1 (Chemistry) 5

6 The Periodic Table shows all of the chemical elements arranged in order of i atomic number. Elements are organised into vertical columns called g and horizontal rows called p. Period 1 contains the elements H and H. Chemical properties are similar for elements that are in the same group because they all have the s number of electrons in the outer shell. The atomic number i as you move from l to r across a period because each element has o more proton than the element before it in the same period. There are trends (or patterns) that r themselves each time you go across a period. For example, each time you go from left to right across a period you go from metals to non- and the atomic radius also decreases. This repeating pattern seen by the elements across a period is called p. Periodicity: the repeating pattern seen by the elements in the periodic table The diagram below shows how the atomic radius d across a period but i down a group. Add the missing electrons to the electron shells using the atomic numbers for the neutral atoms. 3Li 4Be 5B 6C 7N 8O 9F 10Ne 11Na To explain trends in atomic radius (and other trends that you shall meet in this unit), the follow pneumonic shall help you to re-call all of the factors that are responsible for the trends: C A R S Level 3 Applied Science Unit 1 (Chemistry) 6

7 Across a period, the atomic number i as the number of positive protons in the nucleus increases by o each time, so the nuclear charge increases. The increased positive nuclear charge each time you go across a period means even stronger attraction on the electrons; the electrons experience stronger nuclear attraction which draws them to be c to the nucleus. Because the electron shells are closer to the nucleus this means that the atomic radius has d across the period. One last factor is shielding. The inner shell of electrons shields the outer shell of electrons from the positive nuclear charge. The inner shell of electrons also r the outer shell of electrons to be further from the nucleus. Hence, the more inner shells of electrons there are in an atom the greater the shielding. Across a period electrons are added to the s shell, not a different shell. This means that there is the same number of inner shells and therefore the same shielding across a period (and is not a big factor affecting the radius). Each time you go down a group however, there is an additional electron shell added meaning an i in radius. The greater number of inner shells also results in i shielding. Therefore the outermost electrons are l strongly attracted to the nucleus with them being further away with the larger radius, so there is w nuclear attraction on the electrons. Note that the nuclear charge does also increase down a group, and you may think this would draw the electron shells inwards to be closer to the nucleus. However, the increased shielding and larger radius from more electron shells o the increase in nuclear charge, resulting in an overall larger atomic radius. e.g. MODEL ANSWER: explain the difference in atomic radius between C and N. e.g. MODEL ANSWER: explain the difference in atomic radius between Mg and Ca. Complete questions 6-12 Level 3 Applied Science Unit 1 (Chemistry) 7

8 Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7 Group Making ions Cation: ion with positive charge Anion: ion with a negative charge So far we have only looked at neutral atoms, where the number of p protons e the number of n electrons. However, atoms can gain or lose electrons during chemical reactions to form charged particles called i in order to satisfy the octet rule. The octet rule states that elements gain or lose electrons in order to have e electrons in the outermost shell, like the noble gases. The noble gases already have eight electrons in their outer-most shell and they are very stable, existing as atoms only and do not form ions. E.g. Na + has lost o negative electron to leave behind a positive charge; Cl - has g o negative electron and therefore has a negative charge; N 3- has gained t negative electrons. Below are examples of ions careful when working out the number of electrons! Atom/Ion Atomic no. Mass no. No. of protons No. of neutrons No. of electrons 14 N 3-16 O 2-23 Na + 27 Al 3+ We can predict the charge that an ion of a given element shall form by looking at its p in the periodic table. You must learn the charges that elements from each group form common but varied Transition metals E.g. sodium (Na) is in Group 1 of the periodic table and therefore only has o electron in its outer shell. To have a c outer shell containing e electrons, the sodium atom must l this one outer electron (electrons are negatively charged), which would result in a sodium i with a charge of +1 (written as Na + ). All group 1 elements form +1 ions for the same reason. Level 3 Applied Science Unit 1 (Chemistry) 8

9 Note that group 4 elements do not usually form ions; they have f electrons in their outer-most shell and it takes too much energy to gain or lose four more electrons in order to complete the octet. Group 0 (also called Group 8) elements already have e electrons in their outer-most shell and are very s atoms so they do not form ions. The ions below are missing charges predict the charges expected remember to always check the position of the element in the Periodic Table! Mg 2+ F N Al P Br Sr O Cs K Li Cl S Ca Rb Ba I Cu Fe Mn Draw diagrams to show the formation of the following ions: i) Mg to Mg 2+ ii) K to K + iii) O to O 2- Complete questions Level 3 Applied Science Unit 1 (Chemistry) 9

10 1.5 Trends in ionic radius You have seen previously that for NEUTRAL atoms there are trends in radius across a period and down a group. The same patterns also exist with the cations/anions. But you must also know the comparison of the radius of a neutral atom to the radius of its cation/anion across a period. Down a group for cations and anions: the same trend is seen as with the neutral atoms; as more electron shells are added down a group the ionic radius i. Across a period for cations: the same trend is seen as with neutral atoms; across a period the radius of a cation d. The cations have l electrons to have the positive charge and across a period there is a greater nuclear charge and ionic charge on the cation, attracting the remaining electrons even more s resulting in the cations becoming smaller. But you must also know that the ionic radius of a cation compared with the neutral atom is s because there is still the s number of protons in the nucleus of that atom now attracting l electrons (electrons were removed to form the cation). Neutral atom Na Mg Al Cation (smaller) Na + Mg 2+ Al 3+ Note that across a period the cations are isoelectronic they have the s number of electrons. Prove this by completing the electronic structures of the first three cations in period 3 of the Periodic Table: Neutral atom electron structure Na: Na + : Cation electron structure Mg: Mg 2+ : Al: Al 3+ : Isoelectronic: having the same number of electrons Level 3 Applied Science Unit 1 (Chemistry) 10

11 Across a period for anions: the same trend is seen as with neutral atoms; across a period the radius of an anion d. The anions have g electrons to complete the octet with a negative overall charge, but with the extra protons across the period radius still decreases. But you must also know that anions have a l radius than the corresponding neutral atom because the added electron(s) when you form the anion cause extra repulsion. Neutral atom P S Cl Anion (larger) P 3- S 2- Cl - Across a period the anions are also isoelectronic once again prove this by completing the electronic structures of the first three anions in period 3 of the Periodic Table in the table below. Neutral atom electron structure P: P 3- : Anion electron structure S: S 2- : Cl: Cl - : But how are the cations/anions made in the first place? There are two ways of forming ions; removing electrons is the ionisation energy and adding electrons is the electron affinity (see next section). Complete questions Ionisation energy First ionisation energy: the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous +1 ions Because there is a massive number of atoms even in 1 gram of a substance, it is not appropriate to count atoms in millions or even billions. We need a bigger quantity to count atoms - chemists use the mole: One million: One billion: One mole: or x in standard form Level 3 Applied Science Unit 1 (Chemistry) 11

12 When we remove electrons from atoms, we measure the energy required to remove o mole of electrons from one mole of atoms in the g phase. We can write equations to show the process. Removing an electron leaves a p charged ion after the arrow and we also show the electron that has been removed. The state symbol for the gaseous phase must be shown. Na (g) Na + (g) + e first ionisation energy = +496 kjmol -1 You do not need to remember the numerical values, only the trends shown across a period and down a group (see below). For the example shown above, 496 kj of energy are required per mole of electrons removed from one mole of sodium atoms in the g phase. The p sign before the numerical value shows that energy is r for this process; energy is required to break the a between the electron and the positively charged n. More than one electron can be removed in a stepwise process. For example, if you want to form a 2+ ion then you initially remove the first mole of electrons from the n atom in the gaseous phase to form a +1 ion and then go back and remove the second mole of electrons from the +1 ion to form the +2 ion. Hint: if writing the equation for the first ionisation energy then you are going to form the +1 ion after the arrow. If writing the equation for the 7 th ionisation energy then you are going to form the + ion after the arrow. A few examples are shown below: Mg + (g) Mg 2+ (g) + e second ionisation energy = kjmol -1 Al 2+ (g) Al 3+ (g) + e third ionisation energy = kjmol -1 There are periodic trends in the first ionisation energy which can be explained using the pneumonic C which we met earlier. The answer is the exact same as used for explaining trends in atomic radius. Across a period the first ionisation energy i (there are some anomalies e.g. between Be and B and also N and O the reason for this shall be explained in the later section on orbitals). Down a group the ionisation energy d. Level 3 Applied Science Unit 1 (Chemistry) 12

13 e.g. MODEL ANSWER: explain why the first ionisation energy for F (1680 kjmol -1 ) is greater than the first ionisation energy for C (1090 kjmol -1 ). e.g. MODEL ANSWER: explain why the first ionisation energy for K (418 kjmol -1 ) is less than the first ionisation energy for Na (494 kjmol -1 ). Complete questions Electron affinity Electron affinity: the change in energy when one mole of a gaseous atom gains one mole of electrons to form one mole of gaseous negative ions Cl (g) + e Cl - (g) first electron affinity = -349 kjmol -1 When adding electrons to atoms, the negative sign before the numerical value indicates that energy is r (the opposite of ionisation energy). There are also the same general periodic trends in electron affinity values as were seen with ionisation energy values. These can be fully explained using the pneumonic C. Across a period the electron affinity values i as more energy is r when nuclear attraction for the electron is s. Down a group when nuclear attraction for the electron to be captured is w less energy is r. There are some anomalies. Look at the table of electron affinity values for group 6 and group 7 elements below. You should notice that O and F do not fully fit the trend going d a group. This is because these atoms are very s and placing an extra electron in a c area is difficult and there is significant r. This repulsion lessens the attraction for the incoming electron resulting in a l than expected electron affinity value. Level 3 Applied Science Unit 1 (Chemistry) 13

14 Group 6 element First electron affinity/kjmol -1 Group 7 element First electron affinity/kjmol -1 O -141 F -328 S -200 Cl -349 Se -195 Br -324 Te -190 I -295 Note that Group 6 elements can also have a s electron affinity to form the 2- ions. Adding a second electron to the 1- ion shall r energy to overcome the r force, as this time the negative electron to be captured shall be r by the negative 1- ion. The energy change shall therefore be p as energy is required to force the second electron into the 1- ion. E.g. values and equations for oxygen: O (g) + e O - (g) first electron affinity = -141 kjmol -1 (energy released) O - (g) + e O 2- (g) second electron affinity = +844 kjmol -1 (energy required) e.g. MODEL ANSWER: explain why group 6 elements have a lower electron affinity than group 7 elements e.g. MODEL ANSWER: explain why less energy is released for electron affinities down a Group. Complete questions Level 3 Applied Science Unit 1 (Chemistry) 14

15 2. Compounds, Bonding & Intermolecular Forces 2.1 Introduction to bonding in compounds Atoms or ions can come together to form compounds. There are three main types of bonding; i, c and m. You can only determine the type of bonding present if you know where the metals and non-metals are located in the Periodic Table of elements and also what the charges are on ions from the position of the element in the Periodic Table show these on the Periodic Table below: Ionic bonding occurs between m and non-metals (and also when there are molecular ions such as NH4 + present in a compound). Two oppositely charged ions attract each other in the ionic bond. Use the cross-over method to find the formula of the ionic compound. The cross-over method for deducing ionic formulae (i) Write the charge above each ion by checking the position of the element in the Periodic Table. (ii) Make the simplest ratio for the charges. (iii) Swap numbers and use them as a subscript for the other ion. Example 1: what is the formula for aluminium oxide? Al 3+ O 2- Al2O3 Example 2: What is the formula for calcium chloride? Ca 2+ Cl 1- CaCl2 Example 3: what is the formula for magnesium oxide? Mg 2+ O 2- Mg 1 O 1 MgO Example 4: what is the formula for aluminium carbonate (carbonate is the molecular ion CO3 2- )? Al 3+ CO3 2- Al2(CO3)3 *Use brackets for molecular ions Level 3 Applied Science Unit 1 (Chemistry) 15

16 Covalent bonding occurs between two n -metals. Ions are not involved; instead the two nonmetals come together to s one electron each via a single covalent bond. More than one electron may be shared (a multiple bond) and there are also cases where the pair of electrons in the covalent bond has come from only o of the non-metals (dative covalent bond). Examples of covalently bonded molecules are O2, F2 and NH3. Metallic bonding occurs when there are only m present; either a metal on its own or a mixture of metals (an alloy). The negative d electrons are attracted to the p charged metal ions. Metallic bonding exists in e.g. Cu, Fe and the alloy brass (mixture of Zn & Cu). The octet rule states that atoms with e electrons in their outer-most shell are extremely s (apart from the first shell, where two electrons are required). The Noble gases (group 8 elements) exist as single a with 8 electrons in their outer-most shell (except Neon, which has 2 outer electrons in the first shell only). They are therefore extremely stable (and extremely un-reactive). What about atoms that do not have eight electrons in their outer-most shell? These atoms can satisfy the octet rule by either: - S electrons with other elements until both have eight electrons (covalent bonding), or - T electrons until both have eight electrons (to make ions, ionic bonding). We can use dot & cross diagrams to show the number and source of the electrons in a compound. (i) Draw a c to represent the outermost shell only. (ii) Use d to represent the electrons from one of the elements and c to represent electrons from the other element. In cases where there is more than two different types of element involved, use a third symbol (and therefore a symbol key to tell the examiner what your chosen third symbol represents e.g. a triangle, a square, the letter e etc). (iii) Covalent compounds: draw two o circles and put the electrons being shared in the overlapping region. Include those electrons not involved in bonding (l pairs). (iv) Ionic compounds: show two s circles as ions, with s brackets around the circle and the ionic c to the top right outside the brackets. The charge should be the same as the Group number and you should also check the formula with the cross-over method so that both opposite charges cancel out overall due to the correct number of each type of ion. Level 3 Applied Science Unit 1 (Chemistry) 16

17 Example 1: Draw a dot & cross diagram for chlorine, Cl2? X X X X Cl X X X Example 2: Draw a dot & cross diagram for Sodium chloride, NaCl? Na Cl Note: All ionic compounds actually exist as a giant ionic lattice see later. We do not have small molecules of NaCl; each ion attracts the oppositely charged ion in all directions to form a giant ionic lattice. 2.2 Ionic bonding Ionic bonding: the electrostatic force of attraction between two oppositely charged ions Ionic bonding occurs when you have a mixture of m and non-metals bonded together. The Metal l electrons to form a positively charged ion and the non-metal g electrons to form n ions. After electrons have been transferred to give two ions with a stable octet these oppositely charged ions then a each other. Electrostatic attraction is the force experienced by oppositely charged particles and holds particles strongly together. You must also use the cross-over method to find the formula of the ionic compound. Two non-metals, so the bonding is covalent. Draw two overlapping circles. Each Cl atom has 7 electrons in its outer-most shell (group 7). If each Cl atom contributed one electron to the shared area then each Cl atom would now have access to 8 electrons, satisfying the octet rule. Use dots to represent the electrons in one Cl atom and crosses to represent electrons in the second Cl atom. + - A metal and a non-metal, so the bonding is ionic. Draw two separate circles with square brackets. Cl X Na is a Group 1 metal. It forms +1 ions. Cl is a Group 7 non-metal. It can accept the electron that the Na wants to lose so that both now have 8 electrons to satisfy the octet rule. The Cl now has a charge of -1. In the example, the dots represent the 7 electrons in Cl and the cross represents the electron gained from Na. Complete question 30 Name of ionic compound Cross-over working Sodium fluoride Magnesium fluoride Aluminium chloride Magnesium phosphide Formula of ionic compound Level 3 Applied Science Unit 1 (Chemistry) 17

18 When drawing dot and cross diagrams you must draw separate circles as ions; the charge on each ion can be predicted by looking at the position of the element in the Periodic Table. Also ensure that you have the correct number of each type of ion drawn out by using the cross-over method to work out the correct formula. You may also be asked to calculate the relative formula mass for an ionic compound. The relative formula mass is the sum of the relative atomic masses for each element in the formula. There are no units for relative formula mass; the numbers are simply relative to 12 C. Draw dot & cross diagrams for the following and also calculate the relative formula mass for each compound given below. (i) sodium fluoride, NaF (ii) magnesium chloride, MgCl2 (iii) aluminium oxide, Al2O3 R.F.M (iv) Barium oxide, BaO R.F.M (v) Calcium sulphide, CaS R.F.M (vi) Barium nitride, Ba3N2 R.F.M R.F.M R.F.M Level 3 Applied Science Unit 1 (Chemistry) 18

19 Remember that ionic compounds do not exists as simple molecules. Each ion attracts the oppositely charged ion in all d resulting in a giant ionic l, which is a regular arrangement of positive and negative ions. E.g. the giant ionic lattice structure of NaCl: Na+ + Na+ Cl Na+ + Draw in an exam as: Na+ Cl Na+ + Na+ Cl Na+ + Na+ Cl Na+ + Finally, the strength of the electrostatic force and therefore the i bond depends on: (i) the ionic charge: a bigger ionic charge results in a s force of attraction (ii) ionic radius: a larger radius (e.g. when you have more shells of electrons) means the ionic charge is spread over a l surface area, resulting in a w attraction for the oppositely charged ion compared with a smaller radius. Complete questions Covalent bonding Covalent bond: a shared pair of electrons Dative covalent bond: a shared pair of electrons and both electrons have come from the same atom Covalent bonding occurs between two non-metals where the atoms come together to share a p of electrons to form a molecule. This is drawn with overlapping circles and the shared pair of electrons in the covalent bond shown in the overlapped area. One pair of electrons in total in the shared area results in a s bond, two pairs of electrons a d bond and three pairs of electrons a t bond. The term relative molecular mass is used for covalently bonded compounds as these do exist as m. Relative formula mass was used for ionic compounds which do n exist as molecules. Where b electrons have come from the s atom, a d covalent bond is formed. An example is in the reaction of NH3 with a H + ion to form the molecular ion NH4 +. Level 3 Applied Science Unit 1 (Chemistry) 19

20 Dative covalent bond. Two dots to show that both of the electrons have come from the N atom. Because the H + ion is now part of the molecule, the entire molecule now has a charge of +1 (NH 4 + ). Note: only draw the final product, on the right, in an exam, unless asked for the entire reaction. Draw dot & cross diagrams for the following and also calculate the relative molecular mass (covalent compounds exit as MOLECULES) for each compound given below. Remember to include all lone pairs. (i) hydrogen, H2 (ii) chlorine, Cl2 (iii) methane, CH4 R.M.M (iv) oxygen, O2 R.M.M (v) nitrogen, N2 R.M.M R.M.M (vi) H3O + (formed from the reaction of H 2O with H + ) R.M.M R.F.M Level 3 Applied Science Unit 1 (Chemistry) 20

21 Consider the dot & cross diagram drawn above for methane. The diagram would suggest that this is a flat molecule. This is not the case. Organic molecules (compounds that contain one or more carbons in a carbon chain) have a 3D shape. In the 3D diagram below, the bold wedge shows the bond that comes out of the plane of the paper towards you. The dashed wedge represents the bond that goes into the plane of the paper, away from you. The shape around each carbon atom in the alkanes is t with a bond angle of 0. Methane propane Finally, you should know that single bonds are l than double bonds and double bonds are than triple bonds. The shorter the bond the s it is and therefore would require m energy to break. Bond Length (pm) Energy (kj mol -1 ) C-C C=C Shorter bonds Stronger bonds C=C Complete questions Metallic bonding + Find the formula of the other molecular ions: hydroxide, nitrate, sulfate, carbonate Metallic bonding: the electrostatic force of attraction between positive metal ions and negative delocalised electrons. Metallic bonding exits between m only. Metals exist in a giant metallic lattice, which is a 3-D structure of p metal ions surrounded by negative d electrons. There is a very strong attraction between the positively charged metal ions and the negative delocalised electrons, so metallic bonding is s. However, the attraction is not usually as strong as in ionic or covalent bonding. e.g. diagram showing metallic bonding in Na: Positive metal ions Na+ Na+ Na+ Na+ Na+ Na+ Na+ Na+ Na+ delocalised electrons Level 3 Applied Science Unit 1 (Chemistry) 21

22 Because the delocalised electrons are free to move and carry charge, metals conduct e, when solid or molten. This is why copper is used in electrical cables and wires. The delocalised electrons can also a heat energy, which gives them kinetic energy, hence metals are also good t conductors. Copper and aluminium are examples of metals used in saucepans, heat sinks in computers and radiators. Two other properties of metals are: (i) malleable they can be h into shape without breaking. (ii) ductile they can be hammered thin or s out into wires without breaking. These two properties can be explained by the fact that the metal ions are in l, and they can r over each other without breaking the metallic bonding. Aluminium is very malleable and along with its thermal conductivity makes it suitable for use in aluminium foil. Diagram showing how metals are malleable/ductile: Complete questions Electronegativity Consider a H2 molecule. Both atoms are identical and each has an e share of the electron pair in the covalent bond. The electron pair is equally distributed between both atoms. H H H Cl X X Now consider a molecule of HCl. Both atoms are different. One atom is likely to attract the electron pair in the covalent bond m strongly than the other atom (and therefore the electron pair will be closer to this atom). We say that the atom with the greater a for the pair of electrons in the covalent bond is more e than the other atom. Electronegativity: the tendency of an atom to attract a bonded pair of electrons in a molecule Polar molecule: molecule with a partial negative charge on one end and a partial negative charge on another end due to an uneven distribution of electrons Non-polar molecule: a molecule where the electrons are equally distributed throughout the molecule Level 3 Applied Science Unit 1 (Chemistry) 22

23 In the HCl molecule, Cl is more electronegative than H. This means that Cl has the greater attraction for the pair of electrons in the covalent bond and therefore the pair of electrons is c to the Cl atom than the H atom. This difference in electronegativity between the two atoms results in a small charge difference (because electrons are negatively charged) across the H Cl bond called a permanent d, which we show with a δ- and a δ+. These are p charges only, not full ionic charges; the Cl has not captured the electron pair to form a Cl - ion. Quite simply, the electrons are just c to the Cl atom. Permanent dipole means the dipole is always present. δ+ H Cl δ- If we have permanent dipole s across a bond, because the two atoms have a d in electronegativity, then we say that the bond is p. As a result, the molecule may also be polar. H Cl is a polar molecule but H H is a non-polar molecule; in H2, both atoms are identical, there is no difference in electronegativity. How do we work out which atom is the more electronegative atom? We use the Pauling scale to compare the r electronegativity of atoms; the bigger the number, the more electronegative the atom. Increasing electronegativity Increasing electronegativity Do not learn these numbers, they shall be provided to you in an exam. But you should know the general trend; electronegativity i across a period left to right and also up a g. F is the most electronegative atom and Group 0 elements do not have electronegativity that can reliably be determined Group 0 elements exist as a not molecules. Show any dipoles on the examples below and state whether the bond is polar or non-polar: H F C Cl O H S F N H H Cl C O N O Cl F Br Br Level 3 Applied Science Unit 1 (Chemistry) 23

24 You can use the pneumonic C to help you remember the factors that affect the electronegativity of an element. Electronegativity depends on the number of p in the nucleus, the d from the nucleus of the bonding pair of electrons and how much shielding there is from inner electrons. The electronegativity of an element can be used to predict the type of bonding in a compound. It is actually rare to have a wholly ionic or a wholly covalent compound; bonding is a spectrum from ionic to covalent bonding with most compounds sitting somewhere between the two. If the electronegativities are s between both atoms then a c bond forms. As the difference in electronegativity increases the covalent bond will become more p. If the difference in electronegativity is very large then the bond becomes i as one of the atoms has captured the electron pair completely. It is not just covalent bonds that can be polar. Ionic bonds can also show polarity. The extent of polarisation shall depend on whether: either ion is highly charged the cation is relatively small the anion is relatively large e.g. a small cation that is highly charged can draw electrons towards it. A large anion that is highly charged has an electron cloud that is easily distorted. If these two anions attract then the small cation can share some of the negative charge on the anion. This gives the ionic bond some covalent characteristics. Complete questions Introduction to intermolecular forces Intermolecular force: the attractive or repulsive force between molecules When we have covalently bonded molecules, we can also have w attractive forces (compared with covalent and ionic bonds) that exist b the molecules. These are known as intermolecular forces or v d w forces and there are three types: Temporary dipole-induced dipole force weakest Permanent dipole-dipole force Hydrogen bonding strongest Molecule with covalent bonds between atoms Complete question 42 Intermolecular force between molecues Level 3 Applied Science Unit 1 (Chemistry) 24

25 2.7 Permanent dipole-dipole forces P molecules, such as HCl, have permanent dipoles. The permanent dipole of one polar molecule can a the opposite p dipole in a neighbouring molecule. When this happens, we have a weak permanent dipole-dipole force between the neighbouring molecules. δ+ δ- δ+ δ- δ+ δ- Permanent dipoles Permanent dipole-dipole force H Cl H Cl H Cl 2.8 Temporary dipole-induced dipole forces Consider a H2 molecule. The molecule is non-polar as there is no d in electronegativity, no p dipoles and therefore no permanent dipole-dipole force between neighbouring m. However, the electrons are constantly m within the molecules. At any one moment in time, the electrons may be t closer to one of the H atoms in the H2 molecule. This would instantaneously lead to a t dipole on that H atom (not permanent; if the electrons continue moving, this temporary dipole disappears, but can reappear again, temporarily). The temporary dipole can influence and i a neighbouring molecule into also forming a temporary dipole; these temporary dipoles a each other leading to a temporary dipole-induced dipole force of attraction (also called L d forces). London dispersion forces exist in all molecules, whether polar or nonpolar. A non-polar molecule will o have London dispersion forces between neighbouring molecules. A polar molecule with permanent dipoles shall have permanent dipole-dipole forces and also London dispersion forces between neighbouring molecules. Temporary dipole-induced dipole force δ+ δ- δ+ δ- δ+ δ- H H H H H H this molecule has temporary dipoles caused by the movement of electrons this molecule has temporary dipoles which were induced by the first H 2 molecule this molecule has temporary dipoles which were induced by the second H 2 molecule You need to be able to explain how temporary dipole-induced dipole forces arise (model exam answer): Due to the movement of electrons, there is an uneven distribution of electrons throughout the molecule. This causes temporary dipoles on the molecule. The temporary dipole induces a dipole in a neighbouring molecule. The temporary dipoles and induced dipoles attract each other to form weak intermolecular forces called temporary dipole-induced dipole forces (also called London dispersion forces). Level 3 Applied Science Unit 1 (Chemistry) 25

26 Temporary dipole-induced dipole forces i with an increase in the number of electrons. More electrons results in l temporary and induced dipoles which results in s attractive London dispersion forces between the molecules. Stronger attractive forces between the molecules results in an increase in boiling point. e.g., we can explain the following trend in the boiling points of the four halogens listed: F2 Cl2 Br2 I C C 59 0 C C The boiling point of the halogens i as we go down group 7. This is because I2 has m electrons than Br2, which has m electrons than Cl2, which has m electrons than F2; more electrons results in l temporary dipoles. This means there would be s temporary dipole-induced dipole forces between molecules of I2 followed by Br2 followed by Cl2 followed by F2. Note: temporary dipole-induced dipole forces can also exist between atoms, such as the noble gases. Can you predict the trend in boiling points for the noble gases and also explain how the forces arise between the atoms? Complete questions Hydrogen bonding This is a special type of permanent dipole-dipole force when O H, N H or F H bonds are present in a molecule. Because there is a l difference in electronegativity between H and the O/N/F atoms, these bonds are h polar. The permanent dipole-dipole forces between these bonds in neighbouring molecules are particularly strong and are given a special name: h bonding. In hydrogen bonding, the H δ+ in one molecule attracts the l pair of electrons on the O δ-, N δ- or F δ- a the neighbouring molecule. E.g. a diagram showing hydrogen bonding between neighbourng water molecules is shown below. You must include dipoles, lone pairs and a label for the hydrogen bond. δ- δ+ δ+ δ- δ+ δ+ Hydrogen bond Level 3 Applied Science Unit 1 (Chemistry) 26

27 Draw diagrams to show the hydrogen bonding between: (i) neighbouring NH3 molecules (ii) neighbouring HF molecules Complete questions Trends in melting and boiling points across a period There are periodic trends in melting/boiling points across a period which can be explained in terms of the b or i forces present. Melting (solid to liquid) and boiling (liquid to g ) points depend on the strength of the forces between the atoms or molecules that you want to separate to go to a liquid or gaseous phase. During melting, energy is required to overcome the attractive forces between the atoms or molecules and boiling usually means that most of the rest of the attractive forces are broken. The stronger the forces between the atoms/molecules the more e is required to break them and the higher the melting or boiling point. In general there is an i in melting/boiling points across periods 2 & 3 (left to right) up to Group 4. There is then a sharp d in melting/boiling points between Groups 4 and 5. E.g. the trends in boiling points across period 3 is shown in the graph below: Al Si boiling point /K Na Mg atomic number P S Cl Ar general increase groups 1 to 4 sharp decrease groups 4 to 5 generally low for groups 5 to 8 Level 3 Applied Science Unit 1 (Chemistry) 27

28 The table below summarises the reasoning for the trend in melting & boiling points across periods 2 & 3: Period 2 Li Be B C N2 O2 F2 Ne Period 3 Na Mg Al Si P4 S8 Cl2 Ar Giant metallic lattice Simple Giant covalent with strong electrostatic molecular lattice with only weak boiling lattice with strong forces of attraction temporary dipole-induced dipole point covalent bonds to between positive metal forces of attraction between the explained break between the ions and delocalised molecules (atoms for Ne and Ar) atoms electrons to break to break As you go across Periods 2 & 3, left to right, the melting/boiling points are h for the metals (Li, Be, Na, Mg & Al) due to the s metallic bonding present between positive metal ions and delocalised electrons in the g metallic lattice. There is also an increase in melting points for the metals each time you go a the periods (see the graph above) because there is one more p so a greater n charge and also more d electrons with the next metal across the period. This results in a s attraction between the positive metal ions and the delocalised electrons (stronger metallic bonding) which then requires more energy to break for the melting/boiling point. For example, the melting point i going from Na to Mg to Al across Period 3. Positive metal ions Na+ Na+ Na+ Na+ Na+ Na+ Na+ Na+ Na+ A giant metallic lattice delocalised electrons strong metallic bonding Na + Na + Na + Mg 2+ Mg 2+ Mg 2+ Al 3+ Al 3+ Al 3+ Across a period the nuclear charge and number of delocalised electrons increases in the metal resulting in a higher melting point The melting point i again as you move from the metals to the giant covalent lattices across Periods 2 & 3. Covalent compounds usually exist as small molecules. However, C (in the form of diamond and graphite), B and Si exist as a g c lattice structure. A small cross section of the giant covalent lattice of carbon in the form of diamond is shown below. To melt/boil a giant covalent lattice even more energy is required because you have to break s c bonds to achieve this. Level 3 Applied Science Unit 1 (Chemistry) 28

29 Strong covalent bonds between C atoms to break in the giant covalent lattice in diamond There is then a sharp d in melting/boiling points between Groups 4 and 5, and the boiling points for Groups 5-8 is relatively l. This is because the elements now exist as small covalently bonded m between Groups 5-8, with only w temporary dipoleinduced dipole forces of attraction to break between these non-polar molecules. Look at the table above; you must know how the molecules exist, e.g. chlorine exists as Cl2 and phosphorus as P4 molecules. The structure when you have small molecules organised by attractive intermolecular forces is referred to as a simple c lattice or simple m lattice. Note that across Period 3, Groups 5-8, S8 has the h boiling point followed by P4 followed by Cl2 and then Ar. This is because the strength of temporary dipole-induced dipole forces depends on the number of e in the molecule. More electrons means l temporary and induced dipoles resulting in s temporary dipole-induced dipole forces between molecules which require more energy to break. Cl2 molecules Weak temporary dipole-induced dipole forces between molecules to break Complete questions Trends in melting and boiling points down a group Melting points d down Groups 1 & 2 as there are m shells down a group meaning a l radius and m shielding, resulting in w forces of attraction between the particles. The melting points i as you go down group 7 however. This is because the Group 7 elements exist as non-polar molecules and as you go down the Group the diatomic molecules have m electrons resulting in l temporary dipoles and therefore s temporary dipole-induced dipole forces to break between the molecules which require more energy to break. Level 3 Applied Science Unit 1 (Chemistry) 29

30 3. Orbital theory 3.1 Sub-shells and orbitals So far we have assumed that electrons orbit the nucleus of an atom in a similar way the planets orbit the sun. We have also assumed that two electrons are found in the first shell followed by e electrons in the next shells. This was a simplified model of the atom and is not the true picture for the structure of the atom! However, this Bohr model that you have come across at GCSE and earlier in this course is a very useful simplified model and is still widely used. You shall continue to use this model for the previous topics met in this unit. We shall now look at the true model of the atom that was developed from the 1900 s. You shall use this new model when writing out electronic structures for atoms or ions. The actual number of electrons per shell can be found by using the formula 2n 2, where n is the shell number. Shell 1 Electrons Each shell is then broken down into sub-shells within which are found o. There are f types of orbitals; s, p, d & f. Electrons are found in these orbitals 4f 4d 4p 4s Shell 4 contains the sub-shells 4s, 4p, 4d, 4f 3d 3p 3s 2p 2s 1s Shell 3 contains the sub-shells 3s, 3p, 3d Shell 2 contains the sub-shells 2s, 2p Shell 1 contains the sub-shell 1s Nucleus Shell 1 is made up of the sub-shell 1s. Shell 2 is made up of the sub-shells 2s and 2. The number for the sub-shell tells you which s you are in. The letter, s, p, d or f, tells you the type of orbital where the electrons are located. Electrons are actually found in o and each orbital can hold up to t electrons. You should notice that in shell 1 there is only an s orbital. In shell 2, p orbitals come in. In shell 3, d orbitals come in and in shell 4, f orbitals come in. The table below summarises how many of each type of orbital is found in sub-shells. Level 3 Applied Science Unit 1 (Chemistry) 30

31 Orbital: a region of space where there is a 95% probability of locating an electron. Each orbital can hold a maximum of two electrons. Type of orbital Number of these orbitals per sub-shell Total number of electrons held in all of these orbitals s 1 1 x 2 = 2 p 3 3 x 2 = 6 d 5 f 7 Complete the table below to summarise where the electrons are found in each shell in an atom. Complete question 52 + What are the shapes of s and p orbitals? Level 3 Applied Science Unit 1 (Chemistry) 31

32 3.2 Electron in box diagrams and electron configurations In the Bohr model, we used dots or c to show electrons in shells. How do we show electrons in this new model of the atom where there are also subshells and orbitals? We use b to represent the orbitals and a to represent the electrons in the orbitals. Because p orbitals come in threes we use t boxes for p orbitals. In a similar fashion, we use f boxes for d orbitals and s boxes for f orbitals. Each box (orbital) can hold up to t electrons. Electrons are n charged and r each other; we show this by having one arrow pointing up and one pointing d. For the same reason, each orbital is filled s before pairing starts for p, d and f sub-shells. Note that in the electron in box diagram below, the 4 sub-shell is filled with electrons before the 3d subshell. This is because the sub-shells in the 3 rd and 4 th shell are very close together and they can o, which results in the 4s sub-shell coming below the 3d sub-shell. Simply writing out how many electrons appear in each subshell is called the e configuration. This is the most important point and summarises this topic; remember the following order for writing electron configurations: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6. Be very careful when writing out electron configurations for ions remove electrons from the h filled subshell first. The electron in box diagram below is for Mn, which has an atomic number of 25 and so 25 e when a neutral atom. Note how in the 3d sub-shell the orbitals are filled s before pairing starts. 4p Each orbital holds up to two electrons with 3d opposite spins. Show this with one arrow pointing upwards and one pointing downwards 4s 3p Remember to fill from the lowest sub-shell upwards. Each energy level must be full before the next one 3s higher up is filled (Aufbau principle) 2p Each orbital is filled singularly before pairing 2s 1s starts. The electron configuration of Mn is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 Level 3 Applied Science Unit 1 (Chemistry) 32

33 Draw electron in box diagrams and below each write out the electron configuration for: (i) H (ii) Be (iii) O (iv) K + (v) Br (vi) Br - Level 3 Applied Science Unit 1 (Chemistry) 33

34 Write out electron configurations only for the following atoms and ions: He N F Al 3+ P Blocks in the periodic table If you write out the electronic structure (using s, p, d notation) for any element in Groups 1 or 2, you ll find that the o electron is always in a s orbital. Similarly, if you write out the electronic configuration for any element in groups 3-8, you ll find that the highest energy electron is always in a p-orbital. For this reason, we can assign certain blocks of the Periodic Table as being either s, p or d blocks H s d p E.g. C is a p-block element because it has the electronic structure 1s 2 2s 2 2p 2. The highest energy electron is in a p orbital. Prove this fact by writing out the electron configurations for the following elements. Li s-block Na Ca B p-block P Cl Level 3 Applied Science Unit 1 (Chemistry) 34

35 Prove that the cations Na + and Mg 2+ are isoelectronic by writing out their electronic configurations. Complete questions Ionisation energies re-visited We have seen previously seen that there are trends in ionisation energies across a period; in general there is an i in first ionisation energy across a period. However, small decreases can also be seen when going across a period from the graph above. For example, across Period 2 (Li Ne), there is a decrease between Be and B and also between N and O. These slight anomalies are linked to the filling of s and p sub-shells, which shall be explained using the electron in box diagrams below. Comparing Be and B: Beryllium Boron 2p 2s 1s 2p is empty for Be 2p 2s 1s Electron in 2p for B, which is higher in energy and also experiences shielding from the 2s, so it is easier to remove. So B has a lower first ionisation energy than Be Comparing N and O: Nitrogen Oxygen 2p 2s 1s 2p electrons are unpaired for N 2p 2s 1s A paired 2p electron in O is removed. The repulsion between the paired electrons means it is easier to remove. So O has a lower first ionisation energy than N Level 3 Applied Science Unit 1 (Chemistry) 35

36 4.1 balancing equations 4. Balanced Equations and Chemical Reactions A chemical equation for a reaction is always written: REACTANTS PRODUCTS A chemical equation may also include the s symbols below: (s) solid (l) liquid (g) gas (aq) aqueous In a chemical reaction, atoms are never created or destroyed; an equal number of the same atoms must appear on b sides of an equation. Equations are balanced by putting n before the f to ensure that there is the same number of atoms of each element on both sides of the equation. Worked Example 1: write an equation for the reaction of hydrogen gas with nitrogen gas to make gaseous ammonia. Nitrogen exists and N2, hydrogen as H2. Put these before the arrow with a small (g) to show they are in the gaseous state. After the arrow write the formula for ammonia: N2 (g) + H2 (g) NH3 (g) The equation must now be balanced. There are two nitrogen atoms on the left hand side of the equation but only o on the right hand side. Putting a 2 before NH3 (g) balances the number of nitrogen atoms. NEVER CHANGE THE FORMULA OR ANY OF THE SUBSCRIPT NUMBERS, ONLY WRITE NUMBERS IN FRONT OF A FORMULA! THE MULTIPLYING NUMBER MULTIPLIES EVERYTHING THAT APPEARS IN THE FORMULA. IF YOU HAVE BRACKETS, THE SUBSCRIPT APPLIES TO EVERYTHING INSIDE THE BRACKETS. N2 (g) + H2 (g) 2 NH3 (g) There are now 2 x 3 = 6 H atoms on the right hand side of the equation, but only 2 H atoms on the left hand side. Putting a 3 before H2 (g) b the hydrogen s: N2 (g) + 3 H2 (g) 2 NH3 (g) There is now an e number of each atom on each side of the equation; it is fully balanced. Example 2: write an equation for the reaction between sodium and oxygen to form sodium oxide. Sodium exists as Na, oxygen as O2. Sodium is a m so must be a solid at room temperature, oxygen is a gas. Sodium oxide is an i compound (which are usually solids) because it is made up of a m and a non-metal; use the c -over method to work out its formula. ALWAYS DO THIS FOR IONIC COMPOUNDS! Na (s) + O2 (g) Na2O (s) Level 3 Applied Science Unit 1 (Chemistry) 36

37 Balance the oxygens on the right hand side of the equation: Na (s) + O2 (g) 2 Na2O (s) The oxygens are balanced but there are now 2 x 2 = 4 Na atoms on the right but only one on the left. Placing a number 4 before Na in the equation shall balance it: 4 Na (s) + O2 (g) 2 Na2O (s) Now have a go at balancing the following equations. 1. Mg + N2 Mg3N2 2. Ba + HCl BaCl2 + H2 3. C + O2 CO2 4. C2H6 + O2 CO2 + H2O 5. Fe + Cl2 FeCl3 6. Mg + HNO3 Mg(NO3)2 + H2 7. CuCO3 + HCl CuCl2 + + CO2 + H2O 4.2 Reactions of period 2 and 3 elements with oxygen You have previously seen how as you move across periods 2 and 3 l to r, there is an i in melting/boiling points from Groups 1 to 4 followed by a sharp decrease with relatively low boiling/melting points for elements in Groups 5-8. This is because you move from giant m lattice (stronger metallic bonds to break) to giant c (stronger covalent bonds to break) to simple m lattice (weaker intermolecular forces to break). As you move across periods 2 and 3 from left to right, the p formed when these elements react with o tend to be giant i lattices to giant c structures to simple m structures (see the table on the next page). You get giant ionic when a m reacts with the non-metal O2 to form a metal oxide. The bonding present is responsible for the properties of the products formed. The products made change from solids with the giant ionic and giant covalent lattices to g for the simple molecules. The products made also change from being a (metal oxides are examples of bases) to a to a. Level 3 Applied Science Unit 1 (Chemistry) 37

38 WHEN WRITING OUT EQUATIONS FOR THE FORMATION OF IONIC COMPOUNDS, YOU MUST USE THE CROSS OVER METHOD TO ENSURE YOU HAVE THE CORRECT FORMULA FOR THE IONIC PRODUCTS! Si and SiO2 are examples of giant covalent lattices Oxides are acidic A summary table: Period 2 Li2O BeO B2O3 CO or CO2 Period 3 Na2O MgO Al2O3 SiO2 Properties of oxides Metal oxides are ionic compounds and form alkaline solutions. Al2O3 is amphoteric. CO from incomplete combustion and CO2 from complete combustion. SiO2 is giant covalent. Acidic oxides. Complete questions NO, NO2, N2O5 P4O6 P4O10 O3 SO2, SO3 Simple molecular structures and gases. Nitrogen forms a range of oxides (acidic) with different oxidation states. O2 and O3 are allotropes. Ignore Groups 7 and 8. Alkaline solution: a solution with a ph value above 7 Acidic solution: a solution with a ph value below 7 Amphoteric: a substance that can act as both an acid and a base Oxidation: loss of electrons Allotrope: two or more different physical forms that an element can exist in. Level 3 Applied Science Unit 1 (Chemistry) 38

39 4.3 Reaction of metals with oxygen, water, and dilute acids (i) Oxygen: Metals react with the non-metal oxygen to form i metal oxides. The Group 1 and 2 metal oxides are used as bases because they form a solutions. It is very important that you check the formula of the metal oxide formed using the c -over method and then fully b the equation. When using the cross-over method, the positive charge on the metal ion is the same as its G number and the charge on the oxide ion is -2. E.g: 4Na + O2 2Na2O 2Mg + O2 2MgO 4Al + 3O2 2Al2O3 (the general formula for Group 1 metals is 4M + O2 2M2O) (the general formula for Group 2 metals is 2M + O2 2MO) (the general formula for Group 3 metals is 4M + 3O2 2M2O3) Group 1 metals react rapidly with oxygen and are stored under o to prevent contact with air and the more reactive Group 1 metals are stored in small sealed glass tubes. Be and Al form BeO and Al2O3 coatings respectively, which makes them resistant to further o and makes them behave as unreactive metals. The Group 4 metals lead (Pb) and Tin (Sn) also form oxides. The transition metals are much l reactive with oxygen than Group 1 or 2 metals. When transition metals (also called d- block metals) react with oxygen the oxides are usually brittle; for example, when iron reacts with oxygen to form rust (iron oxide). The transition metals form a range of oxides with different oxidation states. Some d-block metals such as titanium are resistant to corrosion because they quickly form an outer unreactive oxide layer which prevents any further oxidation. We have so far assumed that oxygen only forms the oxide ion, O 2-. However, the p ion, O2 2- and the super-oxide ion O2 - also exist. These molecular ions contain a c bond between the two O atoms; O-O 2- and O-O -. These two negative molecular ions are only stable next to a l positive cation, so they only form compounds with metals the further down a Group you go where the larger positive metals ions are found. For example, Li is a s positive ion and when next to a large negative ion such as O 2- or O2 -, the the electrons in the covalent bond between the two O atoms are strongly attracted to the small positive metal ion; the ionic bond becomes p. This results in the covalent bond between the oxygen atoms breaking. However, as you go down Group 1, you shall find peroxides are formed such as Na2O2 and K2O2 because the positive metal ions become larger and therefore the ionic bond is less polarised. Further down the Group super-oxides are formed such as KO2, RbO2 and CsO2. Write an equation for the reaction of magnesium with oxygen: Level 3 Applied Science Unit 1 (Chemistry) 39

40 (ii) Water: Metals react with water to form solutions of a metal h as well as hydrogen g. The hydroxide molecular ion exists as OH -. Once again ensure you use the cross-over method when working out the formula for the metal hydroxide. Where there is more than one hydroxide molecular ion, use b around it with the small multiplying number outside the brackets. E.g.: 2Na + 2H2O 2NaOH + H2 (general formula for Group 1 metals: 2M + 2H2O 2MOH + H2) Mg + 2H2O Mg(OH)2 + H2 (general formula for Group 2 metals: M + 2H2O M(OH)2 + H2) Note that when in solution, the hydroxides exist as separate M + /M 2+ and OH - ions. The hydroxide ions are responsible for the alkalinity of the solutions formed. Group 1 metals are called the a metals because they form a basic solution when reacted with water. The reactivity of Group 1 and 2 metals i down the Group. In Group 2, Be does not react with water, Mg only reacts with steam but the metals further down the Group react increasingly easier with water. Group 3 metals are not very reactive with water and aluminium appears not to react at all due to the outer o layer. Group 4, 5 and 6 metals do not react with water. Some transition metals do react with water but only very slowly. Write an equation for the reaction of calcium with water: Write an equation for the reaction of lithium with water: (iii) Dilute HCl and H2SO4 In a similar reaction to that with water, metals (those above copper in reactivity series) react with dilute acids to form s in a neutralisation reaction as well as H2 gas. C salts form with HCl and s salts form with sulfuric acid. The sulfate molecular ion exists as SO4 2-. Once again remember to use the cross-over method when working out the formula of the ionic salt. Note that the reaction of Ca, Sr and Ba with H2SO4 leads to a protective sulfate layer that is insoluble, preventing them from reacting any further. 2Na + 2HCl 2NaCl + H2 2Na + H2SO4 Na2SO4 + H2 Mg + 2HCl MgCl2 + H2 Mg + H2SO4 MgSO4 + H2 Write equations for the reaction of lithium with HCl and H2SO4: Complete question 58 Level 3 Applied Science Unit 1 (Chemistry) 40

41 4.4 Redox Oxidation numbers (not to be confused with ionic charge) are a measure of the number of electrons that an atom uses to bond with atoms of a d element. We can assign oxidation numbers to both ionic and covalent compounds. The rules, which you shall use when looking at redox reactions, are listed below. The sign of the oxidation number must be placed before the number. Element Ox. No. Examples Exceptions Un-combined element 0 O2, H2, Fe, S8 Combined H +1 H2O, NH3 Combined O -2 H2O, CaO Combined F -1 HF -1 in metal hydrides e.g. NaH, CaH2-1 in peroxides e.g. H-O-O-H +2 when bonded to F e.g. F2O Ions Charge on ion Na + = +1; Mg 2+ = +2 The sum of the oxidation numbers for each atom must equal the charge Assign oxidation numbers to each i atom that is in the molecule/ion when looking at redox reactions. All of the oxidation numbers when added together must e any charge on an ion. Before we look at redox reactions, you need to know the difference between oxidation and reduction. Oxidation: the loss of electrons (resulting in an increase in oxidation number) Reduction: the gain of electrons (resulting in a decrease in oxidation number) Remember this using: OILRIG X is losing electrons it is being oxidised X 3+ X 2+ X 1+ X 0 X 1- X 2- X 3- X is gaining electrons it is being reduced Now consider the oxidation numbers in the following reaction: 2 Fe + 3 Cl2 2 FeCl3 Fe has gone from Fe = 0 to Fe = +3; it has been oxidised. Cl has gone from Cl = 0 to Cl = -1; it has been reduced. A reaction in which both oxidation and reduction is taking place is called a REDOX reaction. redox reaction: one which involves both oxidation and reduction. Level 3 Applied Science Unit 1 (Chemistry) 41

42 In the above reaction, Fe is a r agent; it reduced Cl (Fe itself was oxidised). Cl is an o agent; it oxidised Fe (Cl itself was reduced). Worked examples: Assign oxidation numbers and say whether or not these are REDOX reactions. i) Mg + 2 HCl MgCl2 + H2 ii) 2 Ca + O2 2 CaO iii) SrO + H2O Sr(OH)2 iv) Mg + H2O MgO + H2 v) BaO + 2 HCl BaCl2 + H2O vi) Ca + 2 HCl CaCl2 + H2 Complete questions Level 3 Applied Science Unit 1 (Chemistry) 42

43 4.5 Oxidation numbers for transition metals and oxyanions Transition elements can have v oxidation numbers, e.g. Fe can exist with oxidation numbers of +2 and +3. To differentiate between the different states of +2 and +3 in Fe, the oxidation number of the transition element is given as a r numeral in brackets. The variable oxidation state is what makes the transition metals useful as c in many important industrial reactions I II III IV V VI VII VIII IX X Formula Name Oxidation no. of transition element FeCl2 Iron(II) chloride Fe = +2 FeCl3 Iron(III) chloride Fe = +3 CuO Copper(II) oxide Cu2O Cu = +1 Copper(I) chloride CuCl2 Oxyanions are: n molecular ions that contain o combined with a second element e.g. SO4 2-, CO3 2-, NO3 -. The name of the oxyanion usually ends in ate (e.g calcium sulphide, CaS vs calcium sulfate, CaSO4). The second element has several oxidation states, with the oxidation number of the s element given in brackets after the name of the oxyanion. The oxidation numbers given in brackets after the name of the oxyanion allows us to distinguish between similar molecular ions. For example, SO4 2- is not the only oxyanion called sulfate; SO3 2- is another sulfate ion. The oxidation number of the second element, in this case sulfur, written after the name allows us to distinguish between the two; SO4 2- is referred to as sulfate(vi) and SO3 2- is referred to as sulfate(iv)! Formula Name Oxidation no. of second element SO4 2- Sulfate(VI) S = +6 SO3 2- Sulfate(IV) S = +4 NO3 - Nitrate(V) N = +5 NO2 - N = +3 KMnO4 (K + MnO4 - ) Potassium manganate(vii) Mn = +7 NaNO3 Sodium nitrate(v) N = +5 NaNO2 CaSO4 Level 3 Applied Science Unit 1 (Chemistry) 43

44 4.6 Reactivity series The reactivity series lists metals in order of reactivity with oxygen, water and acids. The most reactive metals are at the t of the list. The order of reactivity for the metals is Group1, Group2, Group3, Group4, Transition metals. The metals higher up are m reactive as they have a greater tendency to lose an e to form a complete outer shell. In general, for the metals, reactivity decreases across a period and i down a Group. This is because across a period the n charge increases, making it harder to lose the outer electron during a chemical reaction. The most reactive metals are in group 1 and these are found towards the top of the reactivity series. The group 1 metals become more reactive as you go down the group, so Fr is the most reactive metal but it is not usually seen on lists as it is so radioactive and unstable. Most reactivity series have potassium at the top as it is very reactive and in Group 1 and gold and platinum at the bottom as these transition metals are so unreactive. It is useful to place carbon and hydrogen in the list. Carbon is used to extract (or displace) metals from their ores; only metals above carbon in the list can be extracted from their ores with the more reactive carbon. In a similar way, only metals above hydrogen will react with acids or water to displace the hydrogen. This leads us onto the topic of displacement reactions which also tend to be redox reactions. We shall look at redox first before moving on to displacement reactions. Level 3 Applied Science Unit 1 (Chemistry) 44

45 4.7 Displacement reactions A more r metal shall displace a less reactive metal in a metal salt solution. You can predict when a metal can displace another metal from its salt by using the r series. Displacement reactions involving metals are also usually r reactions you can prove this by assigning oxidation numbers to each element in the equation. E.g. Prove that the displacement reaction below is also a redox reaction: Fe (s) + CuSO4 (aq) FeSO4 (aq) + Cu (s) Displacement reactions can also happen with the Halogens. The halogens become m reactive as you go up the Group. A more reactive halogen shall displace a less reactive halogen from its metal salt: Cl2 + 2 KBr Br2 + 2 KCl Note that the potassium ion does not actually take part in the reaction it is a s ion. We can simplify the equation by removing the potassium so we can see more clearly what is happening to the electrons in the reaction: Cl2 + 2 Br - Br2 + 2 Cl - The halogens are o agents this means that they oxidise (remove electrons from) another species. In the above reaction, chlorine is a s oxidising agent than bromine, so it removes electrons from the bromide ions to form the chloride ions. Complete questions Uses and applications of the substances in this unit You may be asked for the uses of some of the substances that you have come across in this unit. Research a few substances and summarise their uses below. Level 3 Applied Science Unit 1 (Chemistry) 45

46 5. Quantitative Chemistry 5.1 Moles and masses Mole: a unit of substance equivalent to the number of atoms in 12g of carbon mole of a compound has a mass equal to its relative atomic mass expressed in grams. Molar mass: the mass of one mole of a substance. We previously introduced the concept of one mole; one mole of molecules is molecules,. Where has this number come from? We have also previously seen how the isotope 12 C is used as the s for weighing atoms when we introduced the idea of relative atomic mass, which is the a mass of an atom compared with 1/12 th of 12 C. If you weigh out exactly 12 grams of the isotope 12 C, you shall have x atoms this number was then used as the quantity for one mole of anything. It is also called Avogadro s constant. We can also introduce the idea of molar mass, which is the mass in g of one mole of any substance. It has units of g mol -1 (grams per one mole). To work out how many grams of any element you need to have o mole, simply find it relative atomic mass. For a compound, it is the same as the relative formula mass or relative molecular mass. The molar mass for 12 C is 12 gmol -1 (12 grams are required for you to have one mole, 6.02 x atoms). The molar mass for CO2 is 44 gmol -1 (44 grams are required for you to have o mole, 6.02 x m ). The molar mass (R), the number of moles (M) and the mass in grams of the substance are linked by the following triangle: amount in Moles = mass in Grams / Relative molar mass G = mass of compound (g) M = number of moles of compound (mol) R = relative molar mass of compound (g mol -1 ) Example 1: What is the mass of 0.8 moles of CO2? We have moles, M = 0.8 and the molar mass, R, of CO2 = 12 + (16 x 2) = 44 Using the above triangle, mass in grams, G = M x R = 0.8 x 44 = 35.2 g Level 3 Applied Science Unit 1 (Chemistry) 46

47 Example 2: Calculate the number of moles of NaOH in 0.2 g of the solid? We have G = 0.2 and R of NaOH = = 40 Using the above triangle, moles, M = G / R = 0.2 / 40 = mol Complete questions Moles and solutions The concentration of a solution tells you how much solute (solid) is dissolved in the solvent. Concentrations are measured in moles per cubic decimetre (units mol dm -3 ). A decimetre is the same as one litre. To go from cm 3 to dm 3, d by To go from dm 3 to cm 3, m by So a solution of NaOH with concentration of 1 mol dm -3 (sometimes written as 1 M) contains 1 mole of NaOH dissolved per 1 dm 3 of the solution ( = 1 mole NaOH per 1000 cm 3 ). A 0.5 mol dm -3 NaOH solution contains moles of NaOH in every 1 dm 3. 2 dm 3 of a 0.5 mol dm -3 NaOH solution contains mol of dissolved NaOH. The amount in moles (M), the concentration (C) and the volume of a solution (V) are linked via the following equation. amount in Moles = Concentration of solution (in mol dm -3 ) x Volume of solution (in dm 3 ) M = amount of solute (mol) C = concentration of solution (mol dm -3 ) V = the volume of solution (dm 3 ) Level 3 Applied Science Unit 1 (Chemistry) 47