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 PCP & PCC Tracker 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 Periodic Table 58 Videos can be found at hccappliedscience.weebly.com under unit 1 and chemistry Level 3 Applied Science Unit 1 (Chemistry) 2

3 PCP & PCC Tracker Unit 1 Chemistry Watch the videos on EDpuzzle. Ask your teacher to re-set your video if you achieve below 80%. Answers to questions will be available on the website make sure you check and mark your answers. At level 3 you should be doing a MINIMUM of 5 hours independent study per week per teacher. Week Lesson Date Objectives / checklist Pre-class preparation Post-class consolidation State the relative charge, mass and position in Watch chem videos 1 and Complete questions 1-15 a Bohr atom of protons, neutrons and electrons 2 in the questions booklet. 1 Define atomic and mass number Read sections 1.1 and 1.2 Check answers online and Determine the number of protons, electrons in the notes booklet mark your own work. and neutrons in an atom or ion from given Make summary notes on Revise theory for atomic and mass numbers videos 1 and 2 upcoming chemistry test Define relative atomic mass 1. Calculate relative atomic mass Identify the Groups, Periods, metals and nonmetals in the Periodic Table State that the elements are arranged in order of increasing atomic number in the Periodic Table Explain why elements in the same Group have similar chemical properties State patterns seen across a period and define Periodicity Explain how the atomic radius changes across a Period and down a Group using CARS State what is meant by cation and anion Explain the octet rule Predict the charge on an ion of an element from its position in the Periodic Table Explain trends across a Period and down a Group for cations and anions Compare and explain the size of a neutral atom with its cation / anion Explain what is meant by the term isoelectronic, giving examples Watch chem video 3 Read section 1.3 in the notes booklet Make summary notes on video 3 Watch chem videos 4 and 5 Read sections 1.4 and 1.5 in the notes booklet Make summary notes on videos 4 and 5 Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 1. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 1. Level 3 Applied Science Unit 1 (Chemistry) 3

4 4 5 Define ionisation energy Write equations showing ionisation energies Explain why energy is required to remove an electron from an atom Explain trends in ionisation energy across a Period and down a Group using CARS Define electron affinity Write equations showing electron affinities Explain why energy is released when adding an electron to an atom for the first electron affinity Explain why energy is required for the second electron affinity Explain trends in electron affinity values across a Period and down a Group using CARS Determine the type of bonding present from the combination of elements in a compound Use the cross-over method to work out the formula of ionic compounds Correctly draw dot & cross diagrams for ionic and covalent compounds State what is meant by the term ionic bond Use the cross-over method and draw dot & cross diagrams for ionic compounds Calculate the relative formula mass for ionic compounds Explain the two factors that affect the strength of an ionic bond Recall the common molecular ions. Watch chem videos 6 and 7 Read sections 1.6 and 1.7 in the notes booklet Make summary notes on videos 6 and 7 Watch chem videos 8 and 9 Read sections 2.1 and 2.2 in the notes booklet Make summary notes on videos 8 and 9 Come prepared for chemistry test 1. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 1. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 2. Level 3 Applied Science Unit 1 (Chemistry) 4

5 PCP & PCC Tracker Unit 1 Chemistry Watch the videos on EDpuzzle. Ask your teacher to re-set your video if you achieve below 80%. Answers to questions will be available on the website make sure you check and mark your answers. At level 3 you should be doing a MINIMUM of 5 hours independent study per week per teacher. Week Lesson Date Objectives / checklist Pre-class preparation Post-class consolidation Explain what is meant by a covalent bond and a dative covalent bond Draw dot & cross diagrams for covalently bonded compounds (single, multiple and dative) Know how the ammonium ion is made Calculate the relative molecular mass for Watch chem videos 10 and 11 Read sections 2.3 and 2.4 in the notes booklet Make summary notes on videos 10 and 11 Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 2. 1 covalently bonded molecules State that organic molecules such as methane have a tetrahedral shape Explain that shorter covalent bonds (double 2 and triple) are stronger bonds State what is meant by the term metallic bonding State why metals conduct heat and electricity State what is meant by the terms malleable and ductile 2 Explain what is meant by the terms electronegativity, polar, non-polar and dipole Explain the trend in electronegativity values in the Periodic Table Explain what factors affect electronegativity of an element using CARS Explain how electronegativity values can be used to predict the type of bonding present Explain that ionic bonds can also show polarisation Watch chem video 12 Read section 2.5 in the notes booklet Make summary notes on video 12 Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 2. Level 3 Applied Science Unit 1 (Chemistry) 5

6 3 4 5 State what is meant by the term intermolecular force State what the three intermolecular forces are in order of strength Explain that permanent dipole-dipole forces exist between polar molecules Explain how temporary dipole-induced dipole forces arise Know that temporary dipole-induced dipole forces exist between all molecules Explain that temporary dipole-induced dipole forces increase in strength with more electrons Explain that hydrogen bonding occurs when there is an O-H, N-H or F-H bond present Draw labelled diagrams showing hydrogen bonding, including lone pairs and dipoles Explain the trends in melting/boiling points across periods 2 and 3, in terms of structure and forces Explain the trend in boiling points down Groups 1, 2 and 7 Watch chem video 13 Read sections 2.6, 2.7 and 2.8 in the notes booklet Make summary notes on video 13 Watch chem video 14 Read section 2.9 in the notes booklet Make summary notes on video 14 Watch chem video 15 Read sections 2.10 and 2.11 in the notes booklet Make summary notes on video 15 Come prepared for chemistry test 2. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 2. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 2. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 3. Level 3 Applied Science Unit 1 (Chemistry) 6

7 PCP & PCC Tracker Unit 1 Chemistry Watch the videos on EDpuzzle. Ask your teacher to re-set your video if you achieve below 80%. Answers to questions will be available on the website make sure you check and mark your answers. At level 3 you should be doing a MINIMUM of 5 hours independent study per week per teacher. Week Lesson Date Objectives / checklist Pre-class preparation Post-class consolidation Determine the number of electrons in each Watch chem video 16 Complete questions shell using the formula 2n 2 Read sections 3.1 and 3.2 in the questions booklet. 1 State what is meant by the term orbital and in the notes booklet Check answers online and know the shapes of s and p orbitals. Make summary notes on mark your own work. Know the order of filling the orbitals and subshells when writing out electron configurations upcoming chemistry test video 16 Revise theory for Draw electron in box diagrams and know the 3. rules for filling the boxes with electrons Identify s, p and d blocks in the Periodic Table Explain the anomalies in ionisation energy trends (Be/B and N/O) Be able to balance chemical equations with the use of state symbols Explain the trend in melting/boiling points for the oxides of Period 2 and 3 elements Explain the properties of the oxides of Period 2 and 3 elements Construct balanced equations for the formation of Period 2 and 3 oxides Watch chem video 17 Read sections 3.3 and 3.4 in the notes booklet Make summary notes on video 17 Watch chem videos 18 and 19 Read sections 4.1 and 4.2 in the notes booklet Make summary notes on videos 18 and 19 Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 3. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 3. Level 3 Applied Science Unit 1 (Chemistry) 7

8 4 5 Construct balanced equations for the reaction of metals with O2, H2O, HCl and H2SO4 Recall the rules for oxidation numbers and assign these to species in a chemical equation Define oxidation, reduction and redox Determine whether a reaction is REDOX from a chemical equation Know that transition metals have variable oxidation states, shown by roman numerals Know that roman numerals can also be used with oxyanions Watch chem video 20 Read section 4.3 in the notes booklet Make summary notes on video 20 Watch chem videos 21 and 22 Read sections 4.4 and 4.5 in the notes booklet Make summary notes on videos 21 and 22 Come prepared for chemistry test 3. Complete question 81 in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 3. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 4. Level 3 Applied Science Unit 1 (Chemistry) 8

9 PCP & PCC Tracker Unit 1 Chemistry Watch the videos on EDpuzzle. Ask your teacher to re-set your video if you achieve below 80%. Answers to questions will be available on the website make sure you check and mark your answers. At level 3 you should be doing a MINIMUM of 5 hours independent study per week per teacher. Week Lesson Date Objectives / checklist Pre-class preparation Post-class consolidation Explain the relative position of metals in the Watch chem video 23 Complete questions reactivity series and order of reactivity Read sections 4.6, 4.7 and in the questions booklet. 1 Be able to construct displacement reactions for 4.8 in the notes booklet Check answers online and the metals and the halogens Make summary notes on mark your own work. Be able to explain why the displacement video 23 Revise theory for reactions happen for metals and the halogens upcoming chemistry test Identify uses for substances covered in this 4. unit Define the terms mole and molar mass Use GMR to calculate grams/moles/molar masses for solids Be familiar with the units for concentration and volumes Use MCV to calculate moles/concentrations/volumes for liquids Understand molar ratios in balanced equations (stoichiometry) Use GMR, MCV and stoichiometry to solve calculations in chemistry Watch chem video 24 Read sections 5.1 and 5.2 in the notes booklet Make summary notes on video 24 Watch chem video 25 Read section 5.3 in the notes booklet Make summary notes on video 25 Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 4. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 4. Level 3 Applied Science Unit 1 (Chemistry) 9

10 4 5 Recall the formulae for percentage yield calculations Calculate the percentage yield of a reaction with use of GMR and stoichiometry Revision of unit 1 chemistry content Watch chem video 26 Read section 5.4 in the notes booklet Make summary notes on video 26 Come prepared for chemistry test 4. Complete questions in the questions booklet. Check answers online and mark your own work. Revise theory for upcoming chemistry test 4. Revise all theory for upcoming mock exam. Level 3 Applied Science Unit 1 (Chemistry) 10

11 1.1 Introduction to the atom 1. The Periodic Table, Atoms & Ions In the nucleus of an atom are found: Protons (+1 charge) Neutrons (neutral) In shells around the nucleus you find electrons (-1 charge). Relatively, protons and neutrons have a similar mass. So if we say the mass of a proton is 1 then the mass of a neutron is also 1. Relative to protons and neutrons, the mass of an electron is very small (1/2000 th the mass). Electron Shell Nucleus Neutron Proton You must be able to recall the relative mass and charge for each of the sub-atomic particles in the table below. You should be able to see that most of the mass is found in the nucleus of the atom. Particle Relative mass Relative charge Position in atom Proton 1 +1 Nucleus Neutron 1 0 Nucleus Electron 1/ Electron shells Note that we use dots (or crosses) to show electrons in shells. Up to two electrons are found in the first shell outside the nucleus and eight 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 must equal 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 1-4 Level 3 Applied Science Unit 1 (Chemistry) 11

12 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 must equal the number of negative electrons. Therefore the atomic number also tells you how many electrons are present in a neutral atom. The number of neutrons can be found by subtracting the atomic number from the mass number. E.g. nitrogen has 14 7 = 7 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 isotopes of chlorine are found; 35 Cl and 37 Cl. Both isotopes 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 isotope of the same element is the number of neutrons and therefore the mass 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 electrons. Chemical reactivity depends on electron movement. You might think to yourself that the mean of 35 and 37 is ( ) / 2 = 36. 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 more 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 pure 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) 12

13 Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7 Group 8 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 standard for weighing atomic masses. Atomic masses are weighed relative to 1/12 th of 12 C. The relative atomic masses do not have any units because they are only relative numbers. To calculate the relative atomic mass for a sample of an element found in nature, simply multiply the mass number by the percentage for each isotope and then divide by 100 (see the example calculation below). Remember to use brackets on your calculator and give your answer to one decimal place. WORKED EXAMPLES 1. A sample of chlorine contains 25% 37 Cl and 75% 35 Cl. Calculate the relative atomic mass? (25 x 37) + (75 x 35) = A sample of magnesium contains 78.6% 24 Mg, 10.1% 25 Mg and 11.3% 26 Mg. Calculate the relative atomic mass? (78.6 x 24) + (10.1 x 25) + (11.3 x 26) = Complete questions Introduction to the Periodic Table of elements Period 1 NON-METALS Period 2 Period 3 METALS Period 4 Level 3 Applied Science Unit 1 (Chemistry) 13

14 The Periodic Table shows all of the chemical elements arranged in order of increasing atomic number. Elements are organised into vertical columns called groups and horizontal rows called periods. Period 1 contains the elements H and He. Chemical properties are similar for elements that are in the same group because they all have the same number of electrons in the outer shell. The atomic number increases as you move from left to right across a period because each element has one more proton than the element before it in the same period. There are trends (or patterns) that repeat 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-metals and the atomic radius also decreases. This repeating pattern seen by the elements across a period is called periodicity. Periodicity: the repeating pattern seen by the elements in the periodic table The diagram below shows how the atomic radius decreases across a period but increases down a group. The two electrons in the first shell have been shown. Add the missing electrons to the other electron shells using the atomic numbers for the neutral atoms to prove: (i) electrons are added to the same shell across a period but (ii) a new shell of electrons is added down a group. 3Li 4Be 5B 6C 7N 8O 9F 10Ne 11Na To explain trends in atomic radius seen above (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: Nuclear Charge Nuclear Attraction Radius Shielding Level 3 Applied Science Unit 1 (Chemistry) 14

15 Across a period, the atomic number increases as the number of positive protons in the nucleus increases by one each time, so the nuclear charge increases (e.g. +3 to +4 to +5). 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 closer to the nucleus. The electron experiences increased nuclear attraction as the nuclear charge increases Because the electron shells are closer to the nucleus across a period due to the increased nuclear charge and attraction, this means that the atomic radius has decreased. Radius decreases across a period as the nuclear charge and attraction increases 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 repel 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 same 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). Attraction Repulsion. Shielding increases as the number of inner shells of electrons increases Level 3 Applied Science Unit 1 (Chemistry) 15

16 Each time you go down a group however, there is an additional electron shell added, resulting in an increase in radius. The greater number of inner shells also results in increased shielding. Therefore the outermost electrons are less strongly attracted to the nucleus with them being further away with the larger radius, so there is weaker 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 outweigh the increase in nuclear charge, resulting in an overall larger atomic radius. More shells down a group: Larger radius Increased shielding, Weaker nuclear attraction on electrons MODEL ANSWERS 1. Explain the difference in atomic radius between C and N. N is to the right of C in the same period (period 2) N has more protons so a greater nuclear charge than C The electrons in N experience greater nuclear attraction than those in C As N and C are in the same period, shielding is the same N therefore has a smaller radius than C 2. Explain the difference in atomic radius between Mg and Ca. Ca is below Mg in Group 2 Ca has more shells so a larger radius Ca has more shells so more shielding The electrons in Ca experience weaker nuclear attraction Ca does have a greater nuclear charge than Mg, but this is outweighed by the increased radius and shielding in Ca (resulting in a weaker overall nuclear attraction) Complete questions Level 3 Applied Science Unit 1 (Chemistry) 16

17 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 positive protons equals the number of negative electrons. However, atoms can gain or lose electrons during chemical reactions to form charged particles called ions in order to satisfy the octet rule. The octet rule states that elements gain or lose electrons in order to have eight 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 one negative electron to leave behind a positive charge; Cl - has gained one negative electron and therefore has a negative charge; N 3- has gained three negative electrons. We can predict the charge that an ion of a given element shall form by looking at its position 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 one electron in its outer shell. To have a complete outer shell containing eight electrons, the sodium atom must lose this one outer electron (electrons are negatively charged), which would result in a sodium ion with a charge of +1 (written as Na + ). All group 1 elements form +1 ions for the same reason. Note that group 4 elements do not usually form ions; they have four 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 eight electrons in their outer-most shell and are very stable atoms, so they do not form ions. Complete questions Level 3 Applied Science Unit 1 (Chemistry) 17

18 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 trends can be explained using the pneumonic CARS. The same patterns also exist with the cations/anions radius decreases across a period and increases down a group due to more shells. But you must also know how the size of a cation/anion compares with the neutral atom across a period. Across a period for cations: across a period, the radius of neutral atoms and cations decreases. But compared with neutral atoms, the cations have lost electrons to gain a positive charge. This means that the cations attract the remaining electrons even more strongly than the neutral atoms because there is still the same number of protons in the nucleus now attracting less electrons in the cation. This means that the electron shells are drawn to be even closer to the nucleus in a cation compared with the neutral atom, resulting in the cations being smaller than the neutral atoms. Neutral atom Na Mg Al Across a period, the radius decreases for both cations and the neutral atom CARS! Cation (smaller) Na + Mg 2+ Al 3+ There is an interesting property with cations across a period they all have the same number of electrons despite being different elements, so they are isoelectronic. This is shown below with the cations of the first three elements in Period 3 as an example: Neutral atom electron structure Cation electron structure Na: 2,8,1 Na + : 2,8 Mg: 2,8,2 Mg 2+ : 2,8 Al: 2,8,3 Al 3+ : 2,8 Isoelectronic: having the same number of electrons Level 3 Applied Science Unit 1 (Chemistry) 18

19 Across a period for anions: across a period, the radius of neutral atoms and anions decreases. But compared with neutral atoms, the anions have gained electrons to complete with a negative overall charge. The added electron(s) when you form the anion cause extra repulsion, with the electrons pushing themselves to be further apart. This results in a larger radius for the anion than the corresponding neutral atom. Neutral atom Anion (larger) P S Cl P 3- S 2- Cl - Across a period, the radius decreases for both anions and the neutral atom CARS! Across a period, the anions are also isoelectronic, as shown with the first three anions in Period 3 below: Neutral atom electron structure Anion electron structure P: 2,8,5 P 3- : 2,8,8 S: 2,8,6 S 2- : 2,8,8 Cl: 2,8,7 Cl - : 2,8,7 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) 19

20 When we remove electrons from atoms, we measure the energy required to remove one mole of electrons from one mole of atoms in the gaseous phase. We can write equations to show the process. Removing an electron leaves a positively 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 gaseous phase. The positive sign before the numerical value shows that energy is required for this process; energy is required to break the attraction between the electron and the positively charged nucleus. 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 neutral 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 +7 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 CARS 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 increases (there are some anomalies e.g. between Be and B and also N and O the reason for this shall be explained later in section 3.4). Down a group the ionisation energy decreases. Ionisation energies decrease down a Group Ionisation energies increase across a Period left to right. There are some anomalies which will be discussed in section 3.4 later Level 3 Applied Science Unit 1 (Chemistry) 20

21 MODEL ANSWERS 1. Explain why the first ionisation energy for F (1680 kjmol -1 ) is greater than the first ionisation energy for C (1090 kjmol -1 ). F is to the right of C in the same period (period 2) F has more protons so a greater nuclear charge than C F has a smaller radius than C As F and C are in the same period, shielding is the same The electrons in F experience greater nuclear attraction than those in C This means more energy is required to overcome the attraction and remove an electron in F 2. Explain why the first ionisation energy for K (418 kjmol -1 ) is less than the first ionisation energy for Na (494 kjmol -1 ). K is below Na in Group 1 K has more shells so a larger radius K has more shells so more shielding The electrons in K experience weaker nuclear attraction This means less energy is required to overcome the attraction and remove an electron from K K does have a greater nuclear charge than Na, but this is outweighed by the increased radius and shielding in K (resulting in a weaker overall nuclear attraction) 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 released (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 also be fully explained using the pneumonic CARS. Across a period the electron affinity values increase as more energy is released when nuclear attraction for the electron is stronger. Down a group when nuclear attraction for the electron to be captured is weaker, less energy is released. 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 down a group. This is because these atoms are very small and placing an extra electron in a crowded area is difficult Level 3 Applied Science Unit 1 (Chemistry) 21

22 and there is significant repulsion. This repulsion lessens the attraction for the incoming electron resulting in a lower than expected electron affinity value. 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 second electron affinity to form the 2- ions. Adding a second electron to the 1- ion shall require energy to overcome the repulsive force, as this time the negative electron to be captured shall be repelled by the negative 1- ion. The energy change shall therefore be positive 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) O - (g) + e O 2- (g) first electron affinity = -141 kjmol -1 (energy released) second electron affinity = +844 kjmol -1 (energy required) Complete questions Level 3 Applied Science Unit 1 (Chemistry) 22

23 Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7 Group 8 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; ionic, covalent and metallic. 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. Charge on ions: Period 1 NON-METALS Period 2 Period 3 METALS Period 4 Metallic bonding occurs when there are only metals present; either a metal on its own or a mixture of metals (an alloy). The negative delocalised electrons are attracted to the positively charged metal ions. Metallic bonding exists in e.g. Cu, Fe and the alloy brass (mixture of Zn & Cu). Covalent bonding occurs between two non-metals. Ions are not involved; instead the two non-metals come together to share 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 one of the non-metals (dative covalent bond). Examples of covalently bonded molecules are O2, F2 and NH3. Ionic bonding occurs between metals 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. Level 3 Applied Science Unit 1 (Chemistry) 23

24 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 The octet rule states that atoms with eight electrons in their outer-most shell are extremely stable (apart from the first shell, where two electrons are required). The Noble gases (group 8 elements) exist as single atoms 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). *Use brackets for molecular ions What about atoms that do not have eight electrons in their outer-most shell? These atoms can satisfy the octet rule by either: - Sharing electrons with other elements until both have eight electrons (covalent bonding), or - Transferring 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 circle to represent the outermost shell only. (ii) Use dots to represent the electrons from one of the elements and crosses 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 overlapping circles and put the electrons being shared in the overlapping region. Include those electrons not involved in bonding (lone pairs). (iv) Ionic compounds: show two separate circles as ions, with square brackets around the circle and the ionic charge 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) 24

25 Example 1: Draw a dot & cross diagram for chlorine, Cl2? X X X X Cl X X X Cl 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. Example 2: Draw a dot & cross diagram for Sodium chloride, NaCl? Na + - 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. 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. Complete question Ionic bonding Ionic bonding: the electrostatic force of attraction between two oppositely charged ions Ionic bonding occurs when you have a mixture of metals and non-metals bonded together. The Metal loses electrons to form a positively charged ion and the non-metal gains electrons to form negative ions. After electrons have been transferred to give two ions with a stable octet these oppositely charged ions then attract 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. 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. Level 3 Applied Science Unit 1 (Chemistry) 25

26 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. WORKED EXAMPLE Draw a dot & cross diagram for NaCl. Find its relative formula mass? Na is a metal, in Group 1, so forms +1 ions. Cl is a non-metal in Group 7, so forms -1 ions. Draw separate circles with brackets and charges for ionic compounds Na + X Cl - Relative formula mass: = 58.5 Remember that ionic compounds do not exists as simple molecules. Each ion attracts the oppositely charged ion in all directions resulting in a giant ionic lattice, 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+ + The strength of the electrostatic force and therefore the ionic bond depends on: (i) the ionic charge: a bigger ionic charge results in a stronger 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 larger surface area, resulting in a weaker attraction for the oppositely charged ion compared with a smaller radius. Finally, it is a good idea to learn the following ions which exist as molecules (molecular ions): +1 charged ions -1 charged ions -2 charged ions Ammonium NH4 + Hydroxide OH - Carbonate CO3 2- Nitrate NO3 - Sulfate SO4 2- Complete questions Level 3 Applied Science Unit 1 (Chemistry) 26

27 2.3 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 pair 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 single bond, two pairs of electrons a double bond and three pairs of electrons a triple bond. The term relative molecular mass is used for covalently bonded compounds as these do exist as molecules. Relative formula mass was used for ionic compounds which do not exist as molecules. Where both electrons have come from the same atom, a dative covalent bond is formed. An example is in the reaction of NH3 with a H + ion to form the molecular ion NH4 +. 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. WORKED EXAMPLE Draw a dot & cross diagram for CH4. Find its relative molecular mass? Both C and H are non-metals. Draw over-lapping circles for the covalent bonds. Both atoms contribute one electron each, so single covalent bonds are formed (not a dative bond). Relative formula mass: 12 + (4 x 1) = 16 Level 3 Applied Science Unit 1 (Chemistry) 27

28 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 tetrahedral with a bond angle of Methane propane Finally, you should know that single bonds are longer than double bonds and double bonds are longer than triple bonds. The shorter the bond the stronger it is and therefore would require more energy to break. Bond Length (pm) Energy (kj mol -1 ) C-C C=C Shorter bonds Stronger bonds C=C Complete questions Metallic bonding Metallic bonding: the electrostatic force of attraction between positive metal ions and negative delocalised electrons. Metallic bonding exits between metals only. Metals exist in a giant metallic lattice, which is a 3-D structure of positive metal ions surrounded by negative delocalised electrons. There is a very strong attraction between the positively charged metal ions and the negative delocalised electrons, so metallic bonding is strong. 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) 28

29 Because the delocalised electrons are free to move and carry charge, metals conduct electricity, when solid or molten. This is why copper is used in electrical cables and wires. The delocalised electrons can also absorb heat energy, which gives them kinetic energy, hence metals are also good thermal 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 hammered into shape without breaking. (ii) ductile they can be stretched out into wires without breaking. These two properties can be explained by the fact that the metal ions are in layers, and they can roll and slide over each other without breaking the metallic bonding, as shown in the diagram below. Aluminium is very malleable and along with its thermal conductivity makes it suitable for use in aluminium foil. Complete questions Electronegativity Consider a H2 molecule. Both atoms are identical and each has an equal 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 more strongly than the other atom (and therefore the electron pair will be closer to this atom). We say that the atom with the greater attraction for the pair of electrons in the covalent bond is more electronegative 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) 29

30 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 closer 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 dipole, which we show with a δ- and a δ+. These are partial 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 closer 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 difference in electronegativity, then we say that the bond is polar. 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 relative 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 increases across a period left to right and also up a group. F is the most electronegative atom and Group 0 elements do not have electronegativity that can reliably be determined Group 0 elements exist as atoms not molecules. You can use the pneumonic CARS to help you remember the factors that affect the electronegativity of an element. Electronegativity depends on the number of prtons in the nucleus, the distance 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 similar between both atoms then a covalent bond forms. As the difference in electronegativity increases the Level 3 Applied Science Unit 1 (Chemistry) 30

31 covalent bond will become more polar. If the difference in electronegativity is very large then the bond becomes ionic 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 weaker attractive forces (compared with covalent and ionic bonds) that exist between the molecules. These are known as intermolecular forces or van der waals 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 Intermolecular force between molecues Complete question 55 Level 3 Applied Science Unit 1 (Chemistry) 31

32 2.7 Permanent dipole-dipole forces Polar molecules, such as HCl, have permanent dipoles. The permanent dipole of one polar molecule can attract the opposite permanent 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 difference in electronegativity, no permanent dipoles and therefore no permanent dipole-dipole force between neighbouring molecules. However, the electrons are constantly moving within the molecules. At any one moment in time, the electrons may be temporarily closer to one of the H atoms in the H2 molecule. This would instantaneously lead to a temporary 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 induce a neighbouring molecule into also forming a temporary dipole; these temporary dipoles attract each other leading to a temporary dipole-induced dipole force of attraction (also called London dispersion forces). London dispersion forces exist in all molecules, whether polar or non-polar. A non-polar molecule will only 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) 32

33 Temporary dipole-induced dipole forces increase with an increase in the number of electrons. More electrons results in larger temporary and induced dipoles which results in stronger attractive London dispersion forces between the molecules. Stronger attractive forces between the molecules results in an increase in boiling point. WORKED EXAMPLE Explain the following trend in the boiling points of the halogens listed? F2 Cl2 Br2 I C C 59 0 C C The boiling point of the halogens increases as you go down group 7. This is because I2 has more electrons than Br2, which has more electrons than Cl2, which has more electrons than F2; more electrons results in larger temporary dipoles. This means there would be stronger 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. 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 large difference in electronegativity between H and the O/N/F atoms, these bonds are highly polar. The permanent dipole-dipole forces between these bonds in neighbouring molecules are particularly strong and are given a special name: hydrogen bonding. In hydrogen bonding, the H δ+ in one molecule attracts the lone pair of electrons on the O δ-, N δ- or F δ- in the neighbouring molecule. E.g. a diagram showing hydrogen bonding between neighbouring 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) 33

34 MODEL ANSWER comparing intermolecular forces present Explain why HCl has a higher boiling point than H2? HCl is a polar molecule. Due to the difference in electronegativity between H and Cl, there are permanent dipoles. This means permanent dipole-permanent dipole forces between the molecules in addition to temporary dipole-induced dipole forces. H2 is a non-polar molecule. There is no difference in electronegativity between the atoms. The only intermolecular force present is temporary dipole-induced dipole forces. Permanent dipole-dipole forces are stronger than temporary dipole-induced dipole forced. This means the intermolecular forces between HCl molecules are stronger than those in H2. The stronger intermolecular forces in HCl require more energy to break resulting in a higher boiling point. 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 bonding or intermolecular forces present. Melting (solid to liquid) and boiling (liquid to gas) 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 energy is required to break them and the higher the melting or boiling point. In general there is an increase in melting/boiling points across periods 2 & 3 (left to right) up to Group 4. There is then a sharp decrease 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) 34

35 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 higher for the metals (Li, Be, Na, Mg & Al) due to the strong metallic bonding present between positive metal ions and delocalised electrons in the giant metallic lattice. There is also an increase in melting points for the metals each time you go across the period (e.g. Na Mg Al) because there is one more proton so a greater nuclear charge and also more delocalised electrons with the next metal across the period. This results in a stronger 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. 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 increases 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, Si and SiO2 exist as a giant covalent 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 stronger covalent bonds to achieve this. Level 3 Applied Science Unit 1 (Chemistry) 35

36 Strong covalent bonds between C atoms to break in the giant covalent lattice in diamond There is then a sharp decrease in melting/boiling points between Groups 4 and 5, and the boiling points for Groups 5-8 are relatively low. This is because the elements now exist as small covalently bonded molecules between Groups 5-8, with only weaker temporary dipole-induced 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 covalent lattice or simple molecular lattice. Note that across Period 3, Groups 5-8, S8 has the highest 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 electrons in the molecule. More electrons means larger temporary and induced dipoles resulting in stronger 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 MODEL ANSWER Explain why Mg has a higher boiling point than P? Mg exists as a giant metallic lattice with stronger metallic bonds to break. P exists as P4 molecules in a simple molecular lattice with weaker temporary dipole-induced dipole forces between the molecules. The electrostatic attraction between positive metal ions and delocalised electrons in Mg is stronger than the weaker temporary dipole-induced dipole forces between P4 molecules. Breaking the metallic bonds in Mg requires more energy hence the higher boiling point. Level 3 Applied Science Unit 1 (Chemistry) 36

37 You also need to know the trend in boiling points for the oxides of the elements in Periods 2 and 3. The formulae of the oxides can be worked out using the cross-over method for the ionic compounds. Some important properties of the oxides are also listed. Period 2 Li2O BeO B2O3 CO or CO2 Period 3 Na2O MgO Al2O3 SiO2 Properties of oxides Metal oxides are ionic compounds, existing in a giant ionic lattice with strong ionic bonds to break the strong electrostatic force of attraction between oppositely charged ions. They form alkaline solutions. Al2O3 is amphoteric. SiO2 is giant covalent, strong covalent bonds to break. CO and CO2 are simple molecules with weaker intermolecular forces to break. CO and CO2 are also acidic oxides. NO, NO2, N2O5 P4O6 P4O10 O3 SO2, SO3 Simple molecular structures and gases with weaker intermolecular forces to break. Nitrogen forms a range of oxides (acidic) with different oxidation states. O2 and O3 are allotropes. Ignore Groups 7 and 8. Complete questions Trends in melting and boiling points down a group Melting points decrease down Groups 1 & 2 as there are more shells down a group meaning a larger radius and more shielding, resulting in weaker forces of attraction between the particles. The melting points increase 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 more electrons resulting in larger temporary dipoles and therefore stronger temporary dipole-induced dipole forces to break between the molecules which require more energy to break. Complete question 68 Level 3 Applied Science Unit 1 (Chemistry) 37

38 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 eight 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 Electrons Each shell is then broken down into sub-shells within which are found orbitals. There are four 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 2p. The number for the sub-shell tells you which shell 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 orbitals and each orbital can hold up to two 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) 38

39 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 Shape Number of these orbitals per sub-shell Total number of electrons held in all of these orbitals s Sphere 1 1 x 2 = 2 p Dumb-bell 3 3 x 2 = 6 d 5 5 x 2 = 10 f 7 7 x 2 = 14 The table below summarises where the electrons are found in each shell in an atom. s s 2 p s p d Complete questions Level 3 Applied Science Unit 1 (Chemistry) 39

40 3.2 Electron in box diagrams and electron configurations In the Bohr model, we used dots or crosses 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 boxes to represent the orbitals and arrows to represent the electrons in the orbitals. Because p orbitals come in threes we use three boxes for p orbitals. In a similar fashion, we use five boxes for d orbitals and seven boxes for f orbitals. Each box (orbital) can hold up to two electrons. Electrons are negatively charged and repel each other; we show this by having one arrow pointing up and one pointing down. For the same reason, each orbital is filled singularly before pairing starts for p, d and f sub-shells. Note that in the electron in box diagram below, the 4s 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 overlap, 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 electron 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 highest filled subshell first. The electron in box diagram below is for Mn, which has an atomic number of 25 and so 25 electrons when a neutral atom. Note how in the 3d sub-shell the orbitals are filled singularly 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) 40

41 WORKED EXAMPLE 1. Write out the electron configuration for P and P 3-? P has an atomic number of 15, so 15 electrons: 1s 2 2s 2 2p 6 3s 2 3p 3 P 3- has gained 3 electrons so now has 18 electrons: 1s 2 2s 2 2p 6 3s 2 3p 6 2. Draw an electron in box diagram for P and explain how the subshells are filled with electrons? 3p 3s 2p 2s 1s Lowest energy sub-shell is filled first. Each energy level must be full before the next one higher up is filled (Aufbau principle) Each orbital is filled singularly before pairing starts. Single pairing in the 3p for this example. Complete questions 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 outer 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. Complete questions Level 3 Applied Science Unit 1 (Chemistry) 41

42 3.4 Ionisation energies re-visited We have seen previously seen that there are trends in ionisation energies across a period; in general there is an increase 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) 42

43 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 state 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 both sides of an equation. Equations are balanced by putting numbers before the formulae to ensure that there is the same number of atoms of each element on both sides of the equation. WORKED EXAMPLES 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 one 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) balances the hydrogen s: N2 (g) + 3 H2 (g) 2 NH3 (g) There is now an equal number of each atom on each side of the equation; it is fully balanced. 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 metal so must be a solid at room temperature, oxygen is a gas. Sodium oxide is an ionic compound (which are usually solids) because it is made up of a metal and a non-metal; use the cross-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) 43

44 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) Complete question Reactions of period 2 and 3 elements with oxygen You have previously seen how as you move across periods 2 and 3, left to right, there is an increase 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 metallic lattice (stronger metallic bonds to break) to giant covalent (stronger covalent bonds to break) to simple molecular lattice (weaker intermolecular forces to break). As you move across periods 2 and 3 from left to right, the products formed when these elements react with oxygen tend to be giant ionic lattices to giant covalent structures to simple molecular structures (see the table on the next page). You get giant ionic when a metal 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 gases for the simple molecules. The products made also change from being alkali (metal oxides are examples of bases) to amphoteric to acidic. Level 3 Applied Science Unit 1 (Chemistry) 44

45 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) 45

46 4.3 Reaction of metals with oxygen, water, and dilute acids (i) Oxygen: Metals react with the non-metal oxygen to form ionic metal oxides. The Group 1 and 2 metal oxides are used as bases because they form alkali solutions. It is very important that you check the formula of the metal oxide formed using the cross-over method and then fully balance the equation. When using the crossover method, the positive charge on the metal ion is the same as its Group 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) Further reading: Group 1 metals react rapidly with oxygen and are stored under oil 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 oxidation and makes them behave as unreactive metals. The Group 4 metals lead (Pb) and Tin (Sn) also form oxides. The transition metals are much less 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 peroxide ion, O2 2- and the super-oxide ion O2 - also exist. These molecular ions contain a covalent bond between the two O atoms; O- O 2- and O-O -. These two negative molecular ions are only stable next to a large 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 small positive ion and when next to a large negative ion such as O 2- or O2 -, the electrons in the covalent bond between the two O atoms are strongly attracted to the small positive metal ion; the ionic bond becomes polarised. 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. Level 3 Applied Science Unit 1 (Chemistry) 46

47 (ii) Water: Metals react with water to form solutions of a metal hydroxides as well as hydrogen gas. 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 brackets 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 alkali metals because they form a basic solution when reacted with water. The reactivity of Group 1 and 2 metals inceases down the Group. This can be explained by CARS. 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 oxide layer. Group 4, 5 and 6 metals do not react with water. Some transition metals do react with water but only very slowly. (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 salts in a neutralisation reaction as well as H2 gas. Chloride salts form with HCl and sulfate 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 Complete question 81 Level 3 Applied Science Unit 1 (Chemistry) 47

48 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 different 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 overall charge Assign oxidation numbers to each individual atom that is in the molecule/ion when looking at redox reactions. All of the oxidation numbers when added together must equal 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) 48

49 In the above reaction, Fe is a reducing agent; it reduced Cl (Fe itself was oxidised). Cl is an oxidising agent; it oxidised Fe (Cl itself was reduced). WORKED EXAMPLE Assign oxidation numbers to the equation below and prove that it is a REDOX reaction. Sr + 2 HCl SrCl2 + H2 Oxidation state for Sr: un-combined element, so 0. Oxidation state for H in HCl: combined H, so +1. Oxidation state for Cl in HCl: sum of oxidation numbers must equal the charge. The charge is 0, and the oxidation state of H is +1. So Cl must be -1 as +1-1 = 0. Oxidation state for Sr in SrCl2: SrCl2 is an ionic compound (mixture of metal with non-metal). The charge on any ion is its oxidation number. Sr is in Group 2 so forms 2+ ions. Oxidation state is +2. Oxidation state of Cl in SrCl2: ionic compound and each Cl has a 1- charge. Oxidation numbers must be assigned to each individual ion. Also the sum of the oxidation numbers equals the overall charge of 0. Cl is -1. Oxidation state of H in H2: un-combined element so 0. The oxidation numbers should be written above the equation as seen below. Either state what has been oxidised and reduced or use arrows such as in the diagram below. Careful when you have more than one of the same element in a compound as is the case with Cl in SrCl2. You must not write -2 above the Cl2 part. Oxidation numbers are assigned to EACH individual atom Sr + 2 HCl SrCl2 + H2 Oxidised Reduced Sr has gone from Sr = 0 to Sr = +2, so it has been oxidised H has gone from H = +1 to H = 0, so it has been reduced Both oxidation and reduction have taken place in this reaction so it is a REDOX reaction Oxidation also means the gain of oxygen. Questions may ask which substance in an equation has been oxidised or reduced. In the equation below, CO has been oxidised to CO2. Fe2O3 + 3 CO 2 Fe + 3 CO2 Complete questions Level 3 Applied Science Unit 1 (Chemistry) 49

50 4.5 Oxidation numbers for transition metals and oxyanions Transition elements can have variable 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 roman numeral in brackets. The variable oxidation state is what makes the transition metals useful as catalysts 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 Cu = +2 Cu2O Copper(I) oxide Cu = +1 Oxyanions are: negative molecular ions that contain oxygen 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 second 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 - Nitrate (III) N = +3 KMnO4 (K + MnO4 - ) Potassium manganate(vii) Mn = +7 NaNO3 Sodium nitrate(v) N = +5 NaNO2 Sodium nitrate (III) N = +3 CaSO4 Calcium sulfate (VI) S = +6 Complete question 86 Level 3 Applied Science Unit 1 (Chemistry) 50

51 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 top of the list. 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 in the next section before moving on to displacement reactions. The order of reactivity for the metals is Group1, Group2, Group3, Group4, Transition metals. The metals higher up are more reactive as they have a greater tendency to lose an electron to form a complete outer shell. In general, for the metals, reactivity decreases across a period and increases down a Group. Across a period nuclear charge and attraction increase, making it more difficult to lose an electron during a chemical reaction, but down a Group nuclear attraction decreases making it easier to lose an electron during a chemical reaction. The full explanation is CARS! Complete question 87 Level 3 Applied Science Unit 1 (Chemistry) 51

52 4.7 Displacement reactions A more reactive 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 reactivity series. Displacement reactions involving metals are also usually redox reactions you can prove this by assigning oxidation numbers to each element in the equation. E.g. the displacement reaction below is a redox reaction: Fe (s) + CuSO4 (aq) FeSO4 (aq) + Cu (s) Oxidised Reduced Displacement reactions can also happen with the Halogens. The halogens become more 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 spectator 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 oxidising agents this means that they oxidise (remove electrons from) another species. In the above reaction, chlorine is a stronger 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) 52

53 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 x molecules. Where has this number come from? We have also previously seen how the isotope 12 C is used as the standard for weighing atoms when we introduced the idea of relative atomic mass, which is the average 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 grams 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 one 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, or 6.02 x atoms). The molar mass for CO2 is 44 gmol -1 (44 grams are required for you to have one mole, or 6.02 x molecules. The molar mass (R), the number of moles (M) and the mass in grams of the substance are related 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 ) WORKED 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) 53

54 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, divide by To go from dm 3 to cm 3, multiply 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 0.5 moles of NaOH in every 1 dm 3. 2 dm 3 of a 0.5 mol dm -3 NaOH solution contains1.0 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) 54

55 WORKED EXAMPLES 1. Calculate the amount, in mol, of HCl in 100 cm 3 of a solution with a concentration of mol dm -3. Using the above triangle, M = C x V (in dm 3 ) = x 0.1 = mol 2. Calculate the concentration of a NaOH solution when mol of NaOH is dissolved in 250 cm 3 of water. Using the above triangle, C = M / V (in dm 3 ) = / 0.25 = 0.1 mol dm -3 Complete questions Moles and equations We shall now link the number of moles to balanced chemical equations. We can use chemical equations to work out the masses of the products we expect to make. Looking at a chemical equation, we do not think about how many molecules are reacting but rather how many moles are reacting. The balancing numbers in a balanced equation gives you the ratio of moles reacting in a given reaction. Stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction. Take for example the following equation: 4 Na + O2 2 Na2O 4 moles of Na reacts with... 1 mole of O2 molecules... to form 2 moles of Na2O We can use these molar ratios to calculate the number of moles of reactants needed or the number of moles of products formed, e.g., using the molar ratios in the above balanced equation; 2 moles of Na reacts with 0.5 moles of O2 to form 1.0 mole of Na2O. 4 moles of Na2O can be formed by reacting 2 moles of O2 with 8.0 moles Na. One mole of Na was reacted with excess oxygen to form 0.5 moles of Na2O. Stoichiometry is usually used together with the two triangles met earlier to help you work out quantities in chemical reactions. You shall be using this idea in unit 2 when you perform titrations. In a titration, a standard solution (who s concentration you know) is used to determine the concentration of a second solution where the concentration is unknown. Level 3 Applied Science Unit 1 (Chemistry) 55

56 WORKED EXAMPLES 1. In a reaction, 0.14 g of calcium oxide, CaO, is reacted with excess hydrochloric acid. CaO + 2 HCl CaCl2 + H2O (i) Calculate the amount, in mol, of CaO used in the reaction. For CaO, we have G = 0.14 and R = = 56.1 So we can calculate M = G / R = 0.14 / 56.1 = mol (ii) Determine the amount, in mol, of CaCl2 produced in the reaction. From the stoichiometry in the equation, 1 mole of CaO produces 1 mole of CaCl2. So mol CaO would also produce mol of CaCl2. (iii) Calculate the mass, in grams, of CaCl2 produced in the reaction. For CaCl2, we now have M = and R = (35.5 x 2) =111.1 So we can calculate G for CaCl2 = M x R = x = g 2. In an experiment, 0.10 g of Li metal was added to 200 cm 3 of water to make a solution of LiOH. 2 Li + 2 H2O 2 LiOH + H2 (i) Calculate the amount, in mol, of Li metal used in this reaction. For Li, G = 0.10, R = 6.9 So we can calculate M = G / R = 0.10 / 6.9 = mol (ii) Calculate the concentration of the LiOH solution formed, in mol dm -3. Molar ratio of Li to LiOH in the equation is 2:2 = 1:1, so M = C = M / V = / 0.2 = mol dm -3 Complete questions Level 3 Applied Science Unit 1 (Chemistry) 56

57 5.4 Percentage yield Once you work out the moles of one reactant in a chemical equation, you can calculate the moles of product expected (also called the theoretical moles) using the molar ratios in the chemical equation. Once you know the moles of product expected you can then calculate the mass in grams of product expected (also referred to as the theoretical mass) using GMR. However, you rarely produce the expected (or the theoretical) mass of product when you do a chemical reaction. Some of the product could have been lost when transferring the product from one vessel to another. Some of the reactant chemical may react with impurities. The reaction may be also be reversible. Chemists need to know how efficient their reaction process is so they calculate the percentage yield. You can calculate the percentage yield based on the number of moles or the mass in grams. WORKED EXAMPLES % yield = actual no. of moles x 100 expected no. of moles % yield = actual mass x 100 expected mass 1. We are expecting 10g of product from a reaction but only produce 8g. What % is this? % yield = actual mass x 100 = 8 x 100 = 80% expected mass We are expecting 0.5mol of product but only produce 0.2mol. What % is this? % yield = actual no. of moles x 100 = 0.2 x 100 = 40% expected no. of moles g of ethanol is oxidised to form 2.4 g of ethanoic acid. Calculate the percentage yield. C2H5OH + 2 [O] CH3COOH + H2O Moles of ethanol used = G / R = 2.3 / 46 = 0.05 mol Expected moles of ethanoic acid = 0.05 mol also (1:1 molar ratio in equation) Actual moles of ethanoic acid produced = G / R = 2.4 / 60 = 0.04 mol Percentage yield = (0.04 / 0.05) x 100 = 80% Complete questions Extension: what is empirical formula and how do you calculate this using moles? Level 3 Applied Science Unit 1 (Chemistry) 57

58 Level 3 Applied Science Unit 1 (Chemistry) 58