1:20a understand how to use electrical conductivity to classify elements as metals or non-metals
Metals
- conduct electricity
Non – Metals
- do not conduct electricity (except for graphite)
1:20 understand how to use electrical conductivity and the acid-base character of oxides to classify elements as metals or non-metals
Metals
- conduct electricity
- have oxides which are basic, reacting with acids to give a salt and water
Non – Metals
- do not conduct electricity (except for graphite)
- have oxides which are acidic or neutral
1:21 identify an element as a metal or a non-metal according to its position in the Periodic Table
Metals on the left of the Periodic Table.
Non-Metals on the top-right, plus Hydrogen.
1:22 understand how the electronic configuration of a main group element is related to its position in the Periodic Table
Elements in the same group have the same number of electrons in their outer shell.
This is why elements from the same group have similar properties.
1:23 Understand why elements in the same group of the Periodic Table have similar chemical properties
Elements in the same group of the periodic table have the same number of electrons in their outer shells, which means they have similar chemical properties.
1:24 understand why the noble gases (Group 0) do not readily react
The noble gases are inert (unreactive) because they have a full outer shell of electrons.
1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided
Example:
Sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).
Word equation:
sodium + water –> sodium hydroxide + hydrogen
Writing the chemical equation
A chemical equation represents what happens in terms of atoms in a chemical reaction.
Step 1: To write a chemical equation we need to know the chemical formulae of the substances.
Na + H2O –> NaOH + H2
Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.
During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.
2Na + 2H2O –> 2NaOH + H2
If asked for an equation, the chemical equation must be given.
State symbols are used to show what physical state the reactants and products are in.
| State symbols | Physical state |
|---|---|
| (s) | Solid |
| (l) | Liquid |
| (g) | Gas |
| (aq) | Aqueous solution (dissolved in water) |
Example:
A solid piece of sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).
2Na(s) + 2H2O(l) –> 2NaOH(aq) + H2(g)
Balancing equations – Tyler de Witt videos
This excellent Tyler de Witt video is an introduction to balancing equations:
And here’s another of the lovely Tyler’s videos with some practice questions and answers on equation balancing:
Balancing equations – interactive
This is useful to help you to practice how to balance equations:
1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)
Relative formula mass (Mr) is mass of a molecule or compound (on a scale compared to carbon-12).
It is calculated by adding up the relative atomic masses (Ar) of all the atoms present in the formula.
Example:
The relative formula mass (Mr) for water (H2O) is 18.
Water = H2O
Atoms present = (2 x H) + (1 x O)
Mr = (2 x 1) + (1 x 16) = 18
Calculation of relative formula mass – Tyler de Witt video
Here’s an excellent Tyler de Witt video explaining how to calculate the relative formula mass of compounds with:
- a simple formula
- a formula which includes brackets
- a formula which includes a dot (water of crystallisation)
1:27 know that the mole (mol) is the unit for the amount of a substance
In Chemistry, the mole is a measure of amount of substance (n).
The abbreviation for mole is mol.
The mass of 1 mole of a substance is the relative formula mass (Mr) of the substance in grams.
Example:
The Mr of water is 18.
Therefore the mass of 1 mol of water equals 18 g.
1:28 understand how to carry out calculations involving amount of substance, relative atomic mass (Aᵣ) and relative formula mass (Mᵣ)
The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:
Example 1:
Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO2).
Step 1: Calculate the relative formula mass (Mr) of carbon dioxide (CO2).
Carbon\,dioxide = CO_2 Atoms\,present = (1 \times C) + (2 \times O) M_r = (1 \times 12) + (2 \times 16) = 44Step 2: Use the formula to calculate the amount in moles.
Amount = \frac{mass(g)}{M_r} Amount = \frac{8.8}{44} Amount = 0.2 molExample 2:
Calculate the mass of 2 mol of copper(II) sulfate (CuSO4).
Step 1: Calculate the relative formula mass (Mr) of copper(II) sulfate (CuSO4).
copper(II)\,sulfate = CuSO_4 Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O) M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5Step 2: Rearrange the formula to calculate the mass.
mass = amount \times M_r mass = 2 \times 159.5 mass = 319gCalculations involving mass (in grams), amount (in moles) and relative atomic mass – Tyler de Witt video
This video shows how to perform calculations involving mass (in grams), amount (in moles) and relative atomic mass:
1:29 calculate reacting masses using experimental data and chemical equations
Example: When calcium carbonate (CaCO3) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.
CaCO3 –> CaO + CO2
Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO3)
M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16) M_r(CaCO_3)= 100 Amount = \frac{mass(g)}{M_r} Amount = \frac{25}{100} Amount = 0.25 molStep 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO3.
This step involves using the ratio of CaCO3 to CaO from the chemical equation.
CaCO3 –> CaO + CO2
From the equation you can see that the ratio of CaCO3 to CaO is 1:1.
Therefore if you have 0.25 mol of CaCO3 this will produced 0.25 mol of CaO.
Step 3: Calculate the mass of 0.25 mol of CaO.
M_r(CaO)=(1 \times 40) + (1 \times 16) M_r(CaO)= 56 Mass(CaO)=amount \times M_r Mass(CaO)= 0.25 \times 56 Mass(CaO)= 14g
A simple format for laying out this method can be used.
Example: What mass of ammonia (NH3) is formed when 7 g of nitrogen (N2) is combined with hydrogen (H2).
Calculate Reacting Masses video
This video steps through a very useful method used to calculate reacting masses.
1:30 calculate percentage yield
Yield is how much product you get from a chemical reaction.
The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.
In most chemical reactions, however, you rarely achieved your theoretical yield.
For example, in the following reaction:
CaCO3 –> CaO + CO2
You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO3.
However, what if the actual yield is only 48 g of CaO.
By using the following formula, the % yield can be calculated:
yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product} yield= \frac{48}{56} yield=0.86 \% yield=86\%1:31a understand how the formulae of simple compounds can be obtained experimentally, including metal oxides
Finding the formula of a metal oxide experimentally
The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)
1:31 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation
Finding the formula of a metal oxide experimentally
The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)
Finding the formula of a salt containing water of crystallisation
When some substances crystallise from solution, water becomes chemically bound up with the salt. This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula CuSO4.5H2O which formula indicates that for every CuSO4 in a crystal there are five water (H2O) molecules.
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO4.5H2O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO4). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.
1:32 know what is meant by the terms empirical formula and molecular formula
The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.
The molecular formula shows the actual number of atoms of each type of element in a molecule.
For example, for ethane:
Molecular formula = C2H6
Empirical formula = CH3
Here are some more examples:
| Name | Molecular formula | Empirical Formula |
|---|---|---|
| pentene | C5H10 | CH2 |
| butene | C4H8 | CH2 |
| glucose | C6H12O6 | CH2O |
| hydrogen peroxide | H2O2 | HO |
| propane | C3H8 | C3H8 |
Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.
1:33 calculate empirical and molecular formulae from experimental data
Calculating the Empirical Formula
Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?
Step 1: Put the symbols for each element involved at the top of the page.
Step 2: Underneath, write down the masses of each element combining.
Step 3: Divide by their relative atomic mass (Ar).
Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.
Step 5: Use this to give the empirical formula.
(If your ratio is 1:1.5 then multiple each number by 2
If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)
Calculating the Molecular Formula
If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.
Example: A compound has the empirical formula CH2, and its relative formula mass is 56. Calculate the molecular formula.
Step 1: Calculate the relative mass of the empirical formula.
Step 2: Find out the number of times the relative mass of the empirical formula goes into the Mr of the compound.
Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.
Empirical formula calculations involving water of crystallisation
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula BaCl2.nH2O where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.
Example: If you heated hydrated barium chloride (BaCl2.nH2O) in a crucible you might end up with the following results.
Step 1: Calculate the mass of the anhydrous barium chloride (BaCl2) and the water (H2O) driven off.
Step 2: Use the empirical formula method to find the value of n in the formula.
Empirical formulae and Molecular formulae – Tyler de Witt video
In this video, the lovely Tyler de Witt explains Empirical Formulae and Molecular formulae.
(Please excuse the horrid error right at the start where the video shows a diagram of prop-2-ene, but it is labelled ‘ethene’ by mistake.)
This second video then explains how, if given an empirical formula and a molecular mass, you can calculate the molecular formula:
1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution
Concentration is a measurement of the amount of substance per unit volume.
In Chemistry, concentration is measured in mol/dm3 (read as moles per cubic decimetre).
The following formula allows for the interconversion between a concentration (in mol/dm3), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm3).
amount(mol) = concentration(mol/dm^3) \times volume(dm^3)
or
concentration(mol/dm^3) = \frac{amount(mol)}{volume(dm^3)}Note: 1 dm3 = 1000 cm3
For example, to convert 250 cm3 into dm3:
250cm^3 = \frac{250}{1000} 250cm^3 = 0.25dm^3
Example 1: 0.03 mol of sodium carbonate (Na2CO3) is dissolved in 300 cm3 of water. Calculate the concentration of the solution.
Step 1: Convert the volume of water from cm3 into dm3.
volume = 300cm^3 volume = \frac{300}{1000}dm^3 volume = 0.300dm^3Step 2: Use the molar concentration formula to calculate of the concentration of the solution.
concentration = \frac{amount}{volume} concentration = \frac{0.03}{0.300} concentration = 0.1 mol/dm^3
Example 2: Calculate the amount, in moles, of 25cm3 of hydrochloric acid (HCl) with a concentration of 2 mol/dm3.
Step 1: Convert the volume of HCl from cm3 into dm3.
volume = 25cm^3 volume = \frac{25}{1000}dm^3 volume = 0.025dm^3Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.
amount= concentration \times volume amount = 2 \times 0.025 amount = 0.050 mol
Titration calculations, example 1
The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.
For example, a titration problem will look like this:
A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution (NaOH) required 23.50 cm3 of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm3 of the acid.
The chemical equation for this reaction is:
NaOH(aq) + HCl(aq) –> NaCl(aq) + H2O(l)
volume 25.0cm3 23.50cm3
concentration 0.100mol/dm3
It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.
Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).
First convert the volume of NaOH from cm3 into dm3.
volume = 25cm^3 volume = \frac{25}{1000}dm^3 volume = 0.025dm^3Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.
amount= concentration \times volume amount = 0.100 \times 0.025 amount = 0.0025 mol
Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).
From the chemical equation the ratio of NaOH to HCl is 1:1
Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.
Step 3: Calculate the concentration of the hydrochloric acid (HCl).
Convert the volume of HCl from cm3 into dm3.
volume = 23.5cm^3 volume = \frac{23.5}{1000}dm^3 volume = 0.0235dm^3Use the molar concentration formula to calculate of the concentration of the HCl.
concentration = \frac{amount}{volume} concentration = \frac{0.0025}{0.0235} concentration = 0.106 mol/dm^3
Titration calculations, example 2
25.0 cm3 of sodium carbonate solution (Na2CO3) of unknown concentration was neutralised by 30.0 cm3 of 0.100 mol/dm3 nitric acid (HNO3).
Na2CO3(aq) + 2HNO3(aq) –> 2NaNO3(aq) + CO2(g) + H2O(l)
Calculate the concentration, in mol/dm3 of the sodium carbonate solution.
Step 1: Calculate the amount, in moles, of nitric acid (HNO3).
volume = 30cm^3 volume = \frac{30}{1000}dm^3 volume = 0.030dm^3 amount= concentration \times volume amount = 0.100 \times 0.030 amount = 0.0030 mol
Step 2: Deduce the amount, in moles of sodium carbonate (Na2CO3).
Using the equation:
Na2CO3(aq) + 2HNO3(aq) –> 2NaNO3(aq) + CO2(g) + H2O(l)
Ratio Na2CO3:HNO3 = 1:2
amount=\frac{0.0030}{2} = 0.0015mol
Step 3: Calculate the concentration of sodium carbonate (Na2CO3).
volume = 25cm^3 volume = \frac{25}{1000}dm^3 volume = 0.025dm^3 concentration = \frac{amount}{volume} concentration = \frac{0.0015}{0.025} concentration = 0.06 mol/dm^3
Converting mol/dm3 into g/dm3
Concentration can also be expressed in g/dm3 (grams per cubic decimetre).
Therefore mol/dm3 can be converted into g/dm3.
Example: Convert 0.06 mol/dm3 of sodium carbonate (Na2CO3) into g/dm3
Step 1: calculate the relative formula mass (Mr) of sodium carbonate (Na2CO3).
M_r= (2 \times 23) + (1 \times 12) (3 \times 16) = 106Step 2: Recall the formula giving the relationship between mass, amount and formula mass.
mass= amount(in\,moles) \times M_rtherefore
mass/dm^3= moles/dm^3 \times M_r mass/dm^3= 0.06 \times 106 mass/dm^3= 6.36 g/dm^3Concentration (molarity) practice problems – Tyler de Witt video
Here is a video to help you with concentration calculations. In Chemistry, concentration is know as molarity (i.e. the amount per volume).
Note that in the video, the lovely Tyler de Witt uses the unit “liter” for volume, but the correct international unit (as used by UK exam boards) is decimeters cubed (dm3) which is the same thing as Tyler’s “liter”.
and a follow up video with some more examples:
1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))
The molar volume of a gas is the volume that one mole of any gas will occupy.
1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm3 or 24,000 cm3.
Note: 1 dm3 = 1000 cm3
The following formulae allows for the interconversion between a volume in dm3 or cm3 and a number of moles for a given gas:
Amount(mol) = \frac{volume(dm^3)}{24} or
Amount(mol) = \frac{volume(cm^3)}{24000}
Example 1:
Calculate the amount, in moles, of 12 dm3 of carbon dioxide (CO2).
Amount = \frac{volume(dm^3)}{24} Amount = \frac{12}{24} Amount = 0.5mol
Example 2:
Calculate the volume at rtp in cubic centimetres (cm3), of 3 mol of oxygen, (O2).
Volume = amount \times 24000
Volume = 3 \times 24000
Volume = 72000cm^3
1:36 practical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)
To determine the formula of a metal oxide by combustion.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
2Mg + O2 –> 2MgO
Method:
- Weigh a crucible and lid
- Place the magnesium ribbon in the crucible, replace the lid, and reweigh
- Calculate the mass of magnesium
- Heat the crucible with lid on until the magnesium burns (lid prevents magnesium oxide escaping therefore ensuring accurate results)
- Lift the lid from time to time (this allows air to enter)
- Stop heating when there is no sign of further reaction
- Allow to cool and reweigh
- Repeat the heating, cooling and reweigh until two consecutive masses are the same (this ensures all Mg has reacted and therefore the results will be accurate)
- Calculate the mass of magnesium oxide formed (mass after heating – mass of crucible)
It is then possible to use this data to calculate the empirical formula of the metal oxide.
To determine the formula of a metal oxide by reduction
Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:
4CuO + CH4 –> 4Cu + CO2 + 2H2O
By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.
As an alternative, hydrogen gas could be used instead of methane:
CuO + H2 –> Cu + H2O
It is then possible to use this data to calculate the empirical formula of the metal oxide.
There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.
1:37 understand how ions are formed by electron loss or gain
Ions are electrically charged particles formed when atoms lose or gain electrons.
They have the same electronic structures as noble gases.
Metal atoms form positive ions (cations).
Non-metal atoms form negative ions (anions).
What is an ion? – Tyler de Witt video
Knowing the answer to “What’s an ion?” is absolutely key to Chemistry.
If you aren’t sure what an ion is, or why the idea of ions is crucial to Chemistry, then you should watch this video:
Dogs teach ionic versus covalent bonding – video
This should give you a ruff idea of what the difference is between ionic and covalent bonding.
In an ion, the number of protons does not equal the number of electrons – Tyler de Witt video
In an electrically neutral ATOM, the number of electrons equals the number of protons.
However, an ION is an atom (or group of atoms) which has either gained or lost some electrons, so for an ION the number of electrons does not equal the number of protons.
This excellent Tyler de Witt video explains this clearly:
1:38a understand how to use the charges of these ions in ionic formulae: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)
When given this information of the following ions, it is possible to work out the formulae of ionic compounds which include these ions.
| Name of Ion | Formula | Charge |
|---|---|---|
| Sulfate | SO42- | -2 |
| Carbonate | CO32- | -2 |
| Nitrate | NO3- | -1 |
| Hydroxide | OH- | -1 |
| Ammonium | NH4+ | +1 |
| Silver ion | Ag+ | +1 |
| Zinc ion | Zn2+ | +2 |
| Hydrogen ion | H+ | +1 |
| Copper (II) ion | Cu2+ | +2 |
| Iron (II) ion | Fe2+ | +2 |
| Iron (III) ion | Fe3+ | +3 |
| Lead (II) ion | Pb2+ | +2 |
Ion charges on the periodic table
1:38b know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)
| Name of Ion | Formula | Charge |
|---|---|---|
| Sulfate | SO42- | -2 |
| Carbonate | CO32- | -2 |
| Nitrate | NO3- | -1 |
| Hydroxide | OH- | -1 |
| Ammonium | NH4+ | +1 |
| Hydrogen ion | H+ | +1 |
Ion charges on the periodic table
1:38 know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)
| Name of Ion | Formula | Charge |
|---|---|---|
| Sulfate | SO42- | -2 |
| Carbonate | CO32- | -2 |
| Nitrate | NO3- | -1 |
| Hydroxide | OH- | -1 |
| Ammonium | NH4+ | +1 |
| Silver ion | Ag+ | +1 |
| Zinc ion | Zn2+ | +2 |
| Hydrogen ion | H+ | +1 |
| Copper (II) ion | Cu2+ | +2 |
| Iron (II) ion | Fe2+ | +2 |
| Iron (III) ion | Fe3+ | +3 |
| Lead (II) ion | Pb2+ | +2 |
Ion charges on the periodic table
1:39 write formulae for compounds formed between the ions listed in 1:38
Writing the electron configuration of an atom allows you to work out the electron configuration of the ion and therefore the charge on the ion.
Examples:
Atom = Mg
Electron configuration = 2,8,2
remove the two electrons from the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)
Ion = Mg2+
[2,8]2+
Atom = O
Electron configuration = 2,6
add two electrons to the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)
Ion = O2- [2,8]2-
Writing ionic formulae – Tyler de Witt video
This excellent video from the lovely Tyler de Witt explains how to write the formula of an ionic compound:
Writing formulae with polyatomic ions – Tyler de Witt video
This video shows how to work out the formula of a compound which contains polyatomic ions:
1:40 draw dot-and-cross diagrams to show the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7 only outer electrons need be shown
Sodium chloride, NaCl
Magnesium chloride, MgCl2
Potassium oxide, K2O
Calcium oxide, CaO
Aluminium oxide, Al2O3
Magnesium nitride, Mg3N2
1:41 understand ionic bonding in terms of electrostatic attractions
Ionic bonding: a strong electrostatic attraction between oppositely charged ions.
1:42 understand why compounds with giant ionic lattices have high melting and boiling points
Ionic compounds have high melting and boiling points because they have a giant structure with strong electrostatic forces between oppositely charged ions that require a lot of energy to break.
Giant 3D lattice of sodium chloride (NaCl)
1:43 Know that ionic compounds do not conduct electricity when solid, but do conduct electricity when molten and in aqueous solution
Ionic compounds do not conduct electricity when solid.
However, ionic compounds do conduct electricity if molten or in solution.
Ionic bonding – Tyler de Witt videos
These videos are a really good introduction to ionic bonding:
1:44 know that a covalent bond is formed between atoms by the sharing of a pair of electrons
A covalent bond is formed between two non-metal atoms by sharing a pair of electrons in order to fill the outer shell.
1:45 understand covalent bonds in terms of electrostatic attractions
Covalent bonding: a strong attraction between a shared pair of electrons and two nuclei.
