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35
1 How many protons, neutrons and electrons are in the following atoms and ions?(a) 11
5B; (b) 188O;
(c) 3014Si; (d) 64
30Zn;(e) 23
11Na+; (f) 7935Br–;
(g) 4521Sc3+.
2 Calculate the relative atomic mass, Ar, to 4 significant figures, for the following.(a) Lithium containing: 7.42%6Li and 92.58%7Li.(b) Lead containing: 204Pb, 1.1%; 206Pb, 24.9%;
207Pb, 21.7%; 208Pb, 52.3%.
3 Use Ar values from the Periodic Table to calculate the relative formula mass of the following:(a) Cr2O3; (b) Rb2S;(c) Zn(OH)2; (d) (NH4)2CO3;(e) Fe3(PO4)2.
4 Calculate the mass, in g, in:(a) 6 mol MnO2; (b) 0.25 mol Ga2O3;(c) 0.60 mol Ag2SO4.
5 Calculate the amount, in mol, in:(a) 8.823 g CaI2; (b) 48.55 g K2CrO4;(c) 7.50 g Cu(NO3)2.
6 Determine the empirical formula of the compound formed when 1.472 g of tungsten reacts with 0.384 g of oxygen.
7 Determine the molecular formula for the compound of carbon, hydrogen and oxygen atoms with the composition by mass: C, 40.0%; H, 6.7%; O, 53.3%; Mr 90.
8 (a) At RTP, what amount, in mol, of gas molecules are in: (i) 84 dm3; (ii) 300 cm3; (iii) 10 dm3?(b) At RTP, what is the volume, in dm3, of: (i) 12 mol CO2(g); (ii) 0.175 mol N2(g); (iii) 2.55 g NO(g)?
9 Find the amount, in mol, of solute dissolved in the following solutions:(a) 2 dm3 of a 0.225 mol dm–3 solution;(b) 25 cm3 of a 0.175 mol dm–3 solution.
10 Find the concentration, in mol dm–3, for the following solutions:(a) 0.75 moles dissolved in 5 dm3 of solution;(b) 0.450 moles dissolved in 100 cm3 of solution.
11 Balance each of the following equations:(a) P4(s) + O2(g) P2O5(s)(b) C5H12(l) + O2(aq) CO2(g) + H2O(l)(c) NaOH(aq) + H3PO4(aq) Na2HPO4(aq) + H2O(l).
12 Define the terms:(a) acid; (b) base; (c) alkali.
13 Write balanced equations for the following acid reactions:(a) nitric acid and iron(III) hydroxide.(b) sulfuric acid and copper(II) carbonate.(c) hydrochloric acid and aluminium.
14 What is the oxidation state of each species in the following?(a) Cu; (b) I2; (c) CH4;(d) MgSO4; (e) PbCO3; (f) Cr2O7
2–;(g) (NH4)3PO4.
15 From the experimental results below, work out the formula of the hydrated salt.Mass of ZnSO4·xH2O = 8.985 gMass of ZnSO4 = 5.047 g
16 Lithium carbonate decomposes with heat: Li2CO3(s) Li2O(s) + CO2(g) What volume of CO2, measured at RTP, is formed by the decomposition of 5.7195 g of Li2CO3?
17 Calcium carbonate reacts with nitric acid: Na2CO3(s) + 2HNO3 2NaNO3(aq) + H2O(l) + CO2(g) 0.371 g of Na2CO3 reacts with an excess of HNO3.The final volume of the solution is 25.0 cm3.(a) What volume of CO2, measured at RTP, is formed?(b) What is the concentration, in mol dm–3, of NaNO3formed?
18 In the redox reactions below, use oxidation numbers to find out what has been oxidised and what has been reduced.(a) N2 + 3H2 2NH3(b) 3Mg + 2Fe(NO3)3 3Mg(NO3)2 + 2Fe(c) MnO2 + 4HCl MnCl2 + Cl2 + 2H2O
Practice questions
34
Salt CO2 H20
Salt H2
Salt H20
Metal
Carbonate Base
Acidproton donor
Reduction• Gain of electrons• Decrease in oxidation
number
Oxidation• Loss of electrons• Increase in
oxidation number
Redox
Base
Alkalisoluble base
OH–
Moles
Redox
Acids
n = mass, mmolar mass, M
Mass
Solutions
n = c V (in dm3) V (in cm3)1 000
= c
Gas volumes
n = V (in dm3)24.0
V (in cm3)24 000
=
Moles
1.1 Atoms and reactions summaryModule 1
Atoms and reactionsPractice questions
935 chemistry.U1 M1.indd 34-35
10/3/08 11:28:15 am
Student Books
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28
29
Questions1 Use the method in Worked example 1 to calculate the unknown concentration below.
In a titration, 25.0 cm3 of 0.125 mol dm–3 aqueous sodium hydroxide reacted exactly
with 22.75 cm3 of hydrochloric acid.
HCl(aq) NaOH(aq) NaCl(aq) H2O(l)
Find the concentration of the hydrochloric acid.
2 Use the method in Worked example 2 to calculate the molar mass of the acid H2X.
A student dissolved 1.571 g of an acid, H2X, in water and made the solution up to
250 cm3. She titrated 25.0 cm3 of this solution against 0.125 mol dm–3 sodium
hydroxide, NaOH(aq). 21.30 cm3 of NaOH(aq) were needed to reach the end point.
The equation for this reaction is:
2NaOH(aq) H2X(aq) Na2X(aq) 2H2O(l)
Module 1Atoms and reactions
Titrations
Notes
For (a), we use:
amount, n c V (in cm3)________
1000
For (b), we use the balanced equation to
work out the reacting quantities of the
acid and alkali.
2 mol NaOH reacts with 1 mol H2SO4
For (c), we rearrange: n c V (in cm3)________
1000
Hence, c n 1000________V
In a titration, 25.0 cm3 of 0.150 mol dm–3 sodium hydroxide NaOH(aq) reacted
exactly with 23.40 cm3 of sulfuric acid, H2SO4(aq).
2NaOH(aq) H2SO4(aq) Na2SO4(aq) 2H2O(l)
(a) Calculate the amount, in mol, of NaOH that reacted.
n(NaOH) c V_____
1000 0.150 2
5.0_____1000
3.75 10–3 mol
(b) Calculate the amount, in mol, of H2SO4 that was used.
equation 2NaOH(aq) H2SO4(aq)
moles from equation 2 mol 1 mol
actual moles 3.75 10–3 mol 1.875 10–3 mol
(c) Calculate the concentration, in mol dm–3 of the sulfuric acid.
c(H2SO4) n1000________V
1.875 10–3 1000___________________
23.40 8.01 10–3 mol dm–3
Worked example 1: Calculating an unknown concentration
A student dissolved 2.794 g of an acid HX in water and made the solution up to
250 cm3. The student titrated 25.0 cm3 of this solution against 0.0614 mol dm–3
sodium carbonate Na2CO3(aq). 23.45 cm3 of Na2CO3(aq) were needed to reach
the end point.
The equation for this reaction is:
Na2CO3(aq) 2HX(aq) 2NaX(aq) CO2(g) H2O(l)
(a) Calculate the amount, in mol, of Na2CO3 that reacted.
n(Na2CO3) c V_____
1000 0.0614 2
3.45______1000
1.44 10–3 mol
(b) Calculate the amount, in mol, of HX that was used in the titration.
equation Na2CO3 (aq) 2 HX(aq)
moles from equation 1 mol 2 mol
actual moles 1.44 10–3 mol 2.88 10–3 mol
(c) Calculate the amount, in mol, of HX that was used to make up the 250 cm3
solution.
25.0 cm3 HX(aq) contains 2.88 10–3 mol
So, the 250 cm3 solution contains 10 2.88 10–3 2.88 10–2 mol
(d) Calculate the concentration, in g mol–1, of the acid HX.
n m__M
so, M m__n
2.794__________
2.88 10–2 97.0 g mol–1
Worked example 2: Calculating an unknown molar mass
Notes
For (a) and (b), the steps are the same
as in the first worked example.
In this worked example, however, there
are two further steps. You would be
helped through these in an AS exam.
For (c), we scale up by a factor of 10 to
find the amount, in mol, of HX in the
250 cm3 solution that was made up.
For (d), we rearrange:
amount, n mass, m
____________ molar mass, M
Hence, M m__n
In this example,
the mass of HX is 2.794 g.
1.1 13 TitrationsBy the end of this spread, you should be able to . . .
✱ Perform acid–base titrations, and carry out structured calculations.
Acid–base titrationsDuring volumetric analysis, you measure the volume of one solution that reacts with a
measured volume of a second, different solution.
An acid–base titration is a special type of volumetric analysis, in which you react a
solution of an acid with a solution of a base.
• You must know the concentration of one of the two solutions. This is usually a
standard solution (see spread 1.1.7).
• In the analysis, you use this standard solution to find out unknown information about
the substance dissolved in the second solution.
The unknown information may be:
• the concentration of the solution
• a molar mass
• a formula
• the number of molecules of water of crystallisation.
You carry out a titration as follows.
• Using a pipette, you add a measured volume of one solution to a conical flask.
• The other solution is placed in a burette.
• The solution in the burette is added to the solution in the conical flask until the reaction
has just been completed. This is called the end point of the titration. The volume of the
solution added from the burette is measured.
You now know the volume of one solution that exactly reacts with the volume of the
other solution.
We identify the end point using an indicator.
• The indicator must be a different colour in the acidic solution than in the basic solution.
Table 1 lists the colours of some common acid–base indicators. It also shows the colour
at the end point. Notice that this end point colour is in between the colours in the acidic
and basic solutions.
Indicator Colour in acid Colour in base End point colour
methyl orange redyellow
orange
bromothymol blue yellowblue
green
phenolphthalein colourless pinkpale pink*
* This assumes that the aqueous base has been added from the burette to the aqueous acid. If acid is added to
base, the titration is complete when the solution goes colourless.
Table 1 Colours of some common acid–base indicators
Calculating unknowns from titration results
Analysis of titration results follows a set pattern, as shown in the worked examples:
• the first two steps are always the same;
• the third step may be different, depending on the unknown that you need to work out.
In AS chemistry, any calculations that you carry out will be structured similarly to the
examples below.
For A2, you may have to work out these steps yourself.
Figure 1 This acid–base titration is using
methyl orange as an indicator. Methyl
orange is coloured red in acidic solutions
and yellow in basic solutions. The end
point is the colour in between – orange.
The solution in the conical flask above
has reached the end point
935 chemistry.U1 M1.indd 28-29
10/3/08 11:27:26 am
4
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
Exclusively endorsed by OCR for GCE Chemistry A
AS
In Exclusive Partnership
We listen to teachers’ needs...
Text is structured in line with the new OCR specification by Unit and Module.
Sample pages from OCR AS Chemistry A Student Book.
Learning objectives are taken from the specification to highlight what students need to know and understand.
Questions students should be able to answer after studying each spread.
Worked examples show students how calculations should be set out.
35
1 How many protons, neutrons and electrons are in the following atoms and ions?(a) 11
5B; (b) 188O;
(c) 3014Si; (d) 64
30Zn;(e) 23
11Na+; (f) 7935Br–;
(g) 4521Sc3+.
2 Calculate the relative atomic mass, Ar, to 4 significant figures, for the following.(a) Lithium containing: 7.42%6Li and 92.58%7Li.(b) Lead containing: 204Pb, 1.1%; 206Pb, 24.9%;
207Pb, 21.7%; 208Pb, 52.3%.
3 Use Ar values from the Periodic Table to calculate the relative formula mass of the following:(a) Cr2O3; (b) Rb2S;(c) Zn(OH)2; (d) (NH4)2CO3;(e) Fe3(PO4)2.
4 Calculate the mass, in g, in:(a) 6 mol MnO2; (b) 0.25 mol Ga2O3;(c) 0.60 mol Ag2SO4.
5 Calculate the amount, in mol, in:(a) 8.823 g CaI2; (b) 48.55 g K2CrO4;(c) 7.50 g Cu(NO3)2.
6 Determine the empirical formula of the compound formed when 1.472 g of tungsten reacts with 0.384 g of oxygen.
7 Determine the molecular formula for the compound of carbon, hydrogen and oxygen atoms with the composition by mass: C, 40.0%; H, 6.7%; O, 53.3%; Mr 90.
8 (a) At RTP, what amount, in mol, of gas molecules are in: (i) 84 dm3; (ii) 300 cm3; (iii) 10 dm3?(b) At RTP, what is the volume, in dm3, of: (i) 12 mol CO2(g); (ii) 0.175 mol N2(g); (iii) 2.55 g NO(g)?
9 Find the amount, in mol, of solute dissolved in the following solutions:(a) 2 dm3 of a 0.225 mol dm–3 solution;(b) 25 cm3 of a 0.175 mol dm–3 solution.
10 Find the concentration, in mol dm–3, for the following solutions:(a) 0.75 moles dissolved in 5 dm3 of solution;(b) 0.450 moles dissolved in 100 cm3 of solution.
11 Balance each of the following equations:(a) P4(s) + O2(g) P2O5(s)(b) C5H12(l) + O2(aq) CO2(g) + H2O(l)(c) NaOH(aq) + H3PO4(aq) Na2HPO4(aq) + H2O(l).
12 Define the terms:(a) acid; (b) base; (c) alkali.
13 Write balanced equations for the following acid reactions:(a) nitric acid and iron(III) hydroxide.(b) sulfuric acid and copper(II) carbonate.(c) hydrochloric acid and aluminium.
14 What is the oxidation state of each species in the following?(a) Cu; (b) I2; (c) CH4;(d) MgSO4; (e) PbCO3; (f) Cr2O7
2–;(g) (NH4)3PO4.
15 From the experimental results below, work out the formula of the hydrated salt.Mass of ZnSO4·xH2O = 8.985 gMass of ZnSO4 = 5.047 g
16 Lithium carbonate decomposes with heat: Li2CO3(s) Li2O(s) + CO2(g) What volume of CO2, measured at RTP, is formed by the decomposition of 5.7195 g of Li2CO3?
17 Calcium carbonate reacts with nitric acid: Na2CO3(s) + 2HNO3 2NaNO3(aq) + H2O(l) + CO2(g) 0.371 g of Na2CO3 reacts with an excess of HNO3.The final volume of the solution is 25.0 cm3.(a) What volume of CO2, measured at RTP, is formed?(b) What is the concentration, in mol dm–3, of NaNO3formed?
18 In the redox reactions below, use oxidation numbers to find out what has been oxidised and what has been reduced.(a) N2 + 3H2 2NH3(b) 3Mg + 2Fe(NO3)3 3Mg(NO3)2 + 2Fe(c) MnO2 + 4HCl MnCl2 + Cl2 + 2H2O
Practice questions
34
Salt CO2 H20
Salt H2
Salt H20
Metal
Carbonate Base
Acidproton donor
Reduction• Gain of electrons• Decrease in oxidation
number
Oxidation• Loss of electrons• Increase in
oxidation number
Redox
Base
Alkalisoluble base
OH–
Moles
Redox
Acids
n = mass, mmolar mass, M
Mass
Solutions
n = c V (in dm3) V (in cm3)1 000
= c
Gas volumes
n = V (in dm3)24.0
V (in cm3)24 000
=
Moles
1.1 Atoms and reactions summaryModule 1
Atoms and reactionsPractice questions
935 chemistry.U1 M1.indd 34-35
10/3/08 11:28:15 am
28
29
Questions1 Use the method in Worked example 1 to calculate the unknown concentration below.
In a titration, 25.0 cm3 of 0.125 mol dm–3 aqueous sodium hydroxide reacted exactly
with 22.75 cm3 of hydrochloric acid.
HCl(aq) NaOH(aq) NaCl(aq) H2O(l)
Find the concentration of the hydrochloric acid.
2 Use the method in Worked example 2 to calculate the molar mass of the acid H2X.
A student dissolved 1.571 g of an acid, H2X, in water and made the solution up to
250 cm3. She titrated 25.0 cm3 of this solution against 0.125 mol dm–3 sodium
hydroxide, NaOH(aq). 21.30 cm3 of NaOH(aq) were needed to reach the end point.
The equation for this reaction is:
2NaOH(aq) H2X(aq) Na2X(aq) 2H2O(l)
Module 1Atoms and reactions
Titrations
Notes
For (a), we use:
amount, n c V (in cm3)________
1000
For (b), we use the balanced equation to
work out the reacting quantities of the
acid and alkali.
2 mol NaOH reacts with 1 mol H2SO4
For (c), we rearrange: n c V (in cm3)________
1000
Hence, c n 1000________V
In a titration, 25.0 cm3 of 0.150 mol dm–3 sodium hydroxide NaOH(aq) reacted
exactly with 23.40 cm3 of sulfuric acid, H2SO4(aq).
2NaOH(aq) H2SO4(aq) Na2SO4(aq) 2H2O(l)
(a) Calculate the amount, in mol, of NaOH that reacted.
n(NaOH) c V_____
1000 0.150 2
5.0_____1000
3.75 10–3 mol
(b) Calculate the amount, in mol, of H2SO4 that was used.
equation 2NaOH(aq) H2SO4(aq)
moles from equation 2 mol 1 mol
actual moles 3.75 10–3 mol 1.875 10–3 mol
(c) Calculate the concentration, in mol dm–3 of the sulfuric acid.
c(H2SO4) n1000________V
1.875 10–3 1000___________________
23.40 8.01 10–3 mol dm–3
Worked example 1: Calculating an unknown concentration
A student dissolved 2.794 g of an acid HX in water and made the solution up to
250 cm3. The student titrated 25.0 cm3 of this solution against 0.0614 mol dm–3
sodium carbonate Na2CO3(aq). 23.45 cm3 of Na2CO3(aq) were needed to reach
the end point.
The equation for this reaction is:
Na2CO3(aq) 2HX(aq) 2NaX(aq) CO2(g) H2O(l)
(a) Calculate the amount, in mol, of Na2CO3 that reacted.
n(Na2CO3) c V_____
1000 0.0614 2
3.45______1000
1.44 10–3 mol
(b) Calculate the amount, in mol, of HX that was used in the titration.
equation Na2CO3 (aq) 2 HX(aq)
moles from equation 1 mol 2 mol
actual moles 1.44 10–3 mol 2.88 10–3 mol
(c) Calculate the amount, in mol, of HX that was used to make up the 250 cm3
solution.
25.0 cm3 HX(aq) contains 2.88 10–3 mol
So, the 250 cm3 solution contains 10 2.88 10–3 2.88 10–2 mol
(d) Calculate the concentration, in g mol–1, of the acid HX.
n m__M
so, M m__n
2.794__________
2.88 10–2 97.0 g mol–1
Worked example 2: Calculating an unknown molar mass
Notes
For (a) and (b), the steps are the same
as in the first worked example.
In this worked example, however, there
are two further steps. You would be
helped through these in an AS exam.
For (c), we scale up by a factor of 10 to
find the amount, in mol, of HX in the
250 cm3 solution that was made up.
For (d), we rearrange:
amount, n mass, m
____________ molar mass, M
Hence, M m__n
In this example,
the mass of HX is 2.794 g.
1.1 13 TitrationsBy the end of this spread, you should be able to . . .
✱ Perform acid–base titrations, and carry out structured calculations.
Acid–base titrationsDuring volumetric analysis, you measure the volume of one solution that reacts with a
measured volume of a second, different solution.
An acid–base titration is a special type of volumetric analysis, in which you react a
solution of an acid with a solution of a base.
• You must know the concentration of one of the two solutions. This is usually a
standard solution (see spread 1.1.7).
• In the analysis, you use this standard solution to find out unknown information about
the substance dissolved in the second solution.
The unknown information may be:
• the concentration of the solution
• a molar mass
• a formula
• the number of molecules of water of crystallisation.
You carry out a titration as follows.
• Using a pipette, you add a measured volume of one solution to a conical flask.
• The other solution is placed in a burette.
• The solution in the burette is added to the solution in the conical flask until the reaction
has just been completed. This is called the end point of the titration. The volume of the
solution added from the burette is measured.
You now know the volume of one solution that exactly reacts with the volume of the
other solution.
We identify the end point using an indicator.
• The indicator must be a different colour in the acidic solution than in the basic solution.
Table 1 lists the colours of some common acid–base indicators. It also shows the colour
at the end point. Notice that this end point colour is in between the colours in the acidic
and basic solutions.
Indicator Colour in acid Colour in base End point colour
methyl orange redyellow
orange
bromothymol blue yellowblue
green
phenolphthalein colourless pinkpale pink*
* This assumes that the aqueous base has been added from the burette to the aqueous acid. If acid is added to
base, the titration is complete when the solution goes colourless.
Table 1 Colours of some common acid–base indicators
Calculating unknowns from titration results
Analysis of titration results follows a set pattern, as shown in the worked examples:
• the first two steps are always the same;
• the third step may be different, depending on the unknown that you need to work out.
In AS chemistry, any calculations that you carry out will be structured similarly to the
examples below.
For A2, you may have to work out these steps yourself.
Figure 1 This acid–base titration is using
methyl orange as an indicator. Methyl
orange is coloured red in acidic solutions
and yellow in basic solutions. The end
point is the colour in between – orange.
The solution in the conical flask above
has reached the end point
935 chemistry.U1 M1.indd 28-29
10/3/08 11:27:26 am
5
Sample pages from OCR AS Chemistry A Student Book.
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
Worked examples show students how calculations should be set out.
Don’t forget
our A2 resources
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Autumn term!
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Practice exam questions provided at the end of each module. Answers are in the back of the book.
In our unique Exam Café, students will find plenty of support to help them prepare for their exams. They can Relax and prepare with handy revision advice, Refresh their memories by testing their understanding and Get That Result through practicing exam-style questions, accompanied by lots of hints and tips.
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l weekly teaching plans and guidance sheets to help save time
l customisable student practical sheets with accompanying teacher and technician notes
l a media bank of all the diagrams and learning objectives in the Student Book, all ready to use in PowerPoint format.
8
Sample lesson plan from OCR AS Teacher Support
CD-ROM.
AS
Exclusively endorsed by OCR for GCE Biology A
01865 888118 In Exclusive Partnership
ISBN 978 0 435961 83 7
Teacher Resource CD
Baldwin • Wood
Series Editor: Rob Ritchie
Sarah Baldwin
Christopher Wood
Series Editor: Rob Ritchie
9
Sample screen from OCR AS Chemistry A Teacher Support CD-ROM.
A2
Exclusively endorsed by OCR for GCE Biology A
01865 888118
In Exclusive Partnership
ISBN 978 0 435961 93 6
Teacher Resource CD
Baldwin • WoodSeries Editor: Rob Ritchie
Sarah BaldwinChristopher Wood Series Editor: Rob Ritchie
Split into 30 weekly plans.
Easy-to-use content menu and search box to help you locate your required content.
24
25
H2O
The oxygen atom has four pairs of electrons, so these will take up a basically tetrahedral
shape. But two of the electron pairs are lone pairs, so the shared pairs are repelled more
than in NH3, and the bond angle in H2O is less than in NH3. The final bond angle is
104.5°. The final shape of the molecule is described as non-linear.Shapes of molecules –
electron-pair repulsion theory
1UNIT
Key words
electron-pair
repulsioncovalent bond
lone pair
Electron pairs repel each other so they are as far apart as possible. In a covalent
compound or ion, the number of electron pairs around the central atom determines the
shape of the molecule.
Mod
ule
2
Module 2
Shapes of molecules – electron-pair repulsion theory
Quick check 1✔
WOrked exaMple
To work out the shape of an ion, for example H3O+:
STep 1 There are two ways to approach this
problem.
either draw a dot-and-cross diagram of the
molecule (see diagram opposite);
or count up the number of electrons at the
central atom.
STep 2 See what shape the electron pairs will adopt. They will be arranged
tetrahedrally because there are four pairs of electrons.
STep 3 Now see if there are any lone pairs. Yes, one lone pair.
STep 4 Electron-pair repulsion theory tells you that the lone pair will repel the
bonding pairs a little more strongly than they repel each other, and the
molecule should have the same shape and bond angle as NH3.
H3O+ has a pyramidal shape with a bond angle of 107°.
atomNumber of
electrons
Oxygen 6
Three hydrogens 3
Positive charge –1
Total number 8
× ×××
×
×OH H
H
1 Sketch the shapes and predict the bond angles in the following molecules:
a PCl3
b OCl2
c SeF6 (note that Se is in Group 6) d CCl4
2 Sketch the shapes and predict the bond angles in the following ions:
a PCl4+
b PCl6–
c NH2–
d NH4+
3 The dot-and-cross diagram for SO2 is shown in the diagram opposite.
a What is the shape of the molecule?
b Give the approximate bond angle.
QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by working out how the electron pairs
arrange themselves in space. The rules of electron-pair repulsion theory tell us that:
electron pairs repel each other so they are as far apart as possible
an electron pair shared between two atoms is called a bonding pair (BP)
a non-bonding pair on one atom only (not shared) is called a lone pair (LP)
LP–LP repulsion > LP–BP repulsion > BP–BP repulsion.
These rules explain the shapes of the ammonia, NH3, and water, H2O, molecules.
NH3
The nitrogen atom has four pairs of electrons, so these will take up a basically
tetrahedral shape. But one electron pair is a lone pair, so it repels the shared pairs more
than a bonding pair would, and the bond angle is decreased from 109.5° to 107°. The
final shape of the molecule is described as pyramidal.
Examiner tip
Don’t confuse the
shape of the molecule
with the shape the
electron pairs adopt.
×× ×
××
H
HH
H
HN
N
H
107
Bond angle 107
×× ×
××
×H
H
HH
O
O
104.5
Bond angle 104.5
Number of pairs Shapeexample
2 electron pairs linearCO2
(in fact two double bonds)
3 electron pairs trigonalBF3
4 electron pairs tetrahedralCH4
NH+4
6 electron pairs octahedralSF6
××
××
××××
××××
O OC
O OC
180
Bond angle 180
×× ××
×× ×
××× ×
×××
F
F ×× ××
× ××F
F
F FB
B
120
Bond angle 120
×××
×H
H
H
HH
H
C
H HC
Bond angle 109.5
109.5
FF
F
F
F F
Bond angle 90
What about multiple bonds?
If a molecule has double (or triple) bonds, the bonding electron pairs are all located
between the bonding atoms and therefore they cound as one bonding pair of electrons
around the central atom for working out the shape.
CO2
A dot-and-cross diagram of CO2 shows that the bonding is O=C=O (see table). The
electron pairs in the double bonds will repel each other as much as possible, so the final
shape of the molecule is linear.
What about a non-linear molecule with double bonds? A good example is sulfur dioxide
(see quick check question 3).
Quick check 2✔
Quick check 3✔
Revision Guides
l Clearly written and well designed to aid revision.
l Written by experienced examiners and tailored to the new specification.
l Packed with examiner tips.
l Targeted at ensuring understanding with quick-check questions on each topic and end of unit exam-style questions.
10
Sample pages from OCR AS Chemistry Revision Guide.
Mike Wooster and Helen EcclesSeries editor: Rob Ritchie
In Exclusive Partnership
This revision guide is tailored to the OCR specification and exclusively
endorsed by OCR for GCE Chemistry A. It is written by experienced
examiners and teachers, giving you: complete coverage of the specification for the exams
content organised by module and unit to follow the structure of the
specification and exams bite-sized chunks of information to make it easier to organise your
revision time quick-check revision questions so that you can test your own
knowledge easily hints and tips from examiners to help you avoid common errors
lots of practice exam-style questions for each unit all the answers to questions so that you can check that you’re on the
right track.Titles in this series:OCR AS Chemistry student book and exam café CD-ROM 978 0 435691 81 3
OCR A2 Chemistry student book and exam café CD-ROM 978 0 435691 98 1
OCR AS Chemistry Teacher Support CD-ROM978 0 435691 83 7
OCR A2 Chemistry Teacher Support CD-ROM978 0 435691 93 6
OCR AS Chemistry revision guide
978 0 435583 71 2
OCR A2 Chemistry revision guide
978 0 435583 74 3
Exclusively endorsed by OCR for GCE Chemistry A
AS AS
Second Edition
01865 888118
www.heinemann.co.uk
In Exclusive PartnershipI S B N 978-0-435583-71-2
9 7 8 0 4 3 5 5 8 3 7 1 2
Exclusively endorsed by OCR for GCE Chemistry A
Clearly linked to the specification.
Provides students with lots of opportunities to see problems worked through and the answers are provided step by step.
Hints and tips help students avoid common errors.
Enables students to check their knowledge and understanding. Answers are provided at the back of the book.
iv
v
IntroductionviUNIT 1 Atoms, bonds and groups (F321)
Module 1 Atoms and reactions 21 The changing atom42 Atomic structure63 Atomic masses84 Amount of substance and the mole 105 Types of formula
126 Moles and gas volumes 147 Moles and solutions168 Chemical equations189 Moles and reactions2010 Acids and bases2211 Salts2412 Water of crystallisation2613 Titrations2814 Oxidation number3015 Redox reactions32Summary and practice questions 34End-of-module examination questions 36Module 2 Electrons, bonding and structure
381 Evidence for shells402 Shells and orbitals423 Sub-shells and energy levels 444 Electrons and the Periodic Table 465 An introduction to chemical bonding 486 Ionic bonding
507 Ions and the Periodic Table 528 Covalent bonding549 Further covalent bonding 5610 Shapes of molecules and ions 5811 Electronegativity and polarity 6012 Intermolecular forces
6213 Hydrogen bonding6414 Metallic bonding and structure 6615 Structure of ionic compounds 6816 Structures of covalent compounds 70Summary and practice questions 72End-of-module examination questions 74
Module 3 The Periodic Table 761 The Periodic Table: searching for order 782 The Periodic Table: Mendeleev and beyond 803 The modern Periodic Table 824 Periodicity: ionisation energies and atomicradii845 Periodicity: boiling points 856 Group 2 elements: redox reactions 887 Group 2 compounds: reactions 908 Group 7 elements: redox reactions 929 Group 7 elements: uses and halide tests 94Summary and practice questions 96End-of-module examination questions 98
UNIT 2 Chains, energy and resources (F322)Module 1 Basic concepts and hydrocarbons
1001 Introduction to organic chemistry 1022 Naming hydrocarbons 1043 Naming compounds with functionalgroups1064 Formulae of organic compounds 1085 Structural and skeletal formulae 1106 Skeletal formulae and functional groups 1127 Isomerism
1148 Organic reagents and their reactions 1159 Hydrocarbons from crude oil 11810 Hydrocarbons as fuels 12011 Fossil fuels and fuels of the future 12212 Substitution reactions of alkanes 12413 Alkenes12614 Reactions of alkenes12815 Further addition reactions of alkenes 13016 Alkenes and bromine13217 Industrial importance of alkenes 13418 Polymer chemistry13619 Polymers – dealing with our waste 13820 Other uses of polymer waste 140Summary and practice questions 142End-of-module examination questions 144
Module 2 Alcohols, halogenoalkanes and analysis1461 Making and using alcohol 1482 Properties of alcohols1503 Combustion and oxidation of alcohols 1524 Esterification and dehydration of alcohols 1545 Introduction to halogenoalkanes 1566 Reactions of halogenoalkanes 1587 Halogenoalkanes and the environment 1608 Percentage yield
1629 Atom economy16410 Infrared spectroscopy16611 Infrared spectroscopy: functional groups 16812 Mass spectrometry17013 Mass spectrometry in organic chemistry 17214 Mass spectrometry: fragmentation patterns 17415 Reaction mechanisms 176Summary and practice questions 178End-of-module examination questions 180
Module 3 Energy1821 Enthalpy1842 Exothermic and endothermic reactions 1863 Enthalpy profile diagrams 1884 Standard enthalpy changes 1905 Determination of enthalpy changes 1926 Enthalpy change of combustion 1947 Bond enthalpies
196
ContentsContents
8 Enthalpy changes from Hc 1989 Enthalpy changes from Hf 20010 Rates of reaction – collision theory 20211 Catalysts20412 Economic importance of catalysts 20613 The Boltzmann distribution 20814 Chemical equilibrium
21015 Equilibrium and industry 212Summary and practice questions 214End-of-module examination questions 216Module 4 Resources
2181 The greenhouse effect – global warming 2202 Climate change2223 Solutions to the greenhouse effect 2244 The ozone layer2265 Ozone depletion2286 Controlling air pollution 2307 Green chemistry2328 CO2 – villain and saviour 234Summary and practice questions 236End-of-module examination questions 238
Answers240Glossary258Periodic Table/Data Sheet 262Index264
935 chemistry.prelims.indd 4-5
10/3/08 11:23:32 am
iv
v
Introductionvi
UNIT 1 Atoms, bonds and
groups (F321)Module 1 Atoms and reactions 2
1 The changing atom4
2 Atomic structure6
3 Atomic masses8
4 Amount of substance and the mole 10
5 Types of formula12
6 Moles and gas volumes 14
7 Moles and solutions16
8 Chemical equations18
9 Moles and reactions20
10 Acids and bases22
11 Salts24
12 Water of crystallisation26
13 Titrations28
14 Oxidation number30
15 Redox reactions32
Summary and practice questions 34
End-of-module examination questions 36
Module 2 Electrons, bonding and
structure38
1 Evidence for shells40
2 Shells and orbitals42
3 Sub-shells and energy levels 44
4 Electrons and the Periodic Table 46
5 An introduction to chemical bonding 48
6 Ionic bonding50
7 Ions and the Periodic Table 52
8 Covalent bonding54
9 Further covalent bonding 56
10 Shapes of molecules and ions 58
11 Electronegativity and polarity 60
12 Intermolecular forces62
13 Hydrogen bonding64
14 Metallic bonding and structure 66
15 Structure of ionic compounds 68
16 Structures of covalent compounds 70
Summary and practice questions 72
End-of-module examination questions 74
Module 3 The Periodic Table 76
1 The Periodic Table: searching for order 78
2 The Periodic Table: Mendeleev and beyond 80
3 The modern Periodic Table 82
4 Periodicity: ionisation energies and atomic
radii84
5 Periodicity: boiling points 85
6 Group 2 elements: redox reactions 88
7 Group 2 compounds: reactions 90
8 Group 7 elements: redox reactions 92
9 Group 7 elements: uses and halide tests 94
Summary and practice questions 96
End-of-module examination questions 98
UNIT 2 Chains, energy and
resources (F322)
Module 1 Basic concepts and
hydrocarbons100
1 Introduction to organic chemistry 102
2 Naming hydrocarbons 104
3 Naming compounds with functional
groups106
4 Formulae of organic compounds 108
5 Structural and skeletal formulae 110
6 Skeletal formulae and functional groups 112
7 Isomerism114
8 Organic reagents and their reactions 115
9 Hydrocarbons from crude oil 118
10 Hydrocarbons as fuels120
11 Fossil fuels and fuels of the future 122
12 Substitution reactions of alkanes 124
13 Alkenes126
14 Reactions of alkenes128
15 Further addition reactions of alkenes 130
16 Alkenes and bromine132
17 Industrial importance of alkenes 134
18 Polymer chemistry136
19 Polymers – dealing with our waste 138
20 Other uses of polymer waste 140
Summary and practice questions 142
End-of-module examination questions 144
Module 2 Alcohols, halogenoalkanes
and analysis146
1 Making and using alcohol 148
2 Properties of alcohols150
3 Combustion and oxidation of alcohols 152
4 Esterification and dehydration of alcohols 154
5 Introduction to halogenoalkanes 156
6 Reactions of halogenoalkanes 158
7 Halogenoalkanes and the environment 160
8 Percentage yield162
9 Atom economy164
10 Infrared spectroscopy166
11 Infrared spectroscopy: functional groups 168
12 Mass spectrometry170
13 Mass spectrometry in organic chemistry 172
14 Mass spectrometry: fragmentation patterns 174
15 Reaction mechanisms 176
Summary and practice questions 178
End-of-module examination questions 180
Module 3 Energy182
1 Enthalpy184
2 Exothermic and endothermic reactions 186
3 Enthalpy profile diagrams 188
4 Standard enthalpy changes 190
5 Determination of enthalpy changes 192
6 Enthalpy change of combustion 194
7 Bond enthalpies196
Contents
Contents
8 Enthalpy changes from Hc198
9 Enthalpy changes from Hf200
10 Rates of reaction – collision theory 202
11 Catalysts204
12 Economic importance of catalysts 206
13 The Boltzmann distribution 208
14 Chemical equilibrium210
15 Equilibrium and industry 212
Summary and practice questions 214
End-of-module examination questions 216
Module 4 Resources 218
1 The greenhouse effect – global warming 220
2 Climate change222
3 Solutions to the greenhouse effect 224
4 The ozone layer226
5 Ozone depletion228
6 Controlling air pollution 230
7 Green chemistry232
8 CO2 – villain and saviour 234
Summary and practice questions 236
End-of-module examination questions 238
Answers240
Glossary258
Periodic Table/Data Sheet 262
Index264
935 chemistry.prelims.indd 4-5
10/3/08 11:23:32 am
24
25
H2O
The oxygen atom has four pairs of electrons, so these will take up a basically tetrahedral
shape. But two of the electron pairs are lone pairs, so the shared pairs are repelled more
than in NH3, and the bond angle in H2O is less than in NH3. The final bond angle is
104.5°. The final shape of the molecule is described as non-linear.Shapes of molecules –
electron-pair repulsion theory
1UNIT
Key words
electron-pair
repulsioncovalent bond
lone pair
Electron pairs repel each other so they are as far apart as possible. In a covalent
compound or ion, the number of electron pairs around the central atom determines the
shape of the molecule.
Mod
ule
2
Module 2
Shapes of molecules – electron-pair repulsion theory
Quick check 1✔
WOrked exaMple
To work out the shape of an ion, for example H3O+:
STep 1 There are two ways to approach this
problem.
either draw a dot-and-cross diagram of the
molecule (see diagram opposite);
or count up the number of electrons at the
central atom.
STep 2 See what shape the electron pairs will adopt. They will be arranged
tetrahedrally because there are four pairs of electrons.
STep 3 Now see if there are any lone pairs. Yes, one lone pair.
STep 4 Electron-pair repulsion theory tells you that the lone pair will repel the
bonding pairs a little more strongly than they repel each other, and the
molecule should have the same shape and bond angle as NH3.
H3O+ has a pyramidal shape with a bond angle of 107°.
atomNumber of
electrons
Oxygen 6
Three hydrogens 3
Positive charge –1
Total number 8
× ×××
×
×OH H
H
1 Sketch the shapes and predict the bond angles in the following molecules:
a PCl3
b OCl2
c SeF6 (note that Se is in Group 6) d CCl4
2 Sketch the shapes and predict the bond angles in the following ions:
a PCl4+
b PCl6–
c NH2–
d NH4+
3 The dot-and-cross diagram for SO2 is shown in the diagram opposite.
a What is the shape of the molecule?
b Give the approximate bond angle.
QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by working out how the electron pairs
arrange themselves in space. The rules of electron-pair repulsion theory tell us that:
electron pairs repel each other so they are as far apart as possible
an electron pair shared between two atoms is called a bonding pair (BP)
a non-bonding pair on one atom only (not shared) is called a lone pair (LP)
LP–LP repulsion > LP–BP repulsion > BP–BP repulsion.
These rules explain the shapes of the ammonia, NH3, and water, H2O, molecules.
NH3
The nitrogen atom has four pairs of electrons, so these will take up a basically
tetrahedral shape. But one electron pair is a lone pair, so it repels the shared pairs more
than a bonding pair would, and the bond angle is decreased from 109.5° to 107°. The
final shape of the molecule is described as pyramidal.
Examiner tip
Don’t confuse the
shape of the molecule
with the shape the
electron pairs adopt.
×× ×
××
H
HH
H
HN
N
H
107
Bond angle 107
×× ×
××
×H
H
HH
O
O
104.5
Bond angle 104.5
Number of pairs Shapeexample
2 electron pairs linearCO2
(in fact two double bonds)
3 electron pairs trigonalBF3
4 electron pairs tetrahedralCH4
NH+4
6 electron pairs octahedralSF6
××
××
××××
××××
O OC
O OC
180
Bond angle 180
×× ××
×× ×
××× ×
×××
F
F ×× ××
× ××F
F
F FB
B
120
Bond angle 120
×××
×H
H
H
HH
H
C
H HC
Bond angle 109.5
109.5
FF
F
F
F F
Bond angle 90
What about multiple bonds?
If a molecule has double (or triple) bonds, the bonding electron pairs are all located
between the bonding atoms and therefore they cound as one bonding pair of electrons
around the central atom for working out the shape.
CO2
A dot-and-cross diagram of CO2 shows that the bonding is O=C=O (see table). The
electron pairs in the double bonds will repel each other as much as possible, so the final
shape of the molecule is linear.
What about a non-linear molecule with double bonds? A good example is sulfur dioxide
(see quick check question 3).
Quick check 2✔
Quick check 3✔
11
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Summary of contents
Samples pages from OCR AS Chemistry A Student Book.
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
Exclusively endorsed by OCR for GCE Chemistry A
AS
In Exclusive Partnership
In Exclusive Partnership
Dave Gent and Rob Ritchie
Series Editor: Rob Ritchie
Exclusively endorsed by OCR for GCE Chemistry A
AS
In Exclusive PartnershipDave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
Evaluation PacksEach OCR Chemistry A Evaluation Pack is available on 60 days free evaluation and contains:
l Student Book and Exam Café CD-ROM
l Teacher Support CD-ROM
l Revision Guide (New Edition)
Course componentsOCR AS Chemistry A Student Book and CD-ROM 978 0 435691813 | £17.99* | January 2008
OCR A2 Chemistry A Student Book and CD-ROM 978 0 435691981 | £17.99* | December 2008
OCR AS Chemistry A Teacher Support CD-ROM 978 0 435691837 | £149.00* (+VAT) | April 2008
OCR A2 Chemistry A Teacher Support CD-ROM 978 0 435691936 | £149.00* (+VAT) | December 2008
OCR AS Revise Chemistry A (New Edition) 978 0 435583712 | £6.99* (+VAT) | April 2008
OCR A2 Revise Chemistry A (New Edition) 978 0 435583743 | £6.99* (+VAT) | January 2009
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OCR AS Chemistry A Evaluation Pack 978 0 435692063 | £149.00 | April 2008
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