Essentials Pearson 2016
Essentials Pearson 2016
Essentials Pearson 2016
Essential idea:
Subject vocabulary Physical and chemical properties depend on the ways in which
physical properties a property different atoms combine.
of a substance that can be
measured without it changing into
another substance. Melting points,
boiling points, appearance, and
density are examples of physical
Understanding: Atoms of different elements combine
properties in fixed ratios to form compounds, which have different
chemical properties a property
of a substance that becomes
properties from their component elements.
•
evident when it reacts and
changes into another substance. Elements:
Acidity, reactivity with water
and enthalpies of reactions are • cannot be changed into simpler substances by chemical reactions
examples of chemical properties
element a substance that cannot
• are identified by their name and chemical symbol, e.g.
helium has the symbol He
be broken down into a simpler
substance by chemical means carbon has the symbol C
chemical symbol a one-or
two-letter representation of an
• are organized in the periodic table based on their atomic number, chemical
properties, and the number of valence electrons.
element’s name
periodic table a table that There are over 100 chemical elements, and about 90 of these occur naturally.
organizes the elements by
increasing atomic number and the Hints for success: Section 5 in the IB data booklet alphabetically lists the
number of valence electrons elements and their chemical symbols.
atomic number defined as the
number of protons in the nucleus.
• Atoms:
•
It has the symbol Z
are the simplest unit of an element that can exist on its own
•
valence electrons the electrons
that are in the outermost level of are regarded as the building blocks of matter as all substances are made up
an atom of atoms of one or more element.
atom the smallest unit of an
element that can exist on its own The atoms of one element are all the same but are different to atoms of
compound a pure substance other elements.
that is made up of one or more
• Compounds:
•
elements that are present in a
fixed ratio are substances that contain two or more elements
bond (as a verb) be held together
• are made when the atoms of different elements bond
by strong attractive forces
• have fixed ratios of the component elements
chemical formula a shorthand
representation of a compound.
It uses element symbols and
• are described by chemical formulas that use the symbols of the elements in
the compound, e.g. glucose has the chemical formula C6H12O6, so a unit of
subscripts to show how many glucose contains 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms
•
atoms of each element are in the
compound have different chemical and physical properties from the elements they
are made from, e.g. sodium chloride (NaCl) is made from sodium (Na) and
General vocabulary chlorine (Cl2):
component one part of the 2Na + Cl2 ➝ 2NaCl
whole
•
happens in a chemical reaction
must be balanced, i.e. both sides of the equation must have the same
chemical reaction a process
number of each atom, e.g. in which one set of substances
is turned into another set of
CH4 + 2O2 ➝ CO2 + 2H2O substances
Skill: Application of the state symbols (s), (l), (g), and (aq) in equations.
• States of matter:
Subject vocabulary
aqueous solution a solution
that is formed by dissolving a
substance in water solid liquid gas
sublimation the change of state
• particles close together
• particles more spaced
• particles spread out
• • •
that occurs when a solid changes
directly into a gas particles cannot move particles can move particles move freely
condensing the change of state
• fixed shape
• no fixed shape
• no fixed shape
that occurs when a gas changes
into a liquid
vaporization the change of
• fixed volume
• fixed volume
• no fixed volume
increasing temperature
state that occurs when a liquid
changes into a gas. There are two
types of vaporization: boiling and
evaporation increasing kinetic energy of particles
evaporation the change of state
that occurs when a liquid changes
into a gas at a temperature below
Worked example
the boiling point. Evaporation Lithium (Li) reacts with water (H2O) at room temperature and pressure to give
occurs at the surface of the liquid
lithium hydroxide (LiOH) and hydrogen (H2). Write a balanced equation for this
boiling the change of state that reaction that includes state symbols.
occurs when a liquid changes into
a gas when it has been heated to Solution
the boiling point. Boiling occurs
throughout the liquid
The balanced equation is 2Li + 2H2O ➝ 2LiOH + H2.
General vocabulary At room temperature and pressure lithium is a solid, water is a liquid, and
hydrogen is a gas. Lithium hydroxide is soluble in water so it will dissolve and be
reversibly changes that happen
in one direction can be reversed in an aqueous solution.
and the opposite changes happen
in the other direction Including state symbols the balanced equation is
2Li(s) + 2H2O(l) ➝ 2LiOH(aq) + H2(g).
• Substances can change reversibly between different states of matter if they are
heated or cooled.
sublimation
melting evaporating/boiling
solid liquid gas
freezing condensing
deposition
Figure 1.2 The reversible changes that happen between different states of matter.
• Vaporization:
• is called evaporation when the change happens at the surface of the liquid at
temperatures below the boiling point of the liquid
d
100 °C e
temperature
liquid
b
0 °C c
solid
a
time heating
Worked example
Explain the changes that are occurring in the graph shown in Figure 1.3.
Solution
As the solid water (ice) is heated the temperature increases as the average kinetic
energy of the water particles increases.
When the temperature reaches 0 °C at b in Figure 1.3, the temperature remains
constant even though extra heat is being added. Instead of increasing the kinetic
energy of the water particles the heat is overcoming the attractive forces between
the water molecules in the solid ice.
When all the attractive forces between the solid water molecules have been
overcome the solid water particles are able to move. The solid has melted and
become a liquid.
Further heating of the liquid water (from c) increases the kinetic energy of the
liquid water particles and increases the temperature.
When the temperature reaches 100 °C (at d) it remains constant even though extra
heat is being added. Instead of increasing the kinetic energy of the water particles
the heat is overcoming the attractive forces between the molecules in the liquid.
When all the attractive forces between the liquid water molecules have been
overcome the liquid water particles are able to move independently. The liquid
has boiled and become a gas.
• •
same state and are spread equally
through the mixture are spread evenly are not spread evenly
heterogeneous mixture a through the mixture through the mixture
mixture in which the substances
are present in different states and
are not spread equally through
• cannot be individually identified
Examples:
• can be individually identified
Examples:
• •
the mixture
air (gases are evenly mixed) oil and water (two layers observed)
Essential idea:
The mole makes it possible to correlate the number of particles with
the mass that can be measured.
•
12 g of carbon-12. It has the
is defined as the amount of substance that contains the same number of symbol mol
particles as there are atoms in exactly 12 g of carbon-12 (12C). carbon-12 the most common
isotope of carbon; it has six
protons and six neutrons in the
nucleus
Understanding: Masses of atoms are compared on a scale relative atomic mass the
relative to 12C and are expressed as relative atomic mass (Ar) mass of an atom of an element
relative to the mass of an atom
and relative formula/molecular mass (Mr). of carbon-12. The relative atomic
mass has the symbol Ar and it has
• Ar = 1
mass of carbon-12 atom
12
symbol Mr and it has no units
molecule the smallest unit of a
covalent compound
Hints for success: The relative atomic masses for all the elements are given in covalent compound a
section 6 of the IB data booklet. compound that is formed by
covalent bonding between the
•
atoms of different elements
Relative molecular mass and relative formula mass both have the symbol Mr.
ionic compound a compound
They do not have units as they are ratios. that is formed by ionic bonding
• Relative molecular mass is the relative mass of a molecule (the smallest unit of a
covalent compound):
between the atoms of different
elements
•
It has the symbol M and has the
units g mol–1 has the same value as the relative atomic mass
molar mass the mass of one
mole of a substance. It has the
• has units of g mol–1.
symbol M and has the units For example, the relative atomic mass of chlorine is 35.45 and the atomic mass of
g mol–1 chlorine is 35.45 g mol–1.
Avogadro’s number the number
of particles in one mole of a • Molar mass:
substance, which is 6.02 × 1023.
• has the symbol M
Avogadro’s number is given the
symbol NA or L and has units • is the mass of a substance in grams that contains one mole of the substance
of mol–1
• has the units g mol–1
• is equal to the sum of the atomic masses of the atoms in the substance.
Skill: Calculation of the molar masses of atoms, ions, and molecules, and
of formula mass.
The molar mass of any compound is calculated by adding the atomic masses of all
the elements present in the compound, e.g.
M(CuSO4) = M(Cu) + M(S) + 4 × M(O)
= 63.55 g mol–1 + 32.07 g mol–1 + 4 × 16.00 g mol–1
= 159.62 g mol–1
One mole of CuSO4 has a mass of 159.62 g.
• Avogadro’s number:
• The mass (m) and number of moles (n) of a substance are related by the
molar mass (M).
Figure 1.4 Summary of the vogadro con y by molar mass
by A st a ltipl ,M
central role of the number of de nt mu
moles, n, in converting between iv i ,
d
m
u lt t,
L
i p ly ta n d iv
b y Avogadro co ns id e ,M
b y m o la r m a s s
m
number of particles = nL n=
M
•
of the total mass
gives the exact number of atoms of each element present in a molecule
• is not always the same as the empirical formula for the compound
Propane Glucose
Molecular formula C3H8 C6H12O6
Empirical formula C3H8 CH2O
Multiplier 1 6
(C6H12O6 = 6 × CH2O)
Ca P O
Step 1: Write the mass of 38.76 g 19.97 g 41.27 g
each element in 100 g of the
compound.
Step 2: Divide by the atomic ÷ 40.08 g mol–1 ÷ 30.97 g mol–1 ÷ 16.00 g mol–1
mass to give the number of
= 0.9671 mol = 0.6448 mol = 2.579 mol
moles of each element.
Step 3: Divide by the smallest ÷ 0.6448 mol ÷ 0.6448 mol ÷ 0.6448 mol
value to give the relative
= 1.500 = 1.000 = 4.000
ratios of each element.
Step 4: Convert to the 3 2 8
simplest whole number ratio.
The empirical formula of the compound is Ca3(PO4)2.
Skill: Obtaining and using experimental data for deriving empirical formulas
from reactions involving mass changes.
Worked example
In an experiment a sample of magnesium metal was heated in a crucible using
a Bunsen burner. The product of the reaction was a grey powder known to be
magnesium oxide.
The masses that were recorded in the experiment are given below. Use these
masses to determine the empirical formula of magnesium oxide.
Mass of crucible: 25.623 g Mass of crucible and magnesium: 25.714 g
Mass of crucible and grey powder (magnesium oxide): 25.773 g
Solution
Mg O
Step 1: Write the mass of each element. 0.091 g 0.059 g
Step 2: Divide by the atomic mass to give ÷ 24.31 g mol–1 ÷ 16.00 g mol–1
the number of moles of each element.
= 3.7 × 10–3 mol = 3.7 × 10–3 mol
Step 3: Divide by the smallest value to give ÷ 3.7 × 10–3 mol ÷ 3.7 × 10–3 mol
the relative ratios of each element.
= 1.0 = 1.0
Step 4: Convert to the simplest whole 1 1
number ratio.
The empirical formula of the product is MgO.
Essential idea:
Mole ratios in chemical equations can be used to calculate reacting
ratios by mass and gas volume.
• The stoichiometric coefficients are the molar ratios of products and reactants:
• If 1.5 moles of H2 were reacted with 1.0 moles of Cl2 then it would not be
possible to change all of the reactants into HCl product. There would be too
much H2 (it is in excess) and not enough Cl2 (it is limiting).
•
used up in the reaction
is still present after the reaction is finished.
Worked example
The rocket fuel hydrazine (N2H4) can be formed from ammonia (NH3) and
chlorine (Cl2) as shown in the following balanced equation:
4NH3(g) + Cl2(g) ➝ N2H4(g) + 2NH4Cl(s)
If 12.0 g of NH3 reacts with 25.0 g of Cl2 what is the mass of N2H4 that is produced?
Solution
• is calculated from the amount of the limiting reactant. product that would be formed
if all of the limiting reactant is
• is the mass of product that is actually made in the reaction experimental yield the mass
•
of product that is obtained
is often less than the theoretical yield because of factors such as incomplete when the reaction is carried out
reaction, loss of product, and presence of impurities. experimentally
• The percentage yield of a reaction compares the experimental yield with the
theoretical yield:
percentage yield the
experimental yield expressed as a
percentage of the theoretical yield
experimental yield Avogadro’s law the same
percentage yield = × 100%
theoretical yield volume of any gas at the same
temperature and pressure will
Skill: Solution of problems relating to theoretical, experimental, and contain the same number of
percentage yield. gas particles
Worked example
In the example given above for the reaction of 12.0 g of ammonia (NH3) with
25.0 g chlorine (Cl2) the theoretical yield of hydrazine (N2H4) product was 5.64 g.
The experimental yield was 4.68 g. What was the percentage yield of the reaction?
Solution
experimental yield 4.68 g
percentage yield = × 100% = × 100% = 83.0%
theoretical yield 5.64 g
• states that a fixed volume of all gases at the same temperature and pressure
will contain the same number of particles, e.g.
1.0 dm3 of hydrogen (H2) gas at 298 K and 100 kPa has the same number of
particles as 1.0 dm3 of ammonia gas (NH3) at 298 K and 100 kPa
Worked example
60.0 cm3 of carbon disulfide (CS2) reacts with 210.0 cm3 of oxygen (O2) at a
constant temperature and pressure to form carbon dioxide (CO2) and sulfur
dioxide (SO2) according to the equation below:
CS2(g) + 3O2(g) ➝ CO2(g) + 2SO2(g)
What is the volume of SO2(g) that can be formed in the reaction?
Solution
Skill: Explanation of the deviation of real gases from ideal behaviour at low
temperature and high pressure.
• Gases deviate from ideal behaviour and become real gases at high pressure and
low temperature.
•At high pressure there is a large number of gas particles present and
the volume they occupy is significant compared with the volume of
the container.
This means the gas particles will be closer together and there is a high
probability they will interact (collide) with each other.
•At low temperature the gas particles have low kinetic energies.
This means they will be moving more slowly and will interact more
because they are more strongly affected by the attractive forces that
exist between them.
•
has the symbol Vm
is constant at a specified temperature and pressure
•
gas laws equations that define
is equal to 22.7 dm3 mol–1 at standard temperature and pressure (STP). how the properties of gases such
as temperature, pressure, volume,
and amount are related
Hints for success: The molar volume of an ideal gas at STP is given in section 2 of
the IB data booklet.
• The moles of gas in a given volume can be calculated using the molar volume:
V
n=
Vm
Worked example
In the problem given above the volume of SO2(g) formed in the reaction
was 120.0 cm3. If this reaction was conducted at STP how many moles of Synonym
SO2 were formed?
conducted . . . . . undertook
Solution
120.0 cm3 = (
120.0
1000
)
dm3 = 0.1200 dm3
• Gas laws show the relationships between temperature, pressure, and volume for
a fixed mass (or amount) of an ideal gas.
volume 1
volume
1
Plotting P against gives a straight line that passes through (0, 0)
V
volume
temperature/K
absolute zero
Plotting V against T in kelvin gives a straight line that passes through (0, 0).
P∝T Constant volume
pressure
0
0 temperature/K
Plotting P against T in kelvin gives a straight line that passes through (0, 0).
It has the symbol R and has a Hints for success: The ideal gas equation is given in section 1 of the
value of 8.31 J K–1 mol–1 IB data booklet.
Worked example
An airbag in a car that is fully inflated with nitrogen (N2) has a volume of
100.0 dm3. Determine the number of moles of N2 that are present in a fully
inflated airbag at 20.0 °C and 101.3 kPa.
Solution
Skill: Obtaining and using experimental values to calculate the molar mass
of a gas from the ideal gas equation.
Worked example
A 0.413 g sample of black silver oxide (Ag2O) was heated in a closed boiling tube.
At the end of the reaction 0.385 g of metallic silver (Ag) remained. A colourless gas
was collected that had a volume of 21.3 cm3 at 25 °C and 101.3 kPa.
Determine the molar mass of the gas that was produced in the reaction.
Solution
The volume of the gas must be converted to m3: 21.3 cm3 = 21.3 × 10–6 m3
The temperature of the gas must be converted to K: 25 + 273 = 298 K
mass (gas) = mass of Ag2O – mass of Ag metal = 0.413 g – 0.385 g = 0.028 g
mRT 0.028 g × 8.31 J K–1 mol–1 × 298 K
M= = = 32 g mol–1
PV 101 300 Pa × 21.3 × 10–6 m3
(The molar mass of 32 g mol–1 is consistent with O2 being the colourless gas
produced in the reaction.)
•
solutes) into a solvent
makes the solution less concentrated (i.e. there is less solute relative to the
solute the substance that is
volume of solution). dissolved into a solvent to form a
• The concentration of a solution (C) relates the amount of solute to the amount
of solvent and can be measured in different ways.
solution. The solute is present in a
smaller amount than the solvent
solvent the substance that a
solute dissolves in to form a
Measure of concentration (C) Relevant equation solution. The solvent is present in
a larger amount than the solute
g dm–3 mass of solute in g
C= dilutes increases the volume of
volume of solution in dm3 solution relative to the amount
of solute and decreases the
ppm (parts per million) mass of solute concentration of the solution
C= × 106
mass of solution
Worked example
Determine the molar concentration of a glucose solution (C6H12O6) that
is made by dissolving 3.56 g of glucose in water and making up to a total
volume of 250.00 cm3.
Solution
(i) Calculate the molar mass of glucose and then the moles of glucose
that are dissolved.
M(C6H12O6) = 6 × M(C) + 12 × M(H)+ 6 × M(O)
= (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g mol–1
= 180.18 g mol–1
m 3.56 g
n(C6H12O6) = = = 0.0198 mol
M 180.18 g mol–1
(ii) Convert the volume into dm3.
250.00 cm3 = 0.25000 dm3
(iii) Calculate the molar concentration.
n 0.0198 mol
C= = = 0.0792 mol dm–3
V 0.25000 dm3
Worked example
25.00 cm3 of a 0.750 mol dm–3 glucose solution is diluted to 175.00 cm3. What is
the concentration of the diluted solution in mol dm–3?
Solution
• are made by dissolving a known mass (or moles) of solute in a solvent and
making a precise volume of solution.
that has a known concentration
standardize to determine the
•
concentration of a solution
When a standard solution cannot be directly prepared, or the concentration of a
titration a technique used to
solution is unknown, it is still possible to standardize the solution and determine determine the concentration
its concentration by reacting it with a standard solution. of a solution by reacting it with
another solution of known
Skill: Use of the experimental method of titration to calculate the concentration
concentration of a solution by reference to a standard solution. equivalence point the point in a
•the exact volume of the standard solution that is needed to react with the
solution of unknown concentration.
Worked example
25.00 cm3 of an unknown solution of aqueous oxalic acid (H2C2O4) was titrated
against a standard solution of 0.1000 mol dm–3 aqueous potassium hydroxide
(KOH). The end-point of the reaction was observed after the addition of
19.68 cm3 of KOH(aq).
H2C2O4(aq) + 2KOH(aq) ➝ K2C2O4(aq) + 2H2O(l)
Use this information to determine the concentration of the oxalic acid solution.
Solution
(i) Calculate the moles of the KOH standard used in the titration.
V(KOH) = 19.68 cm3 = 19.68 × 10–3 dm3
n(KOH) = CV = 0.1000 mol dm–3 × 19.68 × 10–3 dm3 = 1.968 × 10–3 mol
(ii) Calculate the number of moles of H2C2O4 using the molar ratio of
H2C2O4:KOH in the balanced equation.
n(H2C2O4) = 1 n(KOH) = 1 × 1.968 × 10–3 mol = 9.840 × 10–4 mol
2 2
(iii) Calculate the concentration of H2C2O4(aq) from the volume that was used
in the titration.
V(H2C2O4) = 25.00 cm3 = 25.00 × 10–3 dm3
n 9.840 × 10–4 mol
C= = = 0.03936 mol dm–3
V 25.00 × 10–3 dm3
Essential idea:
General vocabulary The mass of an atom is concentrated in its minute, positively
minute extremely small charged nucleus.
dense having a relatively large
mass contained in a small volume
negligible so small that its Understanding: Atoms contain a positively charged dense
contribution to the overall total is
minimal nucleus composed of protons and neutrons (nucleons).
concentrated gathered together
in a small region Understanding: Negatively charged electrons occupy the
space outside the nucleus.
Subject vocabulary
proton a positively charged
particle found in the nucleus of
• Protons, neutrons, and electrons are the three sub-atomic particles that
make up an atom.
an atom
neutron a neutral particle found
• Protons and neutrons are found in the dense nucleus of the atom and are
known as nucleons.
•
in the nucleus of an atom
Protons are positively charged and neutrons have no charge (they are neutral) so
electron a negatively charged
particle that occupies the space
the nucleus is positively charged.
outside the nucleus in an atom
sub-atomic particles particles
• Electrons are negatively charged and occupy the space in the atom that is
outside the nucleus.
that are smaller than an atom and
combine to make up an atom • Protons and neutrons have the same relative mass.
nucleus the dense positively
charged core of an atom. Almost
• The mass of an electron is only 0.0005 that of a proton or neutron. This mass is
negligible compared to the mass of protons and neutrons in the nucleus. The
all of the mass of an atom is mass of an atom is concentrated in the nucleus.
contained in the nucleus
nucleons particles found in the Particle Relative mass Relative charge
nucleus of an atom. Protons and
neutrons are both nucleons proton 1 +1
relative mass the mass of an
object or particle expressed as a electron 0.0005 –1
ratio of the mass of another object
or particle. Relative masses have neutron 1 0
no units
atomic number defined as the Skill: Using the nuclear symbol notation AZX to deduce the number of
number of protons in the nucleus. protons, neutrons, and electrons in atoms and ions.
It has the symbol Z
mass number defined as the
mass number
number of protons and neutrons
A
ZX symbol of element
in the nucleus. It has the symbol A
atomic number
• The atomic number of an atom is given the symbol Z and is equal to the
number of protons in the atom.
• The mass number of an atom is given the symbol A and is equal to the total
number of protons and neutrons in the atom.
• An atom contains the same number of protons and electrons. This means
the atomic number is equal to the number of protons and to the number of
electrons in an atom:
number of protons = number of electrons = Z
27
13Al
p = 13
e = 13
• An atom does not have an overall charge because it contains the same number
of positively charged protons and negatively charged electrons.
• The number of protons and neutrons in an ion is the same as the numbers of
protons and neutrons in the atom the ion was made from.
element that have different
numbers of neutrons and different
mass numbers
• The number of electrons can be calculated from the atomic number Z and the
overall charge on the ion.
isotopic composition the
number and abundances of
naturally occurring isotopes for
The atom has lost 3 electrons. an element
27 3+
e = 13 – 3 = 10
13Al
p = 13
n = 27 – 13 = 14
• Atoms of the same element with different numbers of neutrons are called
isotopes. The isotopic composition of an element is the number of isotopes
and the abundance of each. General vocabulary
% abundance
different chemical species present 50
in a sample
Solution 40
mass spectrum a graphical
representation of the results 30
Consider a sample of 100 atoms. 23
obtained by a mass spectrometer. 20
Vertical lines occur at the mass of
each ion present and the height of
total mass of 100 atoms = 10
the line represents the abundance (85 × 77) + (87 × 23) = 8546
0
of that ion 85 86 87 88
relative atomic mass = average mass of atom
mass/charge
General vocabulary total mass 8546
= = = 85.46 Figure 2.1 Mass spectrum for rubidium.
proportional related in size to number of atoms 100
• The atomic mass of a particular isotope can be calculated from the relative
atomic mass of an element if the atomic masses of all other isotopes are known.
Worked example
Boron exists in two isotopic forms, 10B and 11B. 10B is used as a control for nuclear
reactors. Use your periodic table to find the abundances of the two isotopes.
Solution
Essential idea:
The electron configuration of an atom can be deduced from its
atomic number.
• emit photons (light energy) when they move to lower energy levels that are
closer to the nucleus.
are in high energy levels far from
the nucleus
•
photons particles of light
Emission spectra: (electromagnetic radiation)
• are produced when excited electrons move to lower energy levels. wavelengths (or frequencies)
of light emitted by atoms or
compounds that contain excited
Skill: Description of the relationship between colour, wavelength, electrons
frequency, and energy across the electromagnetic spectrum. electromagnetic spectrum the
• is divided into different regions: infrared, visible, ultraviolet, etc. electromagnetic radiation a
form of energy that consists of
• Electromagnetic radiation: perpendicular oscillating electric
• behaves as a wave
and magnetic fields that travel
as waves
Synonyms
emit . . . . . . . . . . . . . give out
Figure 2.2 Snapshot of a wave at a given instant. The distance between successive crests or peaks is called the
wavelength (λ).
• Wavelength, λ:
• has units of metres, m (or other units of distance such as nm, cm).
• Frequency, ν:
• The colour of visible light depends on its wavelength (and frequency), e.g.
Synonym • Because specific wavelengths of light are emitted, this is evidence that electrons
can only exist in discrete energy levels. (If the energy levels were not discrete
specific ......... clearly defined
and all energy levels were possible, then a continuous emission spectrum
would be observed.)
General vocabulary
evidence proof • The lines observed in
the emission spectrum
8 ∞
7
6
of hydrogen converge 5 4 excited
Subject vocabulary (become closer) in energy 3
IR states
discrete energy levels energy as the energy increases
levels that are clearly separated (see Figure 2.4). This can 2
in energy and do not overlap visible
are discrete only happen if the energy
levels converge at higher
converge become closer E
energy and the energy
energy level diagram a diagram difference between them
that shows the energy levels
available to an electron in an atom becomes smaller.
•
integer numbers whole
n = 1 is the lowest energy level. It is also called the ground state. numbers
• n > 1 are higher energy levels. They are called excited states.
• The emission spectrum of the hydrogen atom: ionization the process in which
• is generated when excited electrons move from high energy levels to lower
energy levels
an atom or molecule loses an
electron and becomes a positive
ion
• depends on the energy level that the emissions occur to. that exist within a main energy
level
d
4 p
s d
Understanding: A more detailed model of the atom p
E 3
describes the division of the main energy level into s, p, d, s
• the third energy level (n = 3) contains s, p, and d sub-levels Figure 2.5 The relative energies of
•
finding an electron
can contain one electron or two electrons.
Schrödinger wave equation an
equation that describes the
behaviour of electrons as three- Skill: Recognition of the shape of an s atomic orbital, and the px, py, and pz
dimensional waves moving atomic orbitals.
•
around the nucleus of an atom
The different sub-levels contain different types and numbers of orbitals.
• The s orbital is shaped like a sphere. The higher the energy level the larger the
diameter of the s orbital.
x x x
y y y
z z z
1s 2s 3s
Figure 2.6 The relative sizes of the 1s, 2s, and 3s orbitals.
General vocabulary
• In each p sub-level there are three p orbitals that are shaped like dumb-bells.
All three have the same shape but they have different orientations and are at
90° to each other.
orientations directions in space
that objects are facing or aligned x x x
on y y y
z z z
px pz py
Figure 2.7 The three p orbitals that are in a p sub-level: px, pz, and py.
• The Aufbau principle states that electrons will occupy the lowest energy orbital
available first before filling successively higher energy orbitals.
Subject vocabulary
Aufbau principle ‘Aufbau’ means
‘building up’ in German. Electrons
4p will occupy the lowest energy
3d level available. Only after the
lowest energy level is filled will
4s the next highest energy level be
occupied
3p
electron spin a quantum
3s mechanical property of electrons
Pauli exclusion principle two
energy
• Electrons have a property called spin. There are two possible spin states
an electron can have: ‘spin up’ or ‘spin down’. Arrows are used to represent
electrons with different spins.
means “spin down” means “spin up”
• The Pauli exclusion principle states that if two electrons occupy the same
orbital they must have different spins.
• Hund’s rule states that when more than one orbital is available in a sub-level,
electrons will occupy different orbitals and they will have the same spin.
Px Py Pz Px Py Pz Px Py Pz
•
occupied in an atom
shows the number of electrons in each sub-level using superscripts,
condensed electron
configurations electron e.g. the electron configuration of nitrogen is 1s2 2s2 2p3
configurations that use noble gas which means that nitrogen has 2 electrons in the 1s sub-level, 2 electrons in
configurations as a core the 2s sub-level, and 3 electrons in the 2p sub-level.
1s
• For elements containing 4s and 3d electrons the order can be switched when
writing the electron configuration. The electron configuration of vanadium can
be written as [Ar] 3d3 4s2 or [Ar] 4s2 3d3.
• Two important exceptions to the Aufbau principle are the electron configurations
of chromium and copper.
Cr Cu
Aufbau prediction [Ar] 3d4 4s2 [Ar] 3d9 4s2
Actual configuration [Ar] 3d5 4s1 [Ar] 3d10 4s1
Essential idea:
The arrangement of elements in the periodic table helps to predict
their electron configuration.
• The periodic table arranges the elements by increasing atomic number. It can be
broken into ‘blocks’ of elements based on which sub-levels are occupied by the
s1 electrons
outer s2 d1in thed2atoms
d3of that
d5s1element.
d5 d6 d7 d8 d10s1 d10 p1 p2 p3 p4 p5 p6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
H He
1 hydrogen helium
1 2
Li Be B C N O F Ne
2 lithium beryllium boron carbon nitrogen oxygen fluorine neon
3 4 5 6 7 8 9 10
Na Mg AI Si P S CI Ar
3 sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
11 12 13 14 15 16 17 18
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
4 potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
5 rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Cs Ba Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn
6 caesium barium
57–71
see below
hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
55 56 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103
• f block (green): contains the elements whose outer electrons occupy f orbitals.
The elements in the f block are known as lanthanoids (first row of the f block) and
actinoids elements in the second
row of the f block of the periodic
table
•
to the row in the periodic table
where the element appears contain elements that have the same number of valence electrons and have
groups vertical columns of similar chemical reactivities
elements in the periodic table
• contain elements whose outer electrons occupy the same type of sub-level
reactive can easily undergo
chemical reactions • are numbered from group 1 on the left of the periodic table through to
group 18 on the right.
•
alkali metals the elements that
are in group 1 of the periodic The elements in group 1 are very reactive and are known as the alkali metals.
table Alkali metals have one valence electron that is in the s sub-level, ns1.
halogens the elements that are
in group 17 of the periodic table • The elements in group 17 are very reactive and are known as the halogens.
Halogens have seven valence electrons and the five outermost electrons are in
unreactive does not easily the p sub-level, np5.
•
undergo chemical reaction
The elements in group 18 are unreactive and are known as the noble gases.
noble gases the elements that
are in group 18 of the periodic Noble gases have complete valence shells and the six outermost electrons are in
table the p sub-level, np6.
Worked example
Use the position of the element silicon (Si) in the periodic table to determine:
(i) the number of valence electrons in a silicon atom, (ii) the outer energy level of a
silicon atom, and (iii) the electron configuration of silicon.
Solution
30 Topic 3: Periodicity
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
H He
1 hydrogen helium
1 Metals 2
Li Be Non-metals B C N O F Ne
2 lithium beryllium boron carbon nitrogen oxygen fluorine neon
3 4 5 6 7 8 9 10
Metalloids
Na Mg AI Si P S CI Ar
3 sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
11 12 13 14 15 16 17 18
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
4 potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
5 rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Cs Ba Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn
6 caesium barium
57–71
see below
hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
55 56 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103
Figure 3.2 The periodic table split into metals, non-metals, and metalloids.
• The periodic table can also be broken into families of elements based on their
general properties and reactivity.
• Metals (blue) are elements that lose electrons and form positive ions. Subject vocabulary
• Non-metals (yellow) are elements that gain electrons and form
negative ions.
metals elements that tend to lose
electrons and form positive ions
Essential idea:
Elements show trends in their physical and chemical properties across
periods and down groups.
•
periodicity patterns in physical
and chemical properties observed increase or decrease along a row (period)
in the periodic table
• increase or decrease down a column (group)
effective nuclear charge the
overall attraction that the
electrons have to the nucleus after
• can be explained by how the effective nuclear charge changes for elements
in the period or group.
•
the effect of the nuclear charge
is reduced by the repulsions by Nuclear charge:
other electrons
nuclear charge the total charge
• is a measure of the attraction that an electron has to the nucleus of an atom
when it is not affected by other electrons
•
exerted by the nucleus of an atom.
It depends on the number of depends on the number of protons in the nucleus.
•
protons in the nucleus
Effective nuclear charge:
•
shielding the blocking effect
that electrons in inner energy is a measure of the overall attraction an electron has to the nucleus
levels have on valence electrons. of an atom when it also experiences repulsions and shielding due to
Shielding prevents the valence
electrons from experiencing the
other electrons
full nuclear charge
atomic radius half the distance
• is less than the nuclear charge because repulsions and shielding effects
decrease the overall attraction the electron has to the nucleus
between the nuclei of two bonded
atoms of an element • decreases down a group because more inner energy levels are occupied and
this increases the shielding of the outer electrons
• increases across a period, because the extra attraction to the nucleus that
results when protons are added is greater than the extra repulsion that
occurs when electrons are added to the same energy level.
• Atomic radius:
• is the distance from the centre of the nucleus to the outermost electrons of
an atom
• is defined as half the distance between two nuclei of the element that are
bonded together in a compound
• is determined experimentally
2r
32 Topic 3: Periodicity
• increases down a group because the outer electrons are in energy levels that
are further out from the nucleus.
Element Period No. of occupied Atomic radius/10–12 m
principal energy levels
Li 2 2 152
Na 3 3 186
K 4 4 231
Rb 5 5 244
Cs 6 6 262
Element Na Mg Al Si P S Cl Ar
Atomic radius/10–12 m 186 160 143 117 110 104 99 –
Hints for success: Atomic radius values are given in section 9 of the
IB data booklet.
H He
53 31
Increasing atomic radius
Metals B C N O F Ne
Li Be
Metalloids
167 112 87 67 56 48 42 38
Nonmetals
Na Mg Al Si P S Cl Ar
186 160 143 117 110 104 99 71
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
243 194 184 176 171 166 161 156 152 149 145 142 136 125 114 103 94 88
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
265 219 212 206 198 190 183 178 173 169 165 155 156 145 133 123 115 108
F F–
60 pm 133 pm
Li Li+
130 pm 76 pm
N3– O2– F–
146 pm 140 pm 133 pm
34 Topic 3: Periodicity
Atoms
Ions Metalloids
Na+ Mg2+ Al3+ Si4+ P3– S2– Cl–
Nonmetals
3 Ar
102 72 53.5 40 212 184 181
K+ Ca2+ Sc3+ Ti3+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+Ga3+Ge4+ As3+ Se2– Br–
4 Kr
138 100 74.5 67 79 80 83 78 74.5 69 73 74 62 53 58 198 196
Rb+ Sr2+ Y3+ Zr4+ Nb3+Mo4+ Tc4+ Ru3+ Rh3+ Pd2+ Ag+ Cd2+ In3+ Sn4+ Sb3+ Te2– I–
5 Xe
152 118 90 72 72 65 64.5 68 66.5 86 115 95 80 69 76 221 220
Figure 3.5 A visual comparison of the atomic and ionic radii of the elements.
• decreases down a period because the electrons are in energy levels further
from the nucleus and take less energy to remove
• increases across a group because the effective nuclear charge increases and
this means more energy is needed to remove electrons.
Hints for success: First ionization energy values are given in section 8 of the
IB data booklet.
2000
F
Ar
1500 N
Cl
H C O P
1000 Be
Mg S
B Si Ca
500 Al
Li Na
K
0
element in order of atomic number
Hints for success: Electron affinity values are given in section 8 of the
IB data booklet.
•
Subject vocabulary
Electron affinity:
•
electron affinity the energy
change that occurs when one is the energy change that occurs when one mole of electrons is added to one
mole of electrons is added to one mole of neutral atoms in the gaseous state:
mole of atoms in the gaseous state
X(g) + e– ➝ X–(g)
• is negative for most atoms, which means that energy is released when the
electron is added to the atom
• cannot be measured for group 18 elements as they do not form negative ions
36 Topic 3: Periodicity
•
that occurs when two atoms share
cannot be measured for group 18 elements as they do not form pairs of electrons
covalent bonds
• decreases down a group because the shared pair of electrons are in energy
levels that are further out from the nucleus.
This means that the shared electrons experience more shielding and will be
less attracted to the nucleus of the atom.
Figure 3.7 3D representation of the electronegativity values in section 8 of the IB data booklet.
• have to be stored under oil or in sealed tubes to prevent reaction with oxygen
and water in the air
• have very low ionization energies, which means they easily react to form 1+ ions
• react with water to give hydroxide salts (MOH) and hydrogen gas (H2):
General reaction: 2M(s) + 2H2O(l) ➝ 2MOH(aq) + H2(g) (M = any alkali metal)
Specific example: 2Na(s) + 2H2O(l) ➝ 2NaOH(aq) + H2(g)
• fluorine (F2) and chlorine (Cl2) are yellow-green gases, bromine (Br2) is a brown
liquid, and iodine (I2) is a purple solid
• are reactive
• have high electron affinities and easily gain electrons to form 1– ions
• react with alkali metals (M) to give halide salts (see above)
Subject vocabulary • undergo displacement reactions: a more electronegative halogen will displace
the halide ion of a less electronegative halogen.
displacement reaction a
reaction in which a more reactive General reaction: X2(aq) + 2Y–(s) + ➝ 2X–(aq) + Y2(aq)
element replaces a less reactive
element in a compound (X = more electronegative halogen, Y = less electronegative halogen)
Specific examples:
Cl2(aq) + 2I–(s) + ➝ 2Cl–(aq) + I2(aq)
Br2(aq) + 2Cl–(s) + ➝ no reaction as Cl is more electronegative than Br.
• Metals (blue):
• are found on the left and at the bottom of the periodic table
• have low electronegativities, and their valence electrons are not held to a
single atom so can move from atom to atom throughout the metal solid,
which means that metals have a high conductivity.
• Non-metals (yellow):
• are elements with high ionization energies, high electronegativities, and high
electron affinities
38 Topic 3: Periodicity
• are found between the metals and non-metals in the periodic table
• are usually more similar to metals in their physical properties and more
similar to non-metals in their chemical reactivity.
Worked example
On the basis of their positions in the periodic table predict and explain whether
barium and sulfur will behave as metals or non-metals.
Solution
• are compounds that contain metals and oxygen, e.g. magnesium oxide
(MgO) and sodium oxide (Na2O)
only oxygen and one other
element
• Non-metallic oxides:
• are compounds that contain non-metals and oxygen, e.g. carbon dioxide
(CO2) and sulfur trioxide (SO3)
• The sulfur oxides are sulfur dioxide (SO2) and sulfur trioxide (SO3). They are
both covalent oxides and dissolve in water to give acidic solutions. The pH
is less than 7.
SO2(g) + H2O(l) ➝ H2SO3(aq) (sulfur(IV) acid or sulfurous acid)
SO3(g) + H2O(l) ➝ H2SO4(aq) (sulfur(VI) acid or sulfuric acid)
40 Topic 3: Periodicity
Essential idea:
Ionic compounds consist of ions held together in lattice structures by
ionic bonds.
•
compound
In an ionic solid the positive ions are surrounded by negative ions and the
negative ions are surrounded by positive ions. This results in a three-dimensional
structure known as an ionic lattice.
Skill: Deduction of the formula and name of an ionic compound from its
component ions, including polyatomic ions.
Cl– ion
• When writing the formula of an ionic compound the cation is always put
before the anion.
•
Na+ ion
When an ionic compound is formed the total positive charge of the cations must
Figure 4.1 The NaCl lattice is equal the total negative charge of the anions.
•
built up from oppositely charged
sodium and chloride ions. When naming an ionic compound:
• the name of the cation is always put before the before the name of the anion
• the cation keeps the name of the metal it is made from, e.g. Li+ is the
lithium ion
• the anion starts with the name of the non-metal but ends with ‘-ide’, e.g. S2–
is the sulfide ion, H– is the hydride ion.
Subject vocabulary • Polyatomic ions are ions that contain two or more atoms. The names of
common polyatomic ions, along with their charges and chemical formulas, are
polyatomic ions ions that
contain more than one atom provided in table below.
Hints for success: It is necessary to memorize the names, formulas, and charges
of these polyatomic ions as they are not provided in the IB data booklet.
• Volatility refers to how easily a substance can be converted into a gas. Subject vocabulary
• Substances with a high volatility have low boiling points and are easily
converted into gases.
volatility a measure of how
easily a substance can be
•
converted to a gas
Ionic compounds:
•
ionic compound a chemical
contain strong ionic bonds that require a large amount of energy to break compound that is formed by
•
dipole two regions of opposite
Substances with a high solubility easily dissolve in a solvent and form charges (or partial charges)
concentrated solutions separated by a distance
•
separated by a distance
contain ions that attract the dipoles in polar solvents
•
hydrated surrounded by water
are most soluble in highly polar solvents such as water. molecules
• Most ionic compounds are soluble in water. The polar water molecules are
attracted to the charged ions in the ionic compound. When enough water
molecules surround the ions at the outside of the ionic lattice the ions
become hydrated and break away from the solid and it dissolves.
H H
O O
H H
Na+
H
O O H
H H
H O
H H
Na+ Cl– Na+ O
H
H H
O
Figure 4.2 Dissolving NaCl in water involves the attraction of the polar water molecules to the oppositely
charged ions in the NaCl lattice and the hydration of the separated ions.
Subject vocabulary
electrical conductivity the
• Electrical conductivity refers to the ability of a substance to transport charge.
Substances have a high electrical conductivity if they contain charged particles
ability of a substance to transport (electrons or ions) that are able to move.
charge
• Ionic compounds:
molten liquids liquids made by
heating substances above their
melting point
• have a low electrical conductivity in the solid state because the ions are in
fixed positions and cannot move
Synonym
• have a high electrical conductivity when they are molten liquids because the
ions are able to move
transport ....... movement
• dissolve in water to form solutions that have a high electrical conductivity
because the ions are able to move.
• A covalent bond:
to the nuclei of both atoms.
• forms when two non-metal atoms share a pair of electrons Subject vocabulary
• is the strong electrostatic attraction between the shared pair of electrons and
the positively charged nuclei of both atoms.
covalent bond a chemical bond
that is formed by the electrostatic
attraction between a shared pair
of electrons and the nuclei of two
atoms
Understanding: Single, double, and triple covalent bonds double bond a covalent bond
that is formed by the sharing of
involve one, two, and three shared pairs of electrons, two pairs of electrons
respectively. triple bond a covalent bond that
is formed by the sharing of three
Understanding: Bond length decreases and bond strength pairs of electrons
bond length the distance
increases as the number of shared electrons increases. between the centres of the two
nuclei in a covalent bond
• Single covalent bonds:
• are represented by a single line between the two atoms, e.g. C–C represents a
single bond between two carbon atoms.
• are represented by a double line between the two atoms, e.g. C=C represents
a double bond between two carbon atoms.
• are represented by a triple line between the two atoms, e.g. C≡C represents a
triple bond between two carbon atoms.
• Bond length:
• is longer when bonds are formed from bigger atoms, e.g. Cl2 contains smaller
atoms than Br2 and has a shorter bond length.
• Bond strength:
particular bond in the gaseous Hints for success: Bond enthalpies are given in section 11 of the IB data booklet.
state
non-polar covalent bond a
covalent bond in which the
With more shared electrons there is a greater attraction between the shared
electron pair is equally shared electrons and the nuclei. This results in more energy being needed to break the
between the two nuclei bond and separate the atoms.
polar covalent bond a covalent
bond in which the electron pair is
single bonds < double bonds < triple bonds
not equally shared between the
two nuclei weakest strongest
e.g. C–C > C=C > C≡C
346 kJ mol–1 614 kJ mol–1 839 kJ mol–1
• are bonds in which the electrons are shared equally between the two atoms
• are formed when the two bonded atoms have the same electronegativity;
e.g. H–H, O=O, and N≡N are all non-polar covalent bonds.
• are bonds in which the electrons are not shared equally between the
two atoms
• are formed between atoms with different electronegativities; e.g. H–Cl, C=O,
and C≡N are all polar covalent bonds
• have a partial negative charge (δ–) on the atom with the higher
electronegativity because it exerts a greater attraction on the
shared electrons
• A dipole:
• using vectors (the vector points towards the more electronegative atom)
H Cl
H Cl
Worked example
Rank the polarities of the H–Cl, H–F, H–N, and H–O bonds in order of least polar
to most polar.
Solution
The polarities for each bond can be determined from the electronegativity values
of the atoms in the bond. The most polar bond has the biggest electronegativity
difference between the atoms.
H–Cl: electronegativity difference = 3.2 – 2.2 = 1.0
H–F: electronegativity difference = 4.0 – 2.2 = 1.8
H–N: electronegativity difference = 3.0 – 2.2 = 0.8
H–O: electronegativity difference = 3.4 – 2.2 = 1.2
The ranking of the bonds from least polar to most polar is
H–N < H–Cl < H–O < H–F
Essential idea:
Lewis (electron dot) structures show the electron domains in the
valence shell and are used to predict molecular shape.
Subject vocabulary
• For many atoms a stable electron configuration is achieved when they have eight
electrons in the valence shell, and this is the basis of the octet rule.
octet rule atoms in covalent
compounds will have eight
• The octet rule does not apply to hydrogen and helium atoms as they can only
have a maximum of two electrons in their valence shell.
electrons in their valence shell
Lewis structure a representation
of a molecule that shows the
bonds and lone pairs in the Understanding: Lewis (electron dot) structures show all the
valence shell of the atoms in the
molecule. It uses lines or pairs valence electrons in a covalently bonded species.
•
of crosses (or dots) to represent
electron pairs A Lewis structure is a simple method for showing how valence electrons are
distributed in a molecule. Electrons are arranged in pairs represented by dots or
crosses or as a single line. The electron pairs can be bonding (shared between
two atoms) or non-bonding (belonging to one atom).
• Lewis structures of polyatomic ions must be drawn inside square brackets with
the overall charge on the ion shown at the top right (see example for CN– below).
Steps to follow in drawing a
Lewis structure.
1 Calculate the total number of 2 Draw the skeletal structure
valence electrons in the molecule by of the molecule to show
multiplying the number of valence how the atoms are
electrons of each element by the linked to each other.
number of atoms of the element in
the formula and totalling these.
3 Use a pair of crosses, a
pair of dots, or a single
4 Add more electron pairs to complete line to show one electron
the octets (8 electrons) around the pair and put a pair in each
atoms (other than hydrogen, which bond between atoms.
must have 2 electrons).
Worked example
Deduce the Lewis (electron dot) structures for PCl3 and CN–.
Solution
Cl
Cl
• can occur with stable compounds that have boron (B) or beryllium (Be) as
the central atom, e.g.
has fewer than eight electrons
in its valence shell has an
incomplete octet
BH3 total valence electrons = 3 + 3 × 1 = 6 so the Lewis structure is: resonance structures the
H B H possible Lewis structures that can
be drawn for some compounds
containing a double bond
The central boron atom only has six electrons so it has an incomplete octet.
• have the double bond (or double bonds) in different positions, e.g. ozone
(O3) has two possible Lewis structures. Each Lewis structure contains one
single and one double bond but they are in different positions.
O O O O O O
Worked example
Draw the resonance structures for the carbonate ion CO32–.
Solution
O O O
•
central atom
The electron domain geometry:
4 109.5 tetrahedral
109.5°
A
–
O N
• Bond angles:
• depend on the electron domain geometry and the repulsions exerted by the
different electron domains that are on the central atom.
• Multiple bonds and non-bonded pairs (lone pairs) exert a greater repulsion than
single bonds and this affects the bond angles.
• Non-bonded pairs (lone pairs) will repel more because they are closer to the
atom than the bonded pair in single bonds
• Multiple bonds (double and triple bonds) will repel more because they
contain more electrons than a single bond.
The shapes that are formed from different electron geometries are given in
the table below.
‘A’ represents the central atom and ‘B’ represents the atoms that are bonded to
the central atom.
2 V-shaped B A ≈117°
(bent)
B
A
B
B
B
3 trigonal A ≈107°
pyramidal B
B
B
2 V-shaped A ≈105°
(bent) B B
Skill: The use of VSEPR theory to predict the electron domain geometry
and the molecular geometry for species with two, three, and four
electron domains.
Skill: Prediction of bond angles from molecular geometry and the presence
of non-bonding pairs of electrons.
The steps used to determine the shape (molecular geometry) of a molecule or
polyatomic ion and the relevant bond angles in the molecule or polyatomic ion are:
Step 1: Draw the Lewis structure following the steps on page 48.
Step 2: Count the total number of electron domains on the central ion.
Step 3: Determine the electron domain geometry as follows:
Hints for success: Lewis structures (Step 1) must show all valence electrons. Drawings of the
shape (Step 4) do not require all valence electrons to be shown but the non-bonding pairs
on the central atom should be included. (Drawing all of the electron pairs on the central
atom shows the electron domain geometry used to determine the molecular geometry.)
Step 2: Number of 2 3 3 4 4 4
electron domains
Step 3: Electron linear trigonal planar trigonal tetrahedral tetrahedral tetrahedral
domain geometry planar
Step 4: Number of 2 3 2 4 3 2
bonded electron
domains
Step 5: Molecular linear trigonal planar V-shaped tetrahedral trigonal V-shaped
geometry (shape) (bent) pyramidal (bent)
O C O O – H
O O N O
O N C H
O H H H H
H H
O H
• Polar molecules contain an overall dipole and must contain polar bonds. Subject vocabulary
• Non-polar molecules do not contain an overall dipole. This happens when: polar molecules molecules that
• the molecule has polar bonds but the dipoles cancel. overall dipole the dipole on
a molecule that is made by the
• If there are no polar bonds then the molecule is non-polar. non-polar molecules molecules
• If the dipoles do not cancel there is an overall dipole and the molecule will
be polar.
• the molecule has a regular geometry (linear, trigonal planar, or tetrahedral) and
NH3 H2O
• the molecule has a regular geometry but the bonds have different polarities.
Figure 4.4 Equal and opposite pulls cancel each Figure 4.5 When the pulls are not equal and opposite
other out. there is a net pull.
Subject vocabulary
network covalent structures
covalent compounds in which
all of the atoms are linked by Understanding: Carbon and silicon form giant covalent/
covalent bonds
network covalent structures.
• Network covalent structures:
•
silicon atoms
has each silicon atom covalently bonded to four oxygen atoms
•
oxygen atoms
has each oxygen atom bonded to two silicon atoms
Figure 4.6 The structure of
quartz SiO2.
•has an Si:O ratio of 1:2 so SiO2 is the empirical formula.
• Allotropes are different forms of an element that can exist in the same state, e.g.
oxygen (O2) and ozone (O3) are allotropes as they are both forms of oxygen that
Subject vocabulary
allotropes different forms of
exist as gases. an element that can exist in the
• Carbon has four allotropes that can exist as solids: graphite, diamond, graphene,
and fullerene (C60). Graphite, diamond, and graphene are network solids as all
same state
the carbon atoms in these substances are covalently bonded to two or more
other carbon atoms.
• The structural features and physical properties of the different giant covalent
compounds formed by carbon are outlined in the table below.
Structural features
Each C atom is bonded to three Each C atom bonded to four other Each C atom bonded to three other
other atoms atoms atoms
120° bond angles 109.5° bond angles 120° bond angles
Forms sheets of repeating hexagons Forms repeating tetrahedral units Forms a single sheet of repeating
hexagons
The sheets are held together by Has no delocalized electrons
London dispersion forces because all of the valence electrons Has delocalized valence electrons on
are in covalent bonds the sheet
Has delocalized valence electrons on
each sheet
Physical properties
Is a good electrical conductor Is a poor electrical conductor Is a good electrical conductor
because the delocalized electrons because there are no delocalized because the delocalized electrons
can move electrons that can move can move
Is a good lubricant because the Is very hard as all the atoms are held Is very flexible because the sheet is
sheets can slide over each other in fixed positions only one atom thick
Has a very high melting point Has a very high melting point Is very strong because all of the
because C–C bonds must be broken because C–C bonds must be broken atoms are bonded together
before it melts before it melts
Has a very high melting point
because C–C bonds must be broken
before it melts
Essential idea:
The physical properties of molecular substances result from different
types of forces between their molecules.
•
electrostatic attraction between
permanent dipoles on two London (dispersion) forces:
molecules
• can be called London forces or dispersion forces
hydrogen bonding a strong
intermolecular force that occurs • occur between all molecules
between molecules containing
• are the only intermolecular forces that occur between non-polar molecules
•
a hydrogen atom bonded to a
highly electronegative atom are the electrostatic attraction between an instantaneous dipole and an
instantaneous dipole a dipole
induced dipole
that only exists for a brief time on
a molecule • are stronger between molecules that have a large number of electrons
because there is a higher probability that an instantaneous dipole will form.
induced dipole a dipole that This attraction is a
forms on a molecule when its
London (dispersion) force.
electrons are attracted or repelled
by a dipole on another molecule δ+ δ– δ+ δ–
Cl Cl Cl Cl Cl Cl
London forces are weak forces so most molecules that have only London
forces have low melting and boiling points.
Element Mr Boiling State at room
point / °C temperature
F2 38 –188 gas
boiling point
Cl2 71 –34 gas increases with
increasing number
Br2 160 59 liquid
of electrons
I2 254 185 solid
• exist between polar molecules that each contain a permanent dipole Subject vocabulary
• are the electrostatic attraction
between the permanent dipole on one δ+ δ– δ+ δ–
permanent dipole a dipole
that is always present on a polar
molecule
molecule and the permanent dipole
on another molecule H Cl H Cl
This attraction
Figure 4.8 Dipole–dipole forces in HCl are the attraction
between the permanent dipoles of two molecules. is a dipole–dipole force.
• increase in strength with the size of the overall dipole in the molecule
• are stronger than London forces for molecules of similar size. Compounds
with dipole–dipole forces will have higher melting points and higher
boiling points.
C3H8 CH3OCH3
Molar mass (g mol–1) 44 46
Structure H H H H H
δ+ δ– δ+
H C C C H H C O C H
H H H H H
• Hydrogen bonding:
• there is very little electron density around the nucleus of the hydrogen atom
• the hydrogen nucleus will strongly attract lone pairs of electrons on the
oxygen atoms of neighbouring water molecules.
Hints for success: Do not confuse hydrogen bonds with covalent bonds.
Hydrogen bonds are very strong intermolecular forces. Hydrogen bonds are not
covalent bonds.
Worked example
Determine the strongest intermolecular forces present in the following compounds:
CH3CH2CH3, CH3CH2OH, and CH3CHO.
H H H
• Compounds that can form hydrogen bonds have higher boiling points and are
less volatile than molecules with London (dispersion) forces and dipole–dipole
forces. This is because hydrogen bonding is stronger than London (dispersion)
forces and dipole–dipole forces.
Solubility
• Water is a polar solvent that can form hydrogen bonds. The solutes that have the
highest solubility in water will be polar molecules that can also form hydrogen
bonds, e.g. ethanol (CH3CH2OH).
• Hexane is a non-polar solvent. The solutes that have the highest solubility in
hexane will be non-polar molecules; e.g. Cl2, Br2, and I2 are all non-polar solutes
and dissolve in non-polar solvents such as hexane.
Essential idea:
Metallic bonds involve a lattice of cations with delocalized electrons.
•
Figure 4.11 The electron-sea model
The valence electrons are then present as of metallic bonding.
an electron sea that is delocalized over the
lattice of metal cations.
• the delocalized electrons can still move freely through the new lattice shape
• affects the melting point of the metal: the stronger the metallic bond the
higher the melting point of the metal.
Worked example
Sodium and magnesium are two metals in the third period. Explain why
the melting point of magnesium (650 °C) is higher than the melting point of
sodium (98 °C).
Solution
Magnesium has a higher melting point than sodium so magnesium must have
stronger metallic bonding than sodium.
Magnesium forms Mg2+ ions and has twice as many delocalized electrons per
cation than sodium, which forms Na+ ions.
The attraction between the Mg2+ ions and the larger sea of delocalized electrons
in magnesium will be greater than the attraction between the Na+ ions and the
electron sea in sodium.
Worked example
Explain why sodium (98 °C) has a higher melting point than potassium (63 °C) and
rubidium (39 °C).
Solution
Sodium, potassium, and rubidium are all group 1 metals. The melting points
decrease going down the group so the strength of metallic bonding must also
decrease going down the group.
This is because the radius of the cations increases going down the group and the
charge density of the cations decreases. (The same 1+ charge is spread over a
Subject vocabulary
larger volume with larger cations.)
charge density a measure of
The electron sea is more strongly attracted to cations with a larger charge density the amount of charge that is
so the metallic bond strength and melting points decrease down the group. contained within a volume
•
Usually an alloy is a mixture of
two or more metals. Some alloys have enhanced properties compared with the metals they are made from,
involve a metal mixed with a small e.g. alloys are often stronger than the metals they are made from.
amount of a non-metal
corrosion a natural process in
which a metal reacts with oxygen
• Common alloys include brass (copper and zinc), bronze (copper and tin), and
pewter (tin, antimony, and copper).
in the air to form its oxide
2 2 2
General vocabulary 1 1 1 1
2 2
enhanced improved 2
2
1 2
1
1 1 1 1
2 2 1 2 2
2
1 1 1 1
1 2
1 2
2
1 1 1 1
2
2 2 2
Figure 4.12 Alloys consist of different metal ions and a sea of delocalized electrons.
The smaller ions are able to fit in the spaces between the larger ions in the lattice structure.
Synonym
• The different packing of the two metal cations in an alloy means that alloys have
different properties from those of the parent metals they are made from:
•
deform ......... distort
alloys are often stronger (i.e. harder to deform) than the parent metals
because they are less malleable. (Because the cations have different sizes it is
harder for the layers of cations in an alloy to slide past each other and form a
new shape when pressure is applied.)
• changing the size (atomic radius) of the metal added to make the alloy.
Essential idea:
The enthalpy changes from chemical reactions can be calculated from
their effect on the temperature of their surroundings.
•
surroundings everything that is
Chemical energy is the energy that is stored in the chemical bonds and inter- outside of the system
particle forces that are present in substances. When chemical reactions occur open system can exchange mass
there will be a change in energy as products have different bonds (and different and energy with the surroundings
chemical energies) than reactants.
General vocabulary
proportional related in size to
Understanding: Chemical reactions that involve transfer of
heat between the system and the surroundings are Synonym
described as endothermic or exothermic. conserved . . . . . saved/retained ,
left unchanged
•
closed system can exchange
energy but not mass with the An exothermic reaction transfers heat energy from
surroundings the system to the surroundings. The temperature of
exothermic reaction a reaction the surroundings will increase. Heat (energy)
that gives off heat
endothermic reaction a • An endothermic reaction transfers heat energy
from the surroundings to the system. The
Figure 5.2 A closed system:
reaction that takes in heat
temperature of the surroundings will decrease. gases cannot escape but transfer
specific heat capacity the
amount of heat needed to raise • The specific heat capacity of a substance (c): of energy as heat can occur.
the temperature of 1 g of a
substance by 1 kelvin. Units are
J g–1 K–1
• is the amount of energy needed to raise the temperature
of one gram of the substance by one kelvin
enthalpy (H) the heat that is • has units J g–1 K–1
contained in a system
• The change in heat (q) that occurs in a substance is related to:
enthalpy change (ΔH) the
change in enthalpy that occurs • the mass of the substance (m) in grams
due to a chemical reaction or
• the specific heat capacity (c) of the substance
•
process. It is equal to the amount
of heat energy released or the temperature change observed (ΔT) in kelvin.
absorbed at constant pressure
The equation used to calculate the change in heat (q) is: q = mcΔT.
Worked example
Determine the amount of heat gained when a 3.50 kg aluminium bar is heated
from 298.0 K to 357.3 K. The specific heat capacity of aluminium is 0.900 J g–1 K–1.
Solution
products
reactants
• a concentration of 1 mol dm–3 for all solutions heat that is given off or taken in
•
by reactions that occur under
all reactants in their standard states (the states that the reactants normally standard conditions
exist in at 100 kPa) standard states the state in
• no defined temperature but 298 K (25 °C) is commonly used. which an element or compound
exists under standard conditions
• Enthalpy changes (ΔH) are calculated for one mole of substance so the units are
kJ mol–1 (kJ per mole).
enthalpy of reaction
(ΔH reaction) the enthalpy change
that occurs when one mole of
Skill: A calorimetry experiment for an enthalpy of reaction should be a substance is reacted under
standard conditions
covered and the results evaluated.
• The enthalpy of reaction (ΔH reaction) is the enthalpy change that occurs when
one mole of a substance is reacted under standard conditions.
calorimeter an instrument that
measures the heat changes that
occur during a reaction
Worked example
50.0 cm3 of 1.00 mol dm–3 copper(II) sulfate solution was placed in a polystyrene
coffee cup and excess zinc powder was added to the solution. The temperature
change (ΔT) that occurred due to the reaction was 48.2 K.
Use these results to calculate the enthalpy change for the reaction (ΔHr) in kJ mol–1:
Cu2+(aq) + Zn(s) ➝ Cu(s) + Zn2+(aq)
Solution
Assumptions:
(i) heat capacity (c) of solution = heat capacity of water = 4.18 J g–1 K–1
(ii) density of solution = density of water = 1.00 g cm–3 (so 50.0 cm3 has a
mass of 50.0 g)
q
ΔHr =
n(Cu2+)
q = mcΔT = 50.0 g × 4.18 J g–1 K–1 × 48.2 K = 10.1 kJ
50.0
n(Cu2+) = CV = 1.00 mol dm–3 × dm3 = 0.0500 mol
1000
q –10.1 k J
ΔHr = = = –202 kJ mol–1
n(Cu2+) 0.0500 mol
ΔH is negative as the reaction released heat and is exothermic.
Essential idea:
In chemical transformations energy can be neither created nor
destroyed (the first law of thermodynamics).
Worked example
–47 kJ Figure 5.5 illustrates the enthalpy changes occurring for a set of reactions:
P Q
Use this information to determine the enthalpy change for:
(i) P ➝ S
+45 kJ
(ii) Q ➝ S
Solution
R S
+60 kJ (i) ΔH(P ➝ S) = ΔH(P ➝ R)+ ΔH(R ➝ S) = +45 kJ + 60 kJ = +105 kJ
Hints for success: Enthalpy values are added when following the direction of
arrows in an enthalpy cycle and subtracted when going against an arrow.
• The standard enthalpy of formation (ΔHf ) is the enthalpy change when one
mole of a substance is formed from its elements in their standard states and under
standard conditions, e.g. the standard enthalpies of formation for water (H2O) and
Hints for success: ethanol (C2H5OH) are the enthalpy changes that occur for the reactions:
Standard enthalpies
ΔHf (H2O): H2(g) + 1O2(g) ➝ H2O(l)
of formation are given 2
in section 12 of the ΔHf (C2H5OH): 2C(s) + 3H2(g) + 1O2(g) ➝ C2H5OH (l)
2
IB data booklet.
ΔHf of all elements in their standard states is zero, e.g. ΔHf (O2) = 0.
66 Topic 5: Energetics/thermochemistry
ΣΔHf ΣΔHf
(reactants) Standard states (products)
of elements
Worked example
Use the standard enthalpy of formation data provided to determine the standard
enthalpy change for the reaction of hydrazine (N2H4) and hydrogen (H2) to
form ammonia (NH3):
N2H4(g) + H2(g) ➝ 2NH3(g)
ΔHf (N2H4) = +95 kJ mol–1 ΔHf (NH3) = –46 kJ mol–1
ΔHreaction = ΣΔHf (products) – ΣΔHf (reactants) Hints for success:
The ΔHf values
= 2 × ΔHf (NH3) – [ΔHf (N2H4) + ΔHf (H2)] for a compound
= 2 × –46 kJ mol–1 – [+95 + 0] kJ mol–1 must be multiplied
by the coefficient
= –187 kJ mol–1 of that compound.
Worked example
Determine the standard enthalpy change (ΔH ) for the reaction:
C(graphite) + 1O2(g) ➝ CO(g)
2
Using the information:
C(graphite) + O2(g) ➝ CO2(g) ΔH = –394 kJ (Reaction 1)
CO(g) + 1O2(g) ➝ CO2(g) ΔH = –283 kJ (Reaction 2)
2
Solution
We can rearrange the two equations provided to give the overall equation of the reaction
required. The enthalpy change for the reaction can then be determined using Hess’s Law.
Reaction 1: C(graphite) + O2(g) 1O2(g) ➝ CO2(g) ΔH = –394 kJ
2
Reverse reaction 2: CO2(g) ➝ CO(g) + 1O2(g) ΔH = +283 kJ
2
Hints for success: Reversing a reaction means that +ΔH becomes –ΔH.
Essential idea:
Energy is absorbed when bonds are broken and is released when
bonds are formed.
X–Y(g)
• The enthalpy change for a reaction is equal to the difference between the energy
required to break bonds in the reactants and the energy released when bonds
are formed in the products.
ΔHreaction = ΣE(bonds broken) – ΣE(bonds formed)
Subject vocabulary • The bond enthalpy is the energy required to break one mole of a particular
covalent bond in gaseous molecules.
bond enthalpy the energy
required to break one mole Section 11 of the IB data booklet gives average bond enthalpies for various
of a particular bond in the covalent bonds. These are average values because the enthalpy measured for
gaseous state
a particular bond in various molecules will be affected by the other bonds
that are present in the molecule, e.g. the bond enthalpy for the C–H bonds in
methane (CH4) will not be the same as the bond enthalpy for the C–H bonds in
chloromethane (CH3Cl).
Because the tabulated values are average bond enthalpies they can only be used
to estimate the enthalpy change of a reaction.
The tabulated bond enthalpies give the best estimates for gas reactions as they
do not account for the intermolecular forces present in solids and liquids.
Skill: Calculation of the enthalpy changes from known bond enthalpy values
and comparison of these to experimentally measured values.
68 Topic 5: Energetics/thermochemistry
To answer this question it is necessary to first draw the structures of the reactants
and products. This makes it easier to visualize the bonds that are being broken in
the reactants and the bonds being formed in the products.
H H
O
H C C O H +3 O O 2 O C O +3
H H
H H
• bond enthalpies are measured for gases but the reaction involves liquids
• bond enthalpies are averaged values and will not be the exact values for the
bonds in the reactants and products of this reaction.
• Products are more stable than reactants if they have a lower enthalpy than the
reactants. This happens for exothermic reactions (ΔH is negative).
• For endothermic reactions (ΔH is positive) reactants are more stable than
products.
Figure 5.7 The relative stability
of reactants and products in Reactants (less stable) Products (less stable)
exothermic and endothermic
reactions.
enthalpy (H)
enthalpy (H)
ΔH negative ΔH positive
Skill: Discussion of the bond strength in ozone relative to oxygen and its
importance to the atmosphere.
The combination of oxygen (O2) and ozone (O3) in the atmosphere absorbs most
harmful ultraviolet (UV) radiation. Because O2 contains stronger bonds it absorbs
UV radiation of a shorter wavelength (higher energy) than the UV radiation
absorbed by O3.
Oxygen O2 Ozone O3
O
O O
O O
Ozone levels in the atmosphere have been depleted because of chemicals released
by human activity. This has serious health consequences because the harmful UV
wavelengths that would normally be absorbed by ozone are more likely to reach
the Earth’s surface.
70 Topic 5: Energetics/thermochemistry
Essential idea:
The greater the probability that molecules will collide with sufficient
energy and proper orientation, the higher the rate of reaction. Synonym
rate . . . . . . . . . . . . . . speed
most probable
kinetic energy
kinetic energy, E
• The area below the curve in Figure 6.2 represents the total number of particles
that are in the sample. The peak of the distribution represents the most probable
kinetic energy.
energy
rate of reaction how fast reactants Ea
a reaction happens (how Ea
quickly reactants are changed ΔH
into products) ΔH
reactants
Figure 6.3 Energy path of a products
reaction with the activation
energy for (a) endothermic and
(b) exothermic reactions. (a) progress of (b) progress of
* represents the transition state. reaction reaction
• If the reactants collide with enough energy to reach the transition state then the
reaction can continue on and form products.
kinetic energy, E
•
Figure 6.5 Concentration of
product against time. The rate of reaction is determined by measuring how the concentration of
products increases over time or by measuring how the concentration of
General vocabulary reactants decreases over time.
rate a measure of how quickly a Δ[products] Δ[reactants]
change occurs over time rate of reaction = =–
Δt Δt
1.10
1.00
0.90 tangent to the curve at 90 s
0.80 gives reaction rate at 90 s
–1.18
0.70 = = –7.2 × 10–3 mol dm–3 s–1
164
0.60
0.50
0.40
0.30
0.20
0.10
• The curve shows that the rate of reaction is fastest at the start of the reaction and
decreases as the reaction proceeds.
• The gradient of the curve at any given point gives the rate of reaction at that
time. The gradient of a curve is obtained from the tangent to the curve at that
General vocabulary
gradient the slope (or steepness)
point. In Figure 6.6 this is shown using the blue tangent line at t = 90 s: of a line
Δ[reactants]
rate of reaction (90 s) = – = –(–7.2 × 10–3 mol dm–3 s–1)
Δt = 7.2 × 10–3 mol dm–3 s–1
• When the rate is obtained at t = 0 this is the initial rate of reaction. In Figure 6.6
this is shown using the red tangent line:
Subject vocabulary
initial rate of reaction the rate
Δ[reactants] of reaction that occurs at the start
initial rate of reaction (0 s) = – = –(–2.9 × 10–3 mol dm–3 s–1) of the reaction (when t = 0)
Δt = 2.9 × 10–3 mol dm–3 s–1
The reaction rate is obtained from a graph of the mass of the reaction mixture
against time (Figure 6.8).
Hints for success: This method is not suitable for gases with low molar
time masses (e.g. H2) as the change in mass that occurs may be too small to
measure accurately.
Figure 6.8 Mass against time.
Measuring the change in colour during a reaction
If the reaction involves a change in colour it is possible to measure the change in
Subject vocabulary absorbance of the solution using a spectrophotometer.
absorption the amount of light
Absorbance at 591.0 nm
0.5
taken in when it passes through
a liquid
0.4
absorbance a measure of the
amount of light absorbed by
coloured solutions 0.3
spectrophotometer an
instrument that measures the 0.2
absorbance of light 0 5 10 15 20
Time/min
Figure 6.9 Absorbance against time.
The reaction rate is obtained from a graph of absorbance against time (Figure 6.9).
kinetic energy, E
• At higher temperatures:
• the peak of the curve moves to the right as the most probable kinetic energy
occurs at a higher kinetic energy
• The shaded areas to the right of the activation energy in Figure 6.10 represent
the proportion of particles at each temperature that have sufficient kinetic
energy to overcome the activation energy. At the higher temperature a larger
proportion of particles have enough kinetic energy to react so the rate of
reaction will increase with an increase in temperature.
Effect of concentration and pressure on rate of reaction
• means that there are more reactant particles within the same volume
• means that there are more reactant particles within the same volume (i.e. the
concentration has increased)
• Reactions of solids:
General vocabulary • will have a faster rate if the solid has a bigger surface area
reaction
activation energy
catalysed reaction
reactants
ΔH
products
extent of reaction
Particles with
these low energies
cannot react.
Essential idea:
Many reactions are reversible. These reactions will reach a state of
Subject vocabulary equilibrium when the rates of the forward reaction and reverse
forward reaction reaction that reaction are equal. The position of equilibrium can be controlled by
proceeds from left to right as
written; reactants are converted changing the conditions.
into products
reverse reaction reaction that
proceeds from right to left as Understanding: A state of equilibrium is reached in a closed
written; products are converted
into reactants system when the rates of the forward and reverse reactions
equilibrium a state in which
forward and reverse reactions are
are equal.
occurring at the same rate in a
closed system. There is no overall • In a reaction, reactants are changed into products in the forward reaction.
change in the concentrations of
reactants and products • In some reactions, the reverse (or backward) reaction also occurs, where
products are changed into reactants.
dissociation a reaction in which
a molecule is split into smaller forward reaction
molecules, atoms, or ions
reactants products
backward reaction
increasing concentration of
vapour. Equilibrium is established evaporation = rate of
when the two rates are equal. condensation
rate of
condensation
time
• The dissociation of hydrogen iodide, HI, into hydrogen, H2, and iodine, I2.
2HI(g) H2(g) + I2(g)
colourless gas colourless gas purple gas
78 Topic 7: Equilibrium
concentration
constant. Note the same
equilibrium mixture is reached
starting from (a) a mixture of H2
and I2 or (b) pure HI.
H2, I2 H2, I2
time time
equilibrium equilibrium
•
that is at equilibrium
If the equilibrium mixture has a high proportion of reactants, the equilibrium
position is said to lie to the left. This means that the reverse reaction has been
favoured and the forward reaction has not happened to a great extent.
7.1 Equilibrium 79
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General vocabulary • We can quantify the equilibrium position using the equilibrium expression.
The equilibrium law states that for a reaction:
quantify make a numerical
measurement of a quantity aA + bB cC + dD
numerator the number or when the concentrations of all reactants and products are measured in the
numbers above the line in a equilibrium mixture, the relationship:
fraction
[C]c [D]d
denominator the number or
numbers below the line in a [A]a [B]b
fraction
gives a constant Kc, known as the equilibrium constant. Kc has a constant value
magnitude the relative size of a for a particular reaction at a fixed temperature.
•
measurement, how large it is
The concentration used in the equilibrium expression must be the equilibrium
concentration of reactant or product.
• Where there is more than one reactant or product the terms are
multiplied together.
Worked example
Write the equilibrium expression for the following reactions.
(i) 2H2(g) + O2(g) 2H2O(g)
2+
(ii) Cu (aq) + 4NH3(aq) [Cu(NH3)4]2+(aq)
Solution
[H2O]2
(i) K =
c
[H2]2[O2]
[[Cu(NH3)4]2+]
(ii) Kc =
[Cu2+][NH3]4
80 Topic 7: Equilibrium
• Remember that the value of Kc changes with the temperature of the reaction, so
the temperature must always be stated with any data value for Kc.
7.1 Equilibrium 81
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• Note that Q (unlike Kc) does not have a fixed value, but depends on the point
during the reaction when the concentrations are measured.
General vocabulary • As the reaction proceeds towards equilibrium, the value of Q changes in the
direction of Kc.
•
proceeds moves in a particular
direction We can use the reaction quotient, Q, to predict the direction in which
qualitatively without using the reaction will proceed towards equilibrium. This is summarized in
numerical data the table below.
• Note that the value of the equilibrium constant, Kc, does not change with
changes in any conditions except temperature.
82 Topic 7: Equilibrium
Note again that although the equilibrium position shifts in these cases, the value of
Kc does not change.
Changes in temperature
• Remember that the enthalpy changes of the forward and reverse reactions are
equal and opposite to each other.
Temperature is the only condition that also changes the value of Kc. An increase in
temperature increases the value of Kc for an endothermic reaction and decreases
the value of Kc for an exothermic reaction.
7.1 Equilibrium 83
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The responses of a system at equilibrium to applied changes are summarized in
the table below.
84 Topic 7: Equilibrium
Essential idea:
Many reactions involve the transfer of a proton from an acid to
a base.
Worked example
Deduce the conjugate acid and conjugate base, as well as the conjugate acid–base
pairs, for the following acid–base reaction:
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Solution
NH4+ is the conjugate acid because it donates a proton in the reverse reaction.
OH– is the conjugate base because it accepts a proton in the reverse reaction.
The two conjugate acid–base pairs are NH3/NH4+ and H2O/OH–.
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
base acid conjugate conjugate
acid base
Essential idea:
The characterization of an acid depends on empirical evidence
such as the production of gases in reactions with metals, the colour
changes of indicators, or the release of heat in reactions with metal
oxides and hydroxides.
Worked example
Write balanced equations for the following reactions:
1 Aluminium metal (Al) with sulfuric acid (H2SO4).
2 Sodium oxide (Na2O) with hydrochloric acid (HCl).
3 Calcium carbonate (CaCO3) with nitric acid (HNO3).
Solution
•
neutralization reaction a
reaction between an acid and is highly exothermic and generates heat.
a base
acid + base ➝ salt + water
Examples: HCl(aq) + NaOH(aq) ➝ NaCl(aq) + H2O(l) ΔH –ve
H2SO4(aq) + 2KOH(aq) ➝ K2SO4(aq) + 2H2O(l) ΔH –ve
Skill: Identification of the acid and base needed to make different salts.
The formula of a salt can be used to identify the acid and base that reacted
to make the salt:
Worked example
Deduce the acids and bases that react to make the following salts.
(i) K3PO4
(ii) NH4NO3
(iii) LiCH3COO
(i) K3PO4: K+ comes from the base KOH. PO43– comes from the acid H3PO4.
(ii) NH4NO3: NH4+ comes from the base NH3. NO3– comes from the acid HNO3.
(iii) LiCH3COO: Li+ comes from the base LiOH. CH3COO– comes from
the acid CH3COOH.
conical flask
Hints for success: Figure 8.1 shows an acid–base titration in which acid is added
from the burette to a base in the conical flask. Acid–base titrations can also add
base from the burette to acid in the conical flask.
The equivalence point:
Essential idea:
The pH scale is an artificial scale used to distinguish between acidic,
neutral, and basic/alkaline solutions.
General vocabulary pH and [H+] have an inverse relationship: This means that pH decreases as
[H+] increases.
inverse relationship a
relationship in which the value
of one quantity increases as the Worked example
value of another decreases
An aqueous solution of hydrochloric acid (HCl) has a hydrogen ion concentration
of 5.5 × 10–3 mol dm–3. What is the pH of this solution?
pH = –log10[H+(aq)] = –log(5.5 × 10–3) = 2.26
Hints for success: When An aqueous solution of nitric acid (HNO3) has a pH of 4.78. What is the hydrogen
calculating log values ion concentration of this solution?
the final answer should [H+] = 10–pH = 10–4.78 = 1.7 × 10–5 mol dm–3
have the same number
of decimal places as the
number of significant
figures in the initial Understanding: A change of one pH unit represents a
value. 5.5 × 10–3 has 10-fold change in the hydrogen ion concentration [H+].
2 significant figures,
therefore the answer of In logarithmic scales a difference of one unit represents a change of a factor of 10.
log(5.5 × 10–3) should
have 2 decimal places. pH [H+] /mol dm–3 Change in pH Change in [H+]
3 1.0 × 10–3 decreases by 1 increases by 10
–4
4 1.0 × 10
5 1.0 × 10–5 increases by 1 decreases by 10
Worked example
A sample of blood at 298 K has [H+] = 4.60 × 10–8 mol dm–3.
Calculate the concentration of OH– and state whether the blood is acidic,
neutral, or basic.
Solution
Worked example
An alkaline solution of sodium hydroxide (NaOH) has a pH of 10.45. What is the
concentration of hydrogen ions (H+) and hydroxide ions (OH–) in this solution?
[H+] = 10–pH = 10–10.45 = 3.5 × 10–11 mol dm–3
Kw = [H+][OH–]
K 1.00 × 10–14
[OH–] = W+ = = 2.9 × 10–4 mol dm–3
[H ] 3.5 × 10–11
General vocabulary
progression a gradual change
Figure 8.2 The pH scale and universal indicator. The tubes contain universal indicator added to solutions of
pH 0–14 from left to right.
Essential idea:
The pH depends on the concentration of the solution. The strength
of acids or bases depends on the extent to which they dissociate in
aqueous solution.
• bases dissociate to give OH– ions. weak acids acids that partially
dissociate in aqueous solution
Acids are defined as weak acids or strong acids.
strong acids acids that fully
Weak acids do not dissociate fully and there is an equilibrium reaction: dissociate in aqueous solution
Note that H3O+ is written as a product, not H+. This is because the H+ that
dissociates from the acid bonds to an H2O molecule and really exists as H3O+ in
aqueous solutions.
Bases are defined as weak bases or strong bases.
Weak bases do not dissociate fully and there is an equilibrium reaction:
B(aq) + H2O(l) BH+(aq) + OH–(aq)
Strong bases dissociate fully and there is no equilibrium reaction:
B(aq) + H2O(l) ➝ BH+(aq) + OH–(aq)
Acid Base
common HCl hydrochloric LiOH lithium hydroxide
examples acid
of strong
HNO3 nitric acid NaOH sodium hydroxide
forms
H2SO4 sulfuric acid KOH potassium
hydroxide
Ba(OH)2 barium hydroxide
some CH3COOH ethanoic acid NH3 ammonia
examples of
and other
weak forms
organic
acids
H2CO3 carbonic acid C2H5NH2 ethylamine
H3PO4 phosphoric acid and other
amines
• The rate of reaction of an acid will depend upon the [H+] present in the solution.
Because strong acids dissociate fully their solutions have a larger [H+] in solution
and will react faster than weak acids of the same concentration.
• The rate of reaction of a base will depend upon the [OH–] present in the solution.
Because strong bases dissociate fully their solutions have a larger [OH–] and will
react faster than weak bases of the same concentration.
Essential idea:
Increased industrialization has led to greater production of nitrogen
and sulfur oxides, leading to acid rain, which is damaging our
environment. These problems can be reduced through collaboration
with national and intergovernmental organizations.
• is naturally acidic
• has a pH of 5.6.
Acid deposition:
Subject vocabulary
acid deposition the formation • occurs when rainwater dissolves covalent oxides such as SO2, SO3, and NO2
of acidic solutions that happens
• contains stronger acids than normal rain
•
when acidic substances dissolve in
atmospheric or surface water is more acidic than normal rain
•
a substance
remove sulfur from coal and oil before they are burned
•
post-combustion occuring
remove sulfur from crushed coal by washing with water after the combustion reaction of
• remove sulfur dioxide from exhaust gases after the coal or oil is burned
• react SO2 with calcium oxide (CaO) or calcium carbonate (CaCO3) to form
calcium sulfite:
SO2(g) + CaO(s) ➝ CaSO3(s)
SO2(g) + CaCO3(s) ➝ CaSO3(s) + CO2(g)
Essential idea:
Redox (reduction–oxidation) reactions play a key role in many
chemical and biochemical processes.
+1 –2
Note that the oxidation states apply to each atom and that here the sum of all the
oxidation states must be zero as H2SO4 is electrically neutral.
Therefore, 2(+1) + S + 4(–2) = 0 ⇒ S = +6
(b) Here we start by assigning O: SO32–
–2
Note that here the oxidation states must add up to –2, the charge on the ion.
Therefore, S + 3(–2) = –2 ⇒ S = +4
Worked example
Assign oxidation states to the metal ion in (a) [Co(NH3)6]3+ and (b) [CuCl4]2–.
Solution
(a) NH3 is a neutral ligand, so the charge on the complex is the same as the
charge on the metal ion ∴ Co = +3
(b) Cl has a 1– charge ∴ (charge on Cu) + (4 × 1–) = 2–
∴ Cu = +2
The oxidation state of the transition metal ion is included in the name of
the complex ion.
Worked example
Deduce the names of the following compounds using oxidation numbers.
(a) V2O5 (b) Ni(OH)2 (c) TiCl4
Solution
Subject vocabulary
Skill: Identification of the species oxidized and reduced and the oxidizing
and reducing agents, in redox reactions. half-equation an equation that
shows the changes that happen
A redox equation can be divided into two half-equations showing which species in a redox reaction due to either
oxidation only or reduction only
gain electrons (are reduced) or lose electrons (are oxidized).
•
oxidation half-equation a
half-equation that shows only the The reduction half-equation shows the oxidizing agent being reduced to
chemical changes that happen in its product(s).
a redox reaction due to oxidation
reduction half-equation a Each half-equation must show the number of electrons being gained or lost.
half-equation that shows only the
chemical changes that happen in Worked example
a redox reaction due to reduction
Deduce the two half-equations for the following reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Solution
Assign oxidation states so we can see what is being oxidized and what is reduced.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
0 +2 +2 0
2+
Here we can see that Zn is being oxidized and Cu is being reduced.
oxidation: Zn(s) → Zn2+(aq) + 2e– electrons are lost
reduction: Cu2+(aq) + 2e– → Cu(s) electrons are gained
Note there must be equal numbers of electrons in the two half-equations, so that
when they are added together the electrons cancel out.
Worked example
Balance the redox equation that occurs in acidic solution when sulfurous acid
(H2SO3) is titrated against potassium dichromate (K2Cr2O7):
Cr2O72–(aq) + H2SO3(aq) ➝ Cr3+(aq) + SO42–(aq)
First we need the balanced equation for the reaction, which is solved by the half-
equation method described earlier.
oxidation: Fe2+ ➝ Fe3+ + e–
reduction: MnO4– + 8H+ + 5e– ➝ Mn2+ + 4H2O
overall: 5Fe2+ + MnO4– + 8H+ ➝ 5Fe3+ + Mn2+ + 4H2O
Hints for success: An activity series containing more metals is given in section 25
of the IB data booklet.
Skill: Deduction of the feasibility of a redox reaction from the activity series
or reaction data.
The activity series can be used to predict whether or not a redox reaction that
involves a metal can happen.
(a) This reaction would involve Ag reducing Zn2+ in ZnCl2. But Ag is a weaker
reducing agent than Zn, so this will not occur.
(b) This reaction involves Mg reducing Fe3+ in FeCl3. Mg is a stronger reducing
agent than Fe, so this will occur.
1 Dissolved oxygen reacts with Mn2+ ions in basic conditions to form MnO2: Winkler Method a method that
uses redox reactions to determine
2Mn2+(aq) + O2(aq) + 4OH–(aq) ➝ 2MnO2(s) + 2H2O(l) the concentration of dissolved
oxygen in aqueous solutions
2 MnO2 reacts with I– in acidic conditions to form I2 and Mn2+:
MnO2(s) + 2I–(aq) + 4H+(aq) ➝ Mn2+(aq) + I2(aq) + 2H2O(l)
3 I2 is titrated with S2O32– to give I– and S4O62–:
2S2O32–(aq) + I2(aq) ➝ S4O62–(aq) + 2I–(aq)
The moles of dissolved oxygen in the original solution can be calculated from the
moles of S2O32– used in the final reaction:
n(S2O32–) = 4n(O2)
Worked example
A 500 cm3 sample of water was collected and tested for dissolved oxygen by the
addition of MnSO4 in basic solution, followed by the addition of acidified KI. It
was found that 12.50 cm3 of 0.0500 mol dm–3 Na2S2O3(aq) was required to react
with the iodine produced. Calculate the dissolved oxygen content of the water in
g dm–3, using the equations given above.
Start with calculating the amount of S2O32–, as we are given both its volume and
its concentration.
n = cV
12.50
n(S2O32–) = 0.0500 mol dm–3 × dm3 = 6.25 × 10–4 mol
1000
From the reacting ratio in step 3, S2O32– : I2 = 2 : 1
∴ n(I2) = 0.5 × 6.25 × 10–4 = 3.125 × 10–4 mol
From the reacting ratio in step 2, I2 : MnO2 = 1 : 1
∴ n(MnO2) = 3.125 × 10–4 mol
From the reacting ratio in step 1, MnO2 : O2(g) = 2 : 1
∴ n(O2) = 0.5 × 3.125 × 10–4 mol = 1.5625 × 10–4 mol
(We could also go directly from a ratio of O2 : S2O32– = 1 : 4)
Finally express the amount of O2(g) as g dm–3
m(O2) = n × M = 1.5625 × 10–4 mol × 32.00 g mol–1 = 5.000 × 10–3 g in 500 cm3
∴ dissolved oxygen = 0.0100 g dm–3
Essential idea:
Voltaic cells convert chemical energy to electrical energy and Subject vocabulary
electrolytic cells convert electrical energy to chemical energy. voltaic cells devices that convert
chemical energy to electrical
energy
Understanding: Voltaic cells convert energy from electrolytic cells devices that
convert electrical energy to
spontaneous, exothermic chemical processes to chemical energy
Converts chemical energy into Converts electrical energy into salt bridge a device that
connects the anode and cathode
electrical energy chemical energy in a voltaic cell. It contains a
concentrated solution of an
unreactive salt and allows ions to
flow into the two half-cells
Understanding: Oxidation occurs at the anode (negative
electrode) and reduction occurs at the cathode (positive
electrode) in a voltaic cell.
A voltaic cell contains two half-cells:
Zn2+(aq) Cu2+(aq)
SO42–(aq) SO42–(aq)
• a wire connecting the two electrodes, which allows electrons to flow from the
anode to the cathode
• a salt bridge connecting the two half-cells, which contains a concentrated salt
solution and allows ions to flow into the half-cells.
Oxidation Reduction
occurs at occurs at
the anode. cotton the cathode.
wool Cu2+
zinc Zn2+ copper
sulfate sulfate
solution solution
Subject vocabulary The cell diagram convention is a simple representation of a voltaic cell. The
copper–zinc voltaic cell drawn above is represented as:
cell diagram convention a
method for representing the
components of a voltaic cell salt bridge phase boundary
phase boundary a boundary
that exists between components Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
of a cell that are in different
phases, e.g. a metal electrode and anode cathode
a salt solution in a voltaic half-cell oxidation reduction
spectator ions ions that are Zn(s) ➝ Zn2+(aq) + 2e– Cu2+(aq) + 2e– ➝ Cu(s)
present in solution but do not
participate in any reactions The key features of a cell diagram representation are:
• The anode is normally shown on the left and cathode on the right.
• A double vertical line represents the salt bridge between the anode and cathode.
• Single vertical lines represent phase boundaries and show that the half-cell
contains different phases, for example, a solid metal and an aqueous solution.
• Aqueous solutions in each half-cell are shown next to the salt bridge.
• Spectator ions are not oxidized or reduced in the reactions so are not included
in the representation.
•
being interrupted
the flow of electrons from anode to cathode produces an electric current
battery a device that converts
(electricity). stored chemical energy into
electrical energy
power source an electronic
Understanding: Oxidation occurs at the anode (positive device that is a source of electrical
energy
electrode) and reduction occurs at the cathode (negative electrolyte solution a solution
electrode) in an electrolytic cell. that contains ions and is able to
conduct charge
• a battery or DC power source that forces electrons to flow from the anode
to the cathode
DC stands for
direct current
• an electrolyte solution, which allows ions to flow and balance the changes in
charge that happen at the electrodes because of the redox reactions.
Hints for success: In all electrochemical cells oxidation occurs at the anode and
reduction at the cathode. However, the charge on the cathode and the charge on
the anode are different in voltaic and electrolytic cells.
electrolyte
The electrodes in an electrolytic cell are usually made of inert materials such as
graphite or platinum. These materials can conduct electricity but are not oxidized or
reduced themselves.
• the flow of electrons is controlled by the power source: electrons flow from the
anode to the cathode
• cations (M+) in the electrolyte flow to the cathode and are reduced
• anions (A–) in the electrolyte flow to the anode and are oxidized
Subject vocabulary The pure metals are obtained by electrolysis of their molten salts, e.g. sodium is
obtained from the electrolysis of molten NaCl.
molten melted (of a liquid that is
formed only at high temperatures) + – e–
discharged changed by the e–
electrolysis reaction (by losing its graphite
charge) anode + electrodes – cathode
Cl2(g) bubbles
Cl–
Na+ molten Na collects
Figure 9.4 Electrolysis of molten
sodium chloride. Chloride ions are
oxidized at the anode and sodium
ions are reduced at the cathode. 2Cl–(l) ➝ Cl2(g) + 2e– 2Na+(l) + 2e– ➝ 2Na(l)
The overall equation is
2NaCl(l) ➝ 2Na(l) + Cl2(g). molten NaCl electolyte
Hints for success: The electrolysis products may not be in the same states
as they would be at 25 °C because of the high temperatures needed to
make molten salts.
Essential idea:
Organic chemistry focuses on the chemistry of compounds
containing carbon.
• belong to the same family of compounds (have the same functional group) organic compounds in which the
•
chemical formula of successive
differ by a common structural unit, –CH2– group (the common structural unit) members differs by CH2
H H H H H H H H H H
H C O H H C C O H H C C C O H H C C C C O H
H H H H H H H H H H
General vocabulary • Rule 1: Identify the longest chain in the compound. This gives the stem
of the name.
stem the root or base from which
something larger is made
Number of carbon Stem in IUPAC name Example of compound
atoms in longest chain
1 meth- CH4, methane
2 eth- C2H6, ethane
3 prop- C3H8, propane
• Rule 2: Identify the functional group. This usually gives the suffix of the name.
See the table on pages 114 and 115 for the different functional groups and
General vocabulary
suffix an ending added to a word
their suffixes. substituent something that
• Rule 3: Identify the substituent groups. These give the prefix of the name.
A number is given before the prefix for a substituent group to show where it
has replaced something else. In
organic compounds substituents
have usually replaced a hydrogen
atom
occurs on the longest chain. If a substituent occurs more than once the prefix is
modified to show this: e.g. dimethyl (two methyl substituents), trichloro (three prefix a beginning added to a
word
chloro substituents).
Hints for success: The numbering of the longest chain must start at the end that
gives the lowest numbers for the substituents.
H
2-methylpentane
C6H5
*Esters form when the alkyl group of an alcohol replaces the hydrogen of a
carboxylic acid in a condensation reaction:
R–COOH + R′OH → R–COO–R′ + H2O
They are named in a similar way to salts, which form when a metal has replaced the
hydrogen of a carboxylic acid. Salts take the stem of the name from the parent acid.
For example, C2H5COONa is sodium propanoate. In esters, the alkyl group of the
alcohol is the prefix, so C2H5COOCH3 is methyl propanoate.
H H O H H H O H H H H
H C C C C H H C C C O H H C C C C H
H H H H H
H C H H C H
H H
Solution
H H O H H H H H
H C C C C H H H O H C C C C H
4 3 2 1 1 2 3 4
H H H C C C O H H
3 2 1
H C H H H H C H
H H
H H H H
ethanol, C2H5OH chloroethane, C2H5Cl
H OH H H Cl H
propan-2-ol, CH3CH(OH)CH3 2-chloroethane, CH3CHClCH3
H OH H H Cl H
2-methylpropan-2-ol 2-chloro-2-methylpropane
C(CH3)3OH C(CH3)3Cl
Methanol is also regarded as a primary alcohol even though the hydroxyl Subject vocabulary
group is attached to a carbon that is not bonded to any alkyl chains. primary carbon a carbon atom
that is bonded to one alkyl chain
Skill: Identification of primary, secondary and tertiary nitrogen secondary carbon a carbon
atoms in amines. atom that is bonded to two
alkyl chains
Primary amine Secondary amine Tertiary amine tertiary carbon a carbon
atom that is bonded to three
primary secondary tertiary alkyl chains
nitrogen atom nitrogen atom nitrogen atom primary nitrogen a nitrogen
H H CH3 atom that is bonded to one
alkyl chain
H3C N H3C N H3C N secondary nitrogen a nitrogen
atom that is bonded to two
H CH3 CH3 alkyl chains
tertiary nitrogen a nitrogen
Amine group has a Amine group has a Amine group has a atom that is bonded to three
primary nitrogen that secondary nitrogen that tertiary nitrogen that alkyl chains
is bonded to one alkyl is bonded to two alkyl is bonded to three alkyl
chain chains chains
H C C C H H C C C C H
H H H
propene but-2-yne
C3H6 C4H6
H C H H C H
C C C C
C C C C
H C H H C H
H H
Each point represents a carbon atom. Hydrogen atoms are not shown. The circle
represents the three delocalized double bonds.
benzene 0.139 nm
(b) ΔHhydrogenation for the Theoretical value based on adding Delocalization minimizes the repulsion between
reaction H2 across three localized C==C electrons and so gives benzene a more stable
C6H6 + 3H2 → C6H12 double bonds = −362 kJ mol−1 structure, lowering its internal energy by
152 kJ mol−1.
Experimental value for benzene
= −210 kJ mol−1
Benzene is more stable than
predicted from three localized C==C
bonds by approx. 152 kJ mol−1
(c) Type of reactivity Benzene does not undergo addition Addition reactions are not favoured as they
reactions and is more likely to would involve disrupting the entire cloud of
undergo substitution reactions. delocalized electrons. Benzene can instead
undergo substitution reactions that preserve the
stable ring structure.
(d) Isomers Only one isomer exists As benzene is a symmetrical molecule with no
of compounds such as alternating single and double bonds, all adjacent
1,2-dibromobenzene. positions in the ring are equal.
For example 1,2-dibromobenzene, C6H4Br2
Br
Br
Essential idea:
Structure, bonding, and chemical reactions involving functional group
Synonyms interconversions are key strands in organic chemistry.
key strands .... important
features
initiate .......... start Understanding: Alkanes have low reactivity and undergo
propagate ..... continue free-radical substitution reactions.
terminate ...... end
Alkanes are saturated hydrocarbons with the general formula CnH2n+2.
Subject vocabulary Alkanes have a low reactivity because they contain C–C and C–H bonds which:
alkanes hydrocarbons that only
contain single bonds. They have • are strong bonds and are hard to break
the general formula CnH2n+2
hydrocarbons compounds
• are non-polar bonds and do not attract reactive species such as electrophiles
and nucleophiles.
that contain only carbon and
hydrogen atoms Two reactions of alkanes that do occur are:
combustion the reaction of a
• combustion with oxygen (alkanes are good fuels)
substance with oxygen
free-radical substitution a
• free-radical substitution with halogens.
substitution reaction in which a Skill: Writing equations for the complete and incomplete combustion
bonded atom is replaced by a free
radical
of hydrocarbons.
complete combustion the Complete combustion of alkanes occurs when there is an excess of oxygen. The
combustion reaction of a products are carbon dioxide and water, e.g.
substance that happens when
excess oxygen is available C3H8(g) + 5O2(g) ➝ 3CO2(g) + 4H2O(g)
incomplete combustion the
combustion reaction of a
2C8H18(g) + 25O2(g) ➝ 16CO2(g) + 18H2O(g)
substance that happens when Incomplete combustion of alkanes occurs when oxygen is limiting. The products
limited amounts of oxygen is
available are carbon monoxide and water, e.g.
free radicals species that contain 2C3H8(g) + 7O2(g) ➝ 6CO(g) + 8H2O(g)
an upaired electron
2C8H18(l) + 17O2(g) ➝ 16CO(g) + 18H2O(g)
homolytic fission the breaking
of a covalent bond to form two
free radicals Skill: Explanation of the reaction of methane and ethane with halogens in
terms of a free-radical substitution mechanism involving photochemical
photochemical chemical
changes that happen after light homolytic fission.
energy is absorbed by compounds
Free radicals are species that contain an unpaired electron and they are
initiation the first step in the highly reactive.
free-radical mechanism, where
free radicals are formed Homolytic fission is when a covalent bond breaks and each atom gets one electron:
propagation the steps in the
free-radical mechanism where A–B ➝ A• + B• (A• and B• are free radicals)
free radicals react to form new
free radicals Photochemical homolytic fission happens when the molecule absorbs light energy
and this causes the bond to break.
termination the last step in the
free-radical mechanism, where The free-radical substitution mechanism has three steps:
two radicals combine and end the
reaction Step 1 Initiation: free radicals are made from homoloytic fission and
initiate the reaction.
Step 2 Propagation: the free radicals made in Step 1 propagate (continue
reacting) to create new free radicals.
Step 3 Termination: the free radicals combine to form a covalent bond and the
reaction terminates.
Hints for success: Cl2 and Br2 are good initiators as they have weak bonds that
are easily broken by UV light.
H C C C H + H2 ➝ H C C C H
H H H H
Ni catalyst
CH3CHCH2 + H2 ⎯⎯⎯⎯→ CH3CH2CH3
150 °C
propene propane
Example: H H H H H H
H C C C H + Br2 ➝ H C C C H
H H Br Br
Example: H H H H
H C C H + HCl ➝ H C C H
H Cl
CH2CH2 + HCl ⎯→ CH3CH2Cl
ethene chloroethane
Example: H H H H
H2SO4(conc.)
H C C H + H2O ⎯⎯⎯⎯→ H C C H
H OH
H2SO4(conc.)
CH2CH2 + H2O ⎯⎯⎯⎯→ CH3CH2OH
ethene ethanol
• combustion with oxygen (alcohols are good fuels) compounds that contain the
•
–COOH functional group
reactions with carboxylic acids to make esters
•
esters a family of compounds
oxidation with oxidizing agents such as potassium dichromate(VI) or potassium that contain the –COO– functional
manganate(VII). group
condensation reaction a
Skill: Writing equations for the complete combustion of alcohols. reaction that forms water as
a product
Complete combustion of alcohols occurs when there is an excess of oxygen. The
products are carbon dioxide and water.
Examples:
2C3H7OH(l) + 9O2(g) ➝ 6CO2(g) + 8H2O(g)
C6H13OH (l) + 9O2(g) ➝ 6CO2(g) + 7H2O(g)
Skill: Writing the equation for the condensation reaction of an alcohol with a
carboxylic acid, in the presence of a catalyst (e.g. concentrated sulfuric acid)
to form an ester.
Esterification reactions:
H C C C O H + H O C C H
H H H H
propanol ethanoic acid
H2SO4
H H H O H
H C C C O C C H + H2O
H H H H
propyl ethanoate
secondary ketone H H H H H
+[O], heat
H C C C H H C C C H + H2O
reflux
H OH H H O H
propan-2-ol propanone
secondary alcohol ketone
H OH H
2-methylpropan-2-ol no reaction
tertiary alcohol
•
distillation a process in which
The carboxylic acid is formed if the reaction is in reflux. The initial aldehyde a mixture is separated into
product stays in the reaction mixture and is converted to the carboxylic acid. components by heating the
mixture and selectively boiling off
and condensing the components
into a separate container
Understanding: Halogenoalkanes are more reactive than reflux a process in which a
reaction mixture is kept at the
alkanes. They can undergo (nucleophilic) substitution boiling point of the solvent by
reactions. A nucleophile is an electron-rich species condensing the solvent that boils
off and returning it to the reaction
containing a lone pair that it donates to an electron- mixture
halogenoalkanes alkanes in
deficient carbon. which a hydrogen atom has been
replaced by a halogen atom
Halogenoalkanes are saturated compounds that have the general formula CnH2n+1X, nucleophilic substitution
where X = F, Cl, Br, or I. reactions substitution reactions
in which a nucleophile replaces a
The polar C–X bond makes halogenoalkanes more reactive than alkanes. leaving group
Nucleophilic substitution reactions involve the halogen atom in a halogenoalkane nucleophile an electron-rich
species that can act as a Lewis
being replaced by a nucleophile. base and donate an electron pair
to form a coordinate bond
A nucleophile:
• is an electron-rich species that contains a lone pair of electrons, e.g. OH–, CN–
electron-deficient carbon a
carbon that has a partial positive
n C C C C
n
n(monomer) addition polymer
Hints for success: The brackets contain the repeating unit of the polymer. The
polymer is a chain of these repeating units linked to each other.
n represents the number of monomers used in the reaction and is the number of
Subject vocabulary repeating units which are linked together in the polymer.
repeating units a group of atoms Skill: Relationship of the structure of the monomer to the polymer and
that come from a monomer
and link repeatedly to form the repeating unit.
polymer chain
Monomer Polymer Uses
( ) ( )
Most common plastic:
H H H H used in plastic bags,
bottles, films.
n C C C C
H H H H n
ethene polyethene
( ) ( )
Second most common
H H H H plastic: used in clothing,
ropes, machine parts.
n C C C C
H CH3 H CH3 n
propene polypropene
( ) ( )
Third most common
H H H H plastic: used in pipes,
plastic cards, inflatable
n C C C C toys.
H Cl H Cl n
chloroethene polychloroethene
(vinyl chloride) (polyvinyl chloride, PVC)
( ) ( )
Used in non-stick
F F F F cookware, waterproof
and breathable clothing,
n C C C C medical plastics.
F F F F n
tetrafluoroethene polytetrafluoroethene, PTFE
(Teflon)
H C H Y
H C
C C C C H
addition reaction
+ X Y
C C C C
H C H H C H
no reaction
H H
• have a positive charge or a partial positive charge species that can act as Lewis acids
•
and accept electron pairs to form
are attracted to the electron-rich benzene ring. coordinate bonds
Benzene can undergo electrophilic substitution reactions, in which an electrophile electrophilic substitution
reactions substitution reactions
(E+) replaces a hydrogen atom: in which an electrophile replaces a
hydrogen atom
H E
H H H H
+ E+ 1 H+
H H H H
H H
chlorobenzene
Essential idea:
All measurement has a limit of precision and accuracy, and this must
be taken into account when evaluating experimental results.
•
associated with measurements
and are a result of natural observing that gas bubbles are given off during a reaction
variability due to the procedure
or instrument used to make the
measurement
• observing that a beaker gets hot during a reaction.
•
numbers or values to physical
properties have random errors (uncertainty) associated with the measurement.
numerical including numbers
• Examples of quantitative data:
mL 50
50
40
30 40
20
water
30
10 Close-up view
graduated cylinder
Hints for success: A calculated absolute uncertainty has only one significant
figure. If necessary the calculated value is adjusted so it has the same number of
decimal places as the calculated uncertainty.
• the uncertainty in the measured mass (0.09%) will have a very small effect on the
final propagated uncertainty
• the uncertainty in the measured volume (0.5%) will have a small effect on the
final propagated uncertainty
• the uncertainty in the measured temperature change (8%) will have the largest
effect on the final propagated uncertainty.
Student A Student B
Trial 1 –37.2 kJ mol–1 –43.6 kJ mol–1
Trial 2 –38.1 kJ mol–1 –48.2 kJ mol–1
Trial 3 –37.6 kJ mol–1 –40.2 kJ mol–1
Subject vocabulary
The results obtained by student B are closer to the accepted value of –44.5 kJ mol–1.
accurate the measured value is
similar to the known exact value This means that student B’s results are more accurate.
precise repeated measurements The results obtained by student A have a smaller range. This means that student A’s
give similar results results are more precise.
Essential idea:
Graphs are a visual representation of trends in data.
•
independent variable a variable
Dependent variable: the variable that is measured to determine the effect of (property) that can be changed or
changing the independent variable. modified to see how it changes
•
another variable
Drawing a graph with the independent variable on the x-axis and the dependent
dependent variable the variable
variable on the y-axis shows the relationship between the two variables. (property) that is measured to see
how it is affected by changes in
the independent variable
dependent
0 0
0 1
0 volume volume
pressure / kPa
200.0
pressure
150.0
100.0
50.0
volume 5 10 15 20 25 30 35
volume / dm3
• No scale on axes
• Axes are scaled
• No units on axes
• Axis titles include units
2.50
2.00
y = 0.1876x + 1.7607
pressure / kPa
1.50 R2 = 0.6375
1.00
0.50
0.00
0.00 2.00 4.00 6.00 8.00 10.00 12.00
1/volume / cm–3
-0.50
Figure 11.2 Pressure measured for a sample of air at different volumes.
( 1
volume
increases. )
The relationship does appear to be linear.
3.00
2.50
2.00
pressure / kPa
y = 2.563x – 0.0217
1.50 R2 = 0.9982
1.00
0.50
0.00
0.00 0.20 0.40 0.60 0.80 1.00 1.20
1 / volume / cm–3
Figure 11.3 Pressure plotted against 1 for a sample of air at different volumes.
volume
0: no relationship
The equation for the line of best fit for pressure against
using software analysis and is shown on the graph:
(1
volume )
was determined
y = 2.563x – 0.0217
The gradient and y intercept of the line can be determined from this equation (the
equation of a line is y = mx + c, where m is the gradient and c is the y intercept):
Essential idea:
Analytical techniques can be used to determine the structure of a
compound, analyse the composition of a substance, or determine
the purity of a compound. Spectroscopic techniques are used in the
structural identification of organic and inorganic compounds.
• shows how many H2 molecules are needed to saturate the compound measure of how many H2
•
molecules have to be added to a
provides clues to the structure of an unknown compound. compound to make it saturated
and non-cyclic
Saturated compounds IHD unsaturation multiple bonds
or rings
alkanes, alcohols, For example, H 0 saturate create a non-cyclic
halogenoalkanes, compound that has only
H O H H single bonds
amines, ethers H
C C C C N
H
C C C C
alkyne H H 2
C C
H C C H
cycloalkanes HH 1
CH2
(e.g. cyclopentane) H2C H2C CH2
CH2
H2C H2C CH2
CH2 CH2
nitriles H 2
H
R C N R C N
H
H
H H H H Mr = 15 H H
loss of
100 CH3
31 H Mr = 31
90
loss of OH
80 C+ O H
relative abundance
70
H
60
50
40 H H Mr = 29
45
30
15 H C C+
20
29
10 46
H H
0
0 10 20 30 40 50 60
mass/charge
Figure 11.4 The structure of ethanol and its mass spectrum. Possible fragmentation pattern produced when
ethanol is bombarded with high-energy electrons.
Hints for success: The masses of common fragment ions are given in section 28
of the IB data booklet.
Worked example
A molecule with an empirical formula CH2O has the simplified mass
spectrum below. Deduce the molecular formula and give a possible structure
of the compound.
100
43
90 45
80
relative abundance
70
60 15
50 60
40
30
20
10
0
0 10 20 30 40 50 60
mass/charge
Peaks Explanation
15 presence of CH3+
(60 – 45) loss of COOH from molecule
43 presence of C2H3O+
(60 – 17) loss of OH from molecule
45 presence of COOH+
(60 – 15) loss of CH3 from molecule
A structure consistent with this fragmentation pattern is:
H O
H C C O H
60
40
no broad absorption
in 3200–3600 cm–1
20 region shows that an
O–H bond is not present. C O
0
4000 3000 2000 1000
wavenumber/cm21
100
80
transmittance/%
60
O–H
40
No strong absorption around
20 1600 cm–1 shows that a
C–H C=O bond is not present.
0
4000 3000 2000 1600 1200 800
wavenumber/cm21
C2H6O is an alcohol.
•
environment of the nuclei. It has
units of parts per million (ppm) is the position of a peak in the 1H NMR spectrum
integrated area the total area
• has units of parts per million (ppm)
that is measured under a peak
• depends on the electronegativities of other atoms that are
in the molecule
(A large chemical shift occurs for hydrogen atoms close to
highly electronegative atoms)
Integrated areas The integrated area of a peak is proportional to the number
of hydrogen atoms in the chemical environment
Hints for success: The chemical shifts for various hydrogen chemical
environments are given in section 27 of the IB data booklet.
of protons
are in two different environments. O (1H) CH3
The integrated trace indicates
the relative number of hydrogen H C C
atoms in the two environments.
H CHO TMS (reference)
H
spectrum
trace
10 8 6 4 2 0
chemical shift/ppm
Worked example
The NMR spectrum of a compound that has the molecular formula
C3H8O is shown here.
3H
absorption
2H
1H 2H
10 9 8 7 6 5 4 3 2 1 0
chemical shift/ppm
H C C C O H H C O C C H
H H H
H H H H H H
H C C C H
I II III IV I II III
(3) (2) (2) (1) H O H (3) (2) (3)
propan-1-ol methoxyethane
H
I III I
(3) (1) (3)
propan-2-ol
(b) For each structure, I–IV identifies the different environments of the H atoms
in the molecule. 1–3 represents the number of atoms in each environment.
There are four peaks in the spectrum. Propan-1-ol has four peaks with
the correct areas.
(c) The peak at 0.9 ppm corresponds to the CH3 group.
Essential idea:
Subject vocabulary The quantized nature of energy transitions is related to the energy
quantized occurring with states of electrons in atoms and molecules.
discrete or specific values; not
continuous
transitions movement of an
electron from one energy level to
Understanding: In an emission spectrum, the limit of
another convergence at higher frequency corresponds to the first
emission spectrum the
frequencies of electromagnetic ionization energy.
•
radiation observed when a
high-energy species loses energy The energy levels in ∞
by emitting electromagnetic a hydrogen atom: 6 Ionization
•
radiation 5 4
converge (get closer in energy) 3 energy of
limit a final value hydrogen
as the value of n increases.
∞ the symbol for infinity
limit of convergence the
final value that the emission
• reach a limit in energy when
n = ∞.
2
electron
frequencies converge to
• n = ∞ is the energy level: removed
ground state the lowest energy
level of an atom or molecule • where the electron is so far from
the nucleus that it has been UV
from atom
convergence
ionization energy the energy removed from the atom limit for
required to remove one mole
of electrons from one mole of
gaseous atoms in their lowest
energy state
• where the hydrogen atom has
been ionized: H(g) ➝ H+(g) + e–.
emissions
to n = 1
• converge at higher frequency (because the energy levels are converging the
frequency of the emissions between them must also converge)
• hc c
is related to its wavelength, λ, by the equation E = (because ν = )
Solving problems using these equations requires:
λ λ
Worked example
The frequency of the electromagnetic radiation emitted in the transition
n = 2 ➝ n = 1 for the hydrogen atom is 24.66 × 1014 s–1. What is the energy
difference between the n = 2 and n = 1 energy levels of the hydrogen atom?
Solution
The energy difference between the n = 2 and n = 1 energy levels (ΔE) is equal to
the energy of the electromagnetic radiation emitted:
ΔE = E(radiation) = hν = 6.63 × 10–34 J s × 24.66 × 1014 s–1 = 1.64 × 10–18 J
The energy difference between the n = 2 and n = 1 energy levels is 1.64 × 10–18 J.
Skill: Calculation of the value of the first ionization energy from spectral
data that give the wavelength or frequency of the convergence limit.
Worked example
The convergence limit for emissions from higher energy levels to the n = 1
level of the hydrogen atom occurs at a frequency of 3.29 × 1015 s–1. Use this
convergence limit to:
(i) determine the ionization energy of hydrogen in kJ mol–1
(ii) determine the wavelength of electromagnetic radiation required to ionize a
hydrogen atom.
Solution
General vocabulary • provides evidence for the existence of energy levels and sub-levels in atoms.
• hydrogen (Z = 1) to helium (Z = 2)
• is added to the same energy level
• helium (Z = 2) to lithium (Z = 3)
• is added to an energy level further from the nucleus
• 2 2
beryllium (1s 2s ) to boron (1s 2s 2p ) 2 2 1 the extra electron is in a p orbital
• 2 2 3
nitrogen (1s 2s 2p ) to oxygen (1s 2s 2p ) 2 2 4 the extra electron is in an occupied p orbital
• phosphorus (1s2 2s2 2p6 3s2 3p3) to sulfur Electrons that are paired in p orbitals:
2 2 6 2
(1s 2s 2p 3s 3p )4
•
repel each other more
•
experience a weaker effective nuclear charge
•
are easier to remove from the atom
• the charge on the ion formed increases by one each time (the nth electron
removed forms an n+ ion)
First ionization The energy required to remove one mole of electrons M(g) ➝ M+(g) + e–
energy from one mole of atoms in the gaseous state
Second ionization The energy required to remove one mole of electrons M+(g) ➝ M2+(g) + e–
energy from one mole of 1+ ions in the gaseous state
Third ionization The energy required to remove one mole of electrons M2+(g) ➝ M3+(g) + e–
energy from one mole of 2+ ions in the gaseous state
• Similar definitions and equations are obtained for the fourth and higher
ionization energies.
Trend Explanation
The first ionization energy is The electron being removed:
the lowest
• is furthest from the nucleus
These trends can be observed in Figure 12.3, which shows the successive ionization
energies for aluminium.
the two electrons in the
first energy level closest
to the nucleus are the
these three these eight electrons most difficult to remove
log (ionization energy)
0 1 2 3 4 5 6 7 8 9 10 11 12 13
ionization number
Figure 12.3 Successive ionization energies for aluminium. Note the jumps between the 3rd and 4th and between
the 11th and 12th ionization energies as electrons start to be removed from lower energy levels. The wide range
in values is best presented on a log scale.
Worked example
The graph of successive ionization energies against number of electrons removed
is shown in Figure 12.4 for an unknown element. Not all ionization energies
are shown but the element has Z < 20. Use this graph to determine the identity
of the element.
60000
50000
ioniation energy / kJ mol–1
40000
30000
20000
10000
0
0 2 4 6 8 10 12
number of electrons removed
Figure 12.4 The successive ionization energies of an unknown element.
Solution
• A gradual increase in ionization energy occurs for the first six electrons
removed. A large increase in ionization energy occurs for the seventh electron
removed.
• This means the element has six electrons in its valence energy level and the
seventh electron is removed from an energy level closer to the nucleus. The
element must be from group 16.
• The unknown element cannot be oxygen. The graph shows that 11 electrons
have been removed from the unknown element and oxygen only has
eight electrons.
• The unknown element must be sulfur (Z = 16) as it is the only other group 16
element with Z < 20.
Essential idea:
The transition elements have characteristic properties; these
properties are related to their all having incomplete d sub-levels.
•
d sub-level or can form ions with
a partially filled d sub-level are defined as elements that have partially filled d sub-levels or form ions
ligands species that donate with partially filled d sub-levels
a lone pair of electrons to a
• can have more than one oxidation state
•
transition metal ion and form a
coordinate bond can covalently bond with ligands to form complex ions
complex ions ions that contain a
• are often used as catalysts (have catalytic properties)
transition metal bonded to ligands
• show magnetism when placed in a magnetic field (have magnetic properties).
• can lose the 4s and 3d electrons when forming ions because the 4s and 3d
are the two highest energy sub-levels and they have similar energies
• can lose both 4s electrons to give the +2 oxidation state because the 4s
energy sub-level is lower than the 3d sub-level for transition element ions
• it is easy for titanium in the +1 oxidation state to lose the second 4s electron and
form the +2 oxidation state.
The +5 oxidation state of titanium is not common because:
• the fifth ionization energy of titanium is very high because the fifth electron
must be taken from a 3p subshell that is closer to the nucleus
• it is hard to remove the fifth electron and form the +5 oxidation state.
The common oxidation states of titanium are +2, +3, and +4.
Skill: Explanation of the nature of the coordinate bond within a complex ion. Subject vocabulary
Transition element ions: coordinate bond (also known
• can act as Lewis bases and accept an electron pair from a ligand, which acts
as a Lewis acid
of the electrons that are shared in
the bond
coordination number the
Mn+ L number of ligands that are
bonded to the central transition
Mn+ = transition element ion; L = ligand metal in a complex ion
• form coordinate bonds (dative bonds) with ligands because the ligand donates
both of the electrons that form the covalent bond
• form complex ions when they bond with one or more ligands.
The coordination number of a complex ion is the number of ligands that are
bonded to the transition element ion.
Different coordination numbers can give different shapes for the complex ions.
Figure 13.1 Shapes and
coordination numbers of some
L L complex ions.
L Mn+ L Mn+
L L
shape: linear shape: square planar
coordination number = 2 coordination number = 4
L L
L L
Mn+ Mn+
L L L
L
L L
shape: tetrahedral shape: octahedral
coordination number = 4 coordination number = 6
• include all the ligands bonded to the central transition element ion.
Examples: [Fe(OH2)6]2+, [Cr(NH3)4Cl2]3+, [CuCl4]2–
Skill: Deduction of the total charge [of a complex ion] given the formula of
the ion and ligands present.
Ligands can be negatively charged ions or neutral species.
Hints for success: Common ligands and their charges are given in section 15 of
the IB data booklet.
The total charge on a complex ion will depend on:
(i) the charge on the transition metal ion
(ii) the charge on the ligand(s)
(iii) the number of charged ligands bonded to the transition metal ion.
Worked example
Determine the formula and charge of the complex ion formed when a
chromium(III) ion bonds to four water ligands and two chloride ligands.
Solution
General vocabulary
• for transition metals and transition metal ions that have unpaired electrons.
Paramagnetism is stronger than diamagnetism so transition metals and transition
induces causes to happen metal ions with unpaired electrons will be paramagnetic.
•• Zinc:
has the electron configuration [Kr] 3d10 4s2
• only forms the Zn2+ ion, which has the electron configuration [Kr] 3d10
Essential idea:
All d orbitals have the same energy in an isolated atom, but they
split into two sub-levels in a complex ion. The electric field of ligands
may cause the d orbitals in complex ions to split so that the energy
of an electron transition between them corresponds to a photon of
visible light.
• the energies of the two d orbitals aligned on the x-, y-, and z-axes increase
• the energies of the three d orbitals not aligned on the x-, y-, and z-axes decrease
• the d orbitals are split into two sets that have different energies.
Figure 13.2 The splitting of the
energies of d orbitals in an
octahedral complex.
ΔE
Energy
• the complex moves from the ground state to an excited state the colour spectrum
• the colour that is observed depends on the size of the splitting energy and
the frequency of the visible light that is absorbed
• the colour that is observed is the complementary colour of the light that
is absorbed.
• A colour wheel shows the relationship between the colour that is observed for
a complex ion and the colour of light that is absorbed by the complex ion. The
colour that is observed is on the opposite side of the wheel from the colour
that is absorbed.
Hints for success: Section 17 in the IB data booklet provides a simple colour Figure 13.3 The colour wheel.
There is a colour wheel in
wheel that relates complementary colours. section 17 of the IB data booklet.
Skill: Explanation of the effect of the identity of the metal ion, the oxidation
number of the metal, and the identity of the ligand on the colour of
transition metal ion complexes.
• The colour of light that is observed for any complex ion depends on the size of
the splitting energy (ΔE) between the two sets of d orbitals in the complex.
• The size of the splitting energy depends on the strength of the attraction
between the transition metal ion and the ligands. The greater the attraction:
• the greater the repulsion between d electrons on the metal and the lone
pairs of electrons on the ligands
Subject vocabulary
• the identity of the ligand and where it appears on the
spectrochemical series.
spectrochemical series a
list of ligands that orders Factor Examples Effect on splitting energy (ΔE)
them according to the size
of the splitting energy they Identity and [Cr(OH2)6]3+ The Fe3+ ion (Z = 26) has a greater nuclear
produce when bonding to
nuclear charge charge than the Cr3+ ion (Z = 24)
transition elements [Fe(OH2)6]3+
of the transition
Fe3+ exerts a greater attraction on the ligands
metal ion
and causes a bigger splitting energy
[Fe(OH2)6]3+ absorbs higher energy purple light
and appears yellow-brown
[Cr(OH2)6]3+ absorbs lower energy red light and
appears green
Oxidation state [Fe(OH2)6]2+ Fe3+ has a greater charge than Fe2+
of the transition
[Fe(OH2)6]3+ Fe3+ exerts a greater attraction on the ligands
metal ion
and causes a bigger splitting energy
[Fe(OH2)6]3+ absorbs higher energy purple light
and appears yellow-brown
[Fe(OH2)6]2+ absorbs lower energy red light and
appears pale green
Identity of the [CrF6]3– NH3 is higher on the spectrochemical series for
ligand and its 3+ ligands than F–
[Cr(NH3)6]
position on the
NH3 is more strongly attracted to the Cr3+ ion
spectrochemical
and causes a bigger splitting energy
series
[Cr(NH3)6]3+ absorbs higher energy purple light
and appears yellow
[CrF6]3– absorbs lower energy green light and
appears pink
Essential idea:
Larger structures and more in-depth explanations of bonding systems
often require more sophisticated concepts and theories of bonding.
•
in-depth ........ covered in more
detail or involving a higher level of is formed from the overlap of atomic orbitals on two atoms
understanding
• involves a pair of electrons that move through both overlapping orbitals and
are attracted to the nuclei of both atoms.
Subject vocabulary • There are two different types of covalent bonds: sigma (σ) bonds
and pi (π) bonds.
sigma (σ) bonds covalent bonds
formed by head-on overlap of
atomic orbitals Skill: Prediction of whether sigma (σ) or pi (π) bonds are formed from the
linear combination of atomic orbitals.
pi (π) bonds covalent bonds
formed by side-on overlap of Sigma bonds are formed from the head-on overlap of atomic orbitals. These
atomic orbitals
orbitals can be:
• two s orbitals
• two p orbitals
Later in the chapter we will see that sigma bonds are also formed from the
head-on overlap of hybrid orbitals.
In a double bond one bond is a sigma bond and the other is a pi bond.
In a triple bond one bond is a sigma bond and the other two are pi bonds.
• is the number of valence electrons on the unbonded atom (V) – (half the
number of bonding electrons (B) on the bonded atom in the molecule + the
number of non-bonding electrons (L) on the bonded atom in the molecule):
1
FC = V – (2B + L)
• allows the best Lewis structure to be determined for a molecule when there
is more than one possible Lewis structure.
Hints for success: The sum of the formal charges on all the atoms in a Lewis
structure must equal the overall charge on the Lewis structure.
– – – – –
O C N O O C C N N O O C C N N
Step 1 1
FC(O) = 6 – (2(6) + 2) = +1
1
FC(O) = 6 – (2(4) + 4) = 0
1
FC(O) = 6 – (2(2) + 6) = –1
1 1 1
FC(C) = 4 – 2(8) = 0 FC(C) = 4 – 2(8) = 0 FC(C) = 4 – 2(8) = 0
1 1 1
FC(N) = 5 – (2(2) + 6) = –2 FC(N) = 5 – (2(4) + 4) = –1 FC(N) = 5 – (2(6) + 2) = 0
• Expanded octets:
• occur when an atom has more than eight electrons in its valence shell
• can happen for atoms that have empty d orbitals and are able to accept
electron pairs from other atoms
• can happen for atoms that are in the third or higher periods.
PF6– SF4
Step 1: Total valence electrons 5 + (6 × 7) + 1 = 48 6 + (4 × 7) = 34
Steps 2 and 3: Skeletal F F
structure with bonds shown
F F
as lines
P F S F
F F
F F
• VSEPR theory can also be applied to Lewis structures with five and six electron
domains around the central atom. Subject vocabulary
•
is at 90 degrees to the vertical axis
have three electron domains in the equatorial eq of a molecule is the equatorial
90° plane
plane (eq) at 120° to each other
F
see-saw F 117°
4 1 (unsymmetrical S F
90°
tetrahedron)
F
F
90°
3 2 T-shaped Cl F
–
I
2 3 linear I
octahedral F
F F
6 6 0 octahedral S
F F
F
F
square F F
5 1 Br
pyramidal
F F
F F
4 2 square planar Xe
F F
•
between two atoms
Experimental evidence shows that the O–O bonds in ozone (O3):
•
dissociate to break bonds and
are the same length separate atoms in a molecule
• This experimental evidence shows that the pi bond (and the pair of pi
electrons in the bond):
• is delocalized.
• The true structure that results from the delocalization of the pi electrons is
known as a resonance hybrid.
O O
O
Figure 14.4 In a resonance hybrid the dashed lines represent delocalized pi electrons.
• Bond order:
•
a single bond has a bond order of 1
•
a double bond has a bond order of 2
•
a triple bond has a bond order of 3.
• need UV light of a higher energy (shorter wavelength) to break them than the
bonds in ozone (O3).
• Ozone (O3):
• is formed when nitrogen (N2) reacts with oxygen (O2) in car engines because
of the high temperatures:
N2(g) + O2(g) ➝ 2NO(g)
Subject vocabulary • is a free radical (contains an unpaired electron)
free radical a chemical species
that contains one or more
• is very reactive.
unpaired electrons Hints for success: A dot is added to the chemical formula of a free radical to
chlorofluorocarbons organic signify an unpaired electron, e.g. nitrogen monoxide is written as NO•.
compounds (halogenoalkanes)
that contain both chlorine and
fluorine substituents • Nitrogen monoxide reacts with ozone in a two-step mechanism:
Step 1: NO•(g) + O3(g) ➝ NO2(g) + O2(g)
refrigerants substances used as
coolants in refrigerators Step 2: NO2(g) + O(g) ➝ NO•(g) + O2(g)
aerosols substances used in Overall reaction: O3(g) + O(g) ➝ 2O2(g)
spray cans
• Nitrogen monoxide (NO•):
• can take part in many reactions and destroy many ozone molecules.
• Chlorine atoms:
• can take part in many reactions and destroy many ozone molecules.
Essential idea:
Hybridization results from the mixing of atomic orbitals to form the
same number of new equivalent hybrid orbitals that can have the
Synonym same mean energy as the contributing atomic orbitals.
mean . ........... average
•
orbitals
There are three types of hybrid orbitals: sp3, sp2, and sp hybrids.
•
unhybridized atomic orbitals
that do not mix to form hybrid Sigma bonds are formed from the overlap of hybrid orbitals.
orbitals are unhybridized
• Pi bonds are formed from the overlap of unhybridized p orbitals.
sp3
• forms four sp3 hybrids that are equal in energy and are at 109.5° to
each other.
C 2p
sp3 sp3
H H
sp3 2s
sp3 hybridization
four sp3 hybrid orbitals
H
Figure 14.5 sp3 hybridization.
•
orbitals, which each overlap with
the s orbital of an H atom. All occurs when one s orbital mixes with two p orbitals
bonds are sigma bonds.
• forms three sp2 hybrids that are equal in energy and are at 120° to each other
one unhybridized
2p p orbital
2s
sp2 hybridization three sp2 hybrid
orbitals
Figure 14.7 sp2 hybridization.
• forms two sp hybrids that are equal in energy and are at 180° to each other
two unhybridized
2p p orbitals
2s
sp hybridization two sp hybrid
orbitals
Figure 14.9 sp hybridization.
• The examples given above show the relationship between the electron domain
geometry of an atom and the type of hybridization.
Example Hybridization of Number of Electron domain
carbon atoms electron domains geometry
methane sp3 4 tetrahedral
ethene sp2 3 trigonal planar
ethyne sp 2 linear
CH3COOH CH3CN
Lewis structure H O H
H C C O H H C C N
H H
Atom in structure C C O O C C N
Number of
4 3 3 4 4 2 2
electron domains
Hybridization sp3 sp2 sp2 sp3 sp3 sp sp
Essential idea:
The concept of the energy change in a single-step reaction being
equivalent to the summation of smaller steps can be applied to
changes involving ionic compounds.
standard state
M(g) + ½X2(g)
2
ΔHlat (MX)
M(s) + ½X2(g)
Hints for success: First electron affinities for non-metals are exothermic.
From the Born–Haber cycle we can see that the lattice enthalpy is related to the
enthalpies of the other reactions:
ΔHlat(MX) = –1 + 2 + 3 + 4
= –ΔHf (MX) + ΔHatom(M) + ΔHatom(X) + ΔHi (M) + ΔHe (X).
The total atomization energy (2 in Figure 15.1) includes the atomization of the
metal (M) and the non-metal (X).
Hints for success: Enthalpy values are added when following the direction of
arrows in the Born–Haber cycle and subtracted when going against an arrow.
ΔHe (Cl)
ΔHi (Na) Na+(g) + Cl–(g)
Na(g) + Cl(g)
ΔHatom (Cl)
Enthalpy (H)
Na(g) + ½Cl2(g)
ΔHlat (NaCl)
ΔHatom (Na)
Na(s) + ½Cl2(g)
ΔHf (NaCl)
NaCl(s)
ΔHe (O) +
Mg (g) + 2e + O(g)
2+ – ΔH2e (O)
ΔHi (Mg) +
ΔH2i (Mg)
Mg(g) + O(g)
Enthalpy (H)
ΔHatom (O)
Mg(g) + ½O2(g) ΔHlat (MgO)
ΔHatom (Mg)
Mg(s) + ½O2(g)
ΔHf (NaCl)
MgO(s)
δ+ δ+
b c
δ+ δ–
δ+
δ+ δ+
δ– δ+
δ+
δ– δ+
+ δ– –
δ–
δ+ δ– δ+
δ+
δ+
δ+ δ+
δ+ δ–
172 Topic 15: Energetics/thermochemistry
Hints for success: Values for enthalpies of hydration are given in section 20 of the
IB data booklet.
A dissolution energy cycle can be constructed using the enthalpy of solution,
lattice enthalpy, and the hydration enthalpies of the ions.
Figure 15.5 The dissolution
ΔHlat energy cycle for an ionic solid
MX(s) M+(g) + X–(g) MX. M represents the metal and
(i) X represents the non-metal.
(ii)
ΔHsol M (aq) + X (aq)
+ – ΔHhyd (M+) + ΔHhyd (X–)
Enthalpies of solution:
Start by constructing a dissolution energy cycle for NaOH showing the relationship between
the lattice enthalpy, enthalpy of solution, and hydration enthalpies of the ions.
ΔHlat (NaOH)
NaOH(s) (i) Na+(g) + OH–(g)
(ii)
ΔHsol (NaOH) Na (aq) + OH (aq)
+ – ΔHhyd (Na+) + ΔHhyd (OH–)
From section 20 in the IB data booklet: ΔHhyd(Na+) = –424 kJ mol–1 ΔHhyd(OH–) = –519 kJ mol–1
ΔHsol(NaOH) = ΔHlat(NaOH) + ΔHhyd(Na+) + ΔHhyd(OH–)
= +900 kJ mol–1 + (–424 kJ mol–1) + (–519 kJ mol–1)
= –43 kJ mol–1
Skill: Relate size and charge of ions to lattice and hydration enthalpies.
Lattice enthalpies are related to the strength of the ionic bonding between the
cation and anion. The stronger the ionic bond, the larger the lattice enthalpy.
The size of lattice enthalpies depends on:
(i) the ionic radius of the ions: smaller ions give larger lattice enthalpies
decreasing size of anion
F– Cl– Br– I–
decreasing size of cation
(ii) the charge on the ions: bigger charges give larger lattice enthalpies
NaCl 1049 MgCl2 2540 MgCl2 has more than double the lattice
enthalpy of NaCl, as Mg2+ has double
the charge of Na+ and a smaller ionic
radius.
CaF2 2651 CaO 3401 CaO has a higher lattice enthalpy than
CaF2, as O2– has double the charge
of F–. The value is less than double as
O2– has a larger ionic radius than F–.
(ii) the charge on the ions: bigger charges give larger hydration enthalpies
Na+ −424
Mg2+ −1963
Al3+ −4741
Essential idea:
General vocabulary A reaction is spontaneous if the overall transformation leads to an
transformation change increase in total entropy (system plus surroundings). The direction
of spontaneous change always increases the total entropy of the
universe at the expense of the energy available to do useful work.
This is known as the second law of thermodynamics.
•
units J K–1 mol–1
It increases as temperature increases.
Synonym
disorder ........ randomness
• A greater disorder (and higher entropy) occurs when there are more ways
to distribute energy through the system. When entropy increases, energy is
dispersed ...... spread out
dispersed and is not available to do useful work.
INCREASING ENTROPY
Solid Liquid Gas
• Particles in fixed positions • Particles can move • Particles move freely
• Little distribution of energy • Some distribution of energy • Large distribution of energy
• Lower disorder (highly ordered) • More disordered • Highly disordered
• Low entropy • Higher entropy • Highest entropy
Worked example
Predict the entropy change (ΔS) for the following reactions or processes:
(i) Br2(l) ➝ Br2(g)
(ii) Ag+(aq) + Br–(aq) ➝ AgBr(s)
(iii) CH4(g) + 2O2(g) ➝ CO(g) + 2H2O(l)
(iv) 2SO2(g) + O2(g) ➝ 2SO3(g)
Solution
Hints for success: Standard entropy values are given in section 12 of the
IB data booklet.
Worked example
Calculate the standard entropy change (ΔS ) that occurs for the following reaction
using the information provided and the IB data booklet.
CH4(g) + 2O2(g) ➝ CO2(g) + 2H2O(l) S (O2) = +205.0 J K–1 mol–1
Solution
Hints for success: The S values for a compound must be multiplied by the
coefficient of that compound.
• ΔH is negative
• ΔS is positive.
But some reactions that are spontaneous have a positive ΔH or negative
ΔS, for example:
Subject vocabulary The change in Gibbs free energy (ΔG) that occurs determines whether a reaction is
spontaneous or not:
Gibbs free energy the energy
of a reaction that is available to ΔG = ΔH – TΔS
do work
These are the units that are typically given with each term in the equation.
ΔH ΔS ΔG (= ΔH – TΔS) Comments
Positive Negative Always positive Non-spontaneous at all
(endothermic) (less disordered temperatures
products) (Products are always less
stable and less disordered
than reactants)
Negative Positive Always negative Spontaneous at all
(exothermic) (more temperatures
disordered (Products are always more
products) stable and more disordered
than reactants)
Worked example
Calculate the value of ΔG for the reaction of ethanol (CH3CH2OH) and oxygen
(O2) to produce ethanoic acid (CH3COOH) and water (H2O) at 298 K using the
ΔHf and S values in section 12 of the IB data booklet.
Predict whether the reaction will be spontaneous or non-spontaneous at
this temperature.
CH3CH2OH(l) + O2(g) ➝ CH3COOH(l) + H2O (l) S (O2) = +205.0 J K–1 mol–1
Solution
ΔS = ΣS (products) – ΣS (reactants)
= [S (CH3COOH) + S (H2O(l)] – [S (CH3CH2OH) + S (O2)]
= [+160 + +70.0] J K–1 mol–1 – [+161 + +205.0] J K–1 mol–1
= –136 J K–1 mol–1
= –0.136 kJ K–1 mol–1
Hints for success: Standard Gibbs free energy of formation values are given in
section 12 of the IB data booklet.
• ΔGf of all elements in their standard states is zero, for example ΔGf (N2) = 0.
Worked example
Calculate the value of ΔG for the reaction of ethanol (CH3CH2OH) and oxygen
(O2) to produce ethanoic acid (CH3COOH) and water (H2O) at 298 K using the
ΔGf and S values in section 12 of the IB data booklet.
CH3CH2OH(l) + O2(g) ➝ CH3COOH(l) + H2O(l)
Hints for success: The ΔGf values for a compound must be multiplied by the
coefficient of that compound.
Solution
A+B C+D
Reverse reaction
Essential idea:
Rate expressions can only be determined empirically. They limit
possible reaction mechanisms. In particular cases, such as a linear
chain of elementary reactions, no equilibria, and only one significant
activation barrier, the rate equation is equivalent to the slowest step
of the reaction.
Subject vocabulary Reactions occur via a series of elementary steps involving one or two reactant
particles. The overall sequence of steps is known as the reaction mechanism.
elementary step a single step in
a reaction mechanism Elementary steps in a reaction mechanism add together to give the overall reaction.
reaction mechanism a
description of a reaction as a
A possible two-step reaction mechanism for the reaction
series of elementary steps 2NO(g) + O2(g) ➝ NO2(g) is:
reaction intermediate a species Step 1: NO(g) + NO(g) ➝ N2O2(g)
made in one elementary step and
used in another step. It does not Step 2: N2O2 (g) + O2(g) ➝ 2NO2(g)
appear in the overall equation for
the reaction Overall: 2NO(g) + O2(g) ➝ 2NO2(g)
rate-determining step the
slowest elementary step in a
N2O2 is a reaction intermediate because it is a product in one step and is used
reaction mechanism as a reactant in another step. Reaction intermediates do not appear in the
overall equation.
Experimental evidence can show if a reaction mechanism is possible but cannot
prove that it is the correct mechanism.
Example: if N2O2 is observed experimentally when NO reacts with O2 then the
reaction mechanism above is possible. However, it would not prove that this is
the actual mechanism because other reaction mechanisms might also predict the
formation of N2O2.
Elementary steps in a reaction mechanism occur with different rates. The rate-
determining step (RDS) is the slowest step and it controls the overall rate of the
reaction because it has the largest activation energy.
reactants
ΔH
products
extent of reaction
• k is the rate constant for the elementary step depends on the concentrations of
•
the reactants
A and B represent the reactant particles in the elementary step
•
rate constant a constant that
x and y are equal to the coefficients of the reactants in the elementary step. relates the rate of a reaction to the
concentrations of the reactants
Elementary step Number of reactant Molecularity Rate expression for order of reaction the number of
particles elementary step particles of a particular reactant
that are involved in the rate-
A ➝ products 1 unimolecular rate = k[A] determining step
Worked example
Determine the reaction order for each reactant and the overall reaction order
for the reaction:
NO(g) + H2(g) ➝ H2NO(g)
The experimentally determined rate expression is rate = [NO]2[H2].
Solution
Hints for success: Notice how the coefficients in the rate expression cannot be
predicted from the reaction stoichiometry.
time [A]
rate
[A]
straight line that
goes through (0, 0).
Gradient = k
time [A]
rate
[A]
through (0, 0)
time [A]
first
zero order
order
zero
order
rate
[A]
first
order second
order
second
order
time [A]
Figure 16.2 Concentration–time graphs for Figure 16.3 Rate–concentration graphs for
zero-, first-, and second-order reactions. zero-, first-, and second-order reactions.
Worked example
Use the data in the table below to work out the order of reaction with respect to
reactants A and B, and so write the rate expression for this reaction.
Worked example
Nitrogen dioxide (NO2) reacts with carbon monoxide (CO) to form nitrogen oxide
(NO) and carbon dioxide (CO2):
NO2(g) + CO(g) ➝ NO(g) + CO2(g)
The rate expression determined experimentally is rate = k[NO2]2.
Determine which of the two mechanisms below is possible by determining the
rate expression for each mechanism and comparing it to the experimental result.
Reaction mechanism 1
Step 1: NO2(g) + NO2(g) ➝ NO(g) + NO3(g) (slow)
Step 2: NO3(g) + CO(g) ➝ NO2(g) + CO2(g)
Overall: NO2(g) + CO(g) ➝ NO(g) + CO2(g)
Reaction mechanism 2
Step 1: NO2(g) + NO2(g) ➝ NO(g) + NO3(g)
Step 2: NO3(g) + CO(g) ➝ NO2(g) + CO2(g) (slow)
Overall: NO2(g) + CO(g) ➝ NO(g) + CO2(g)
Solution
Reaction mechanism 1
Rate expression for RDS = k[NO2]2 because RDS is step 1
Overall rate expression = k[NO2]2
Reaction mechanism 2
Step 2 is RDS
Rate expression for RDS = k[NO3][CO]
NO3 is a reaction intermediate; from step 1: [NO3] ∝ [NO2]2
Overall rate expression = k[NO2]2[CO]
Mechanism 1 is a possible reaction mechanism because it gives a rate expression
that is the same as the experimental rate expression.
Ea
reactants
catalysed
ΔH
products
extent of reaction
Essential idea:
The activation energy of a reaction can be determined from the effect
of temperature on reaction rate.
• The units of A are the same as the units of k and depend on the order
of the reaction.
frequency of collisions between
reactant particles that occur with
the correct orientation for the
reaction to happen
Skill: Describing the relationships between temperature and rate constant, reactive site the part of a
frequency factor, and complexity of molecules colliding. molecule where the reaction
happens
From the Arrhenius equation the following relationships can be determined:
•
–1
k ∝ e T : this means that the value of the rate constant (k) increases as the
temperature (T) increases
• k ∝ A: this means that the value of the rate constant (k) increases as the
frequency factor (A) increases.
The value of A decreases as reactants increase in complexity and contain more General vocabulary
atoms. When reactants contain more atoms the probability that the collision complexity level of complication
occurs with the correct orientation and happens at the reactive site of the
molecule decreases.
–Ea
Skill: Using the Arrhenius equation k = Ae RT
Worked example
The reaction
2N2O5(g) ➝ 4NO2(g) + O2(g)
has an activation energy (Ea) of 1.0 × 102 kJ mol–1. At 20 °C the rate constant
k = 2.0 × 10–5 s–1.
Use this information to determine the frequency factor (A) for this
reaction at 20 °C.
e RT e 8.314 J–1
K mol × –1
293 K
( )
–E 1
ln k = a + ln A
R T
1
This is in the form of the equation of a line (y = mx + c) so a graph of ln k vs will
–E T
give a straight line with gradient = a and y intercept = ln A.
R
Skill: Determining and evaluating values of activation energy and frequency
factors from data.
Convert the data in the table and plot ln k vs 1/T. (Note: the temperature must be
converted from °C to kelvin.)
26
24
22
20
18
16
14
12
10
8
6 1 –1
(K )
4 T
2
0 0.0005 0.001 0.0015 0.002
0
–2
–4
–6
ln k
–8
–10
–12
–14
–16
The equation of the line shown in the graph can be determined using
a graphing calculator:
1
ln k = –2.24 × 104 × + 25.3
T
Ea = –R × gradient = –8.31 J K–1 mol–1 × –2.24 × 104 K
= 1.86 × 105 J mol–1
= 186 kJ mol–1
y intercept = ln A = 25.3
A = e25.3 = 9.72 × 1010 mol–1 dm3 s–1
Essential idea:
The position of equilibrium can be quantified by the equilibrium law.
The equilibrium constant for a particular reaction only depends on
the temperature.
First write the equation for the reaction, making sure it is correctly balanced:
CO(g) + H2O(g) H2(g) + CO2(g)
Next write the equilibrium expression:
[H ] [CO2]
Kc = 2
[CO] [H2O]
Now substitute the given values for each component:
(0.200) (0.0200)
Kc = = 1.84
(0.150) (0.0145)
Hints for success: The ICE method can be used to calculate Kc using initial and equilibrium
ICE stands for: concentrations.
Initial 1 Write the balanced equation.
Change 2 Under the equation, write in the values of the concentrations of each
Equilibrium component using three rows: initial, change, and equilibrium.
Worked example
A student placed 0.20 mol of PCl3(g) and 0.10 mol of Cl2(g) into a 1.0 dm3 flask at
350 °C. The reaction, which produced PCl5, was allowed to come to equilibrium,
at which time it was found that the flask contained 0.12 mol of PCl3. What is the
value of Kc for this reaction?
Solution
Worked example
The reaction:
CO(g) + 2H2(g) CH3OH(g)
has Kc = 0.500 at 350 K. If the concentrations at equilibrium are:
[CO] 0.200 mol dm–3
[H2] 0.155 mol dm–3
what is the equilibrium concentration of CH3OH?
Worked example
The thermal decomposition of water has a very small value of Kc. At 1000 °C,
Kc = 7.3 × 10–18 for the reaction
2H2O(g) 2H2(g) + O2(g)
A reaction is set up at this temperature with an initial H2O concentration of
0.10 mol dm–3. Calculate the H2 concentration at equilibrium.
Solution
• the minimum in total Gibbs free energy is positioned closer to the right of the
graph where there are more products than reactants
ΔGnegative
spontaneous
reaction,
ΔG negative
non- spontaneous
spontaneous reaction,
reaction, ΔG negative Gproducts
minimum ΔG positive
free energy
non-spontaneous reaction, ΔG positive
• the minimum in total Gibbs free energy is positioned closer to the left of the
graph where there are more reactants than products
non-
spontaneous
reaction, ΔGpositive
ΔG positive spontaneous
reaction,
spontaneous ΔG negative
reaction,
Greactants
ΔG negative
minimum non-spontaneous
free energy reaction, ΔG positive
reactants equilibrium products
only mixture only
mostly reactants
Hints for success:
When using ΔG = –RTln The relationship between ΔG and Kc is given by the equation:
Kc the value of ΔG ΔG = –RTln Kc
must be converted from
where R is the universal gas constant (8.31 J K–1 mol–1), T is the temperature in kelvin
kJ mol–1 to J mol–1 to be
(K), and Kc is the equilibrium constant for the reaction.
consistent with the units
of R ( J K–1 mol–1). Skill: Calculations using the equation ΔG = –RTln K.
Worked example
The reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3) is
an equilibrium reaction. At 298 K the value of Kc is 1.45 × 10–6.
N2(g) + 3H2(g) ➝ 2NH3(g)
Use this information to determine the value of the standard change in Gibbs free
energy (ΔG ) for this reaction at 298 K.
Solution
ΔG = –RTln Kc
= –8.31 J K–1 mol–1 × 298 K × ln(1.45 × 10–6)
= –33 300 J mol–1
= –33.3 kJ mol–1
Worked example
The esterification reaction that produces ethyl ethanoate has a free energy change
ΔG = –4.38 kJ mol–1.
CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(aq)
Calculate the value of the equilibrium constant of this reaction at 298 K.
Solution
ΔG = –RT ln Kc
ΔG = –4.38 kJ mol–1 = –4.38 × 1000 J mol–1
∴ –4.38 × 1000 J mol–1 = –(8.31 J K–1 mol–1) × (298 K) × ln Kc
–4380 J mol–1
ln Kc = = 1.77
2476 J mol–1
∴ Kc = e1.77 = 5.9
Essential idea:
The acid–base concept can be extended to reactions that do not
involve proton transfer.
B N H F B N H
F F H F H
Brønsted–Lowry acids and bases are also Lewis acids and bases:
+
H H
H+ + N H H N H
H H Subject vocabulary
nucleophiles electron-rich
electron pair acceptor/ electron pair donor/ species that can act as Lewis bases
Lewis acid Lewis base and donate electron pairs to form
coordinate bonds
proton donor/ proton acceptor/
Brønsted–Lowry acid Brønsted–Lowry base electrophiles electron-poor
species that can act as Lewis acids
and accept electron pairs to form
coordinate bonds
Nucleophiles Electrophiles
Electron-rich species Electron-poor species
Donate electron pairs to form Accept electron pairs to form
coordinate bonds coordinate bonds
Are Lewis bases Are Lewis acids
Examples: OH–, NH3, Cl– Examples: BF3, Br+, (CH3)3C+
– – F CH3
H O N Cl +
H H B Br C+
H
F F H3C CH3
Essential idea:
The equilibrium law can be applied to acid–base reactions. Numerical
problems can be simplified by making assumptions about the relative
concentrations of the species involved. The use of logarithms is also
significant here.
Skill: Discussion of the relative strengths of acids and bases using values of
Ka, pKa, Kb, and pKb.
An inverse relationship exists between the strength of a weak acid and the strength
of its conjugate base.
Figure 18.1 The relative
strengths of some acids and their
Acid Base
conjugate bases in strongest acid HClO4 ClO4– weakest base
aqueous solution. HCl Cl–
H2SO4 HSO4–
increasing H3O+ H2O increasing
acid CH3COOH CH3COO– base
strength strength
H2S SH–
NH4+ NH3
H2O OH–
weakest acid C2H5OH C2H5O– strongest base
• The pH (or pOH) of a solution refers to the concentration of H+ ions (or OH–
ions) at equilibrium.
• The concentration values substituted into the expressions for Ka and Kb must be
the equilibrium values for all reactants and products.
• When the extent of dissociation is very small (very low value for Ka or Kb) it is
appropriate to use the approximations:
[acid]initial ≈ [acid]equilibrium
[base]initial ≈ [base]equilibrium
Calculating Ka from initial and equilibrium concentrations
Worked example
Calculate Ka at 298 K for a 0.01 mol dm–3 solution of ethanoic acid (CH3COOH).
It has a pH of 3.4 at this temperature.
Solution
Write the equation for the dissociation of the acid. Insert the data in three rows:
initial, change, and equilibrium. As in Chapter 17, numbers in black are data that
were given in the question, numbers in blue have been derived.
From the pH we get the [H+] at equilibrium:
pH 3.4 ⇒ [H+] = 10–3.4 = 4.0 × 10–4 mol dm–3
From the stoichiometry of the reaction we know that [H+] = [CH3COO–]
CH3COOH(aq) CH3COO–(aq) + H+(aq)
–3
initial (mol dm ) 0.01 0.00 0.00
change (mol dm–3) –4.0 × 10–4 +4.0 × 10–4 +4.0 × 10–4
equilibrium (mol dm–3) 0.01 – (4.0 × 10–4) 4.0 × 10–4 4.0 × 10–4
≈ 0.01
The approximation 0.01 ≈ 0.01 – (4.0 × 10–4) is valid within the
precision of this data.
Write the expression for Ka and substitute the equilibrium values.
[CH3COO–] [H+] (4.0 × 10–4)2
Ka = = = 1.6 × 10–5
[CH3COOH] 0.01
Worked example
Calculate the Kb for a 0.100 mol dm–3 solution of methylamine, CH3NH2, at 25 °C .
Its pH is 11.80 at this temperature.
Solution
Worked example
A 0.75 mol dm–3 solution of ethanoic acid has a value for Ka = 1.8 × 10–5 at a
specified temperature. What is its pH at this temperature?
Solution
Worked example
A 0.20 mol dm–3 aqueous solution of ammonia has Kb of 1.8 × 10–5 at 298 K .
What is its pH?
Solution
Essential idea:
pH curves can be investigated experimentally but are mathematically
determined by the dissociation constants of the acid and base. An
indicator with an appropriate end-point can be used to determine the
Subject vocabulary equivalence point of the reaction.
equivalence point where
stoichiometricallly equal amounts
of acid and base have neutralized
each other
Understanding: The composition and action of a
buffer solutions solutions that buffer solution.
resist any change in pH when
small amounts of acid or base Buffer solutions:
•
are added
show little change in pH when small amounts of acid or base are added to them
General vocabulary
composition make-up,
• contain a conjugate pair of a weak acid/base.
ingredients Skill: While the nature of the acid–base buffer always remains the same,
buffer solutions can be prepared by either mixing a weak acid/base with a
solution of a salt containing its conjugate, or by partial neutralization of a
weak acid/base with a strong acid/base.
Buffers can be prepared in two different ways:
• mixing a weak acid (or base) with the salt of its conjugate base (or acid)
• adding base (acid) to a weak acid (base) and converting some of the weak acid
(base) to its conjugate base (acid).
Both methods will result in a buffer solution of the same nature being formed, as
the solution made will contain the same conjugate acid–base pair.
Buffers
•
pH curves graphs of pH against
strong acid and strong base volume of acid or base added in
pH
7
3
6
5
4
3
2
2 1
1
0
25 50 75
volume of 0.10 mol dm–3 NaOH/cm3
7 region
6
5
pH = pKa 4 2
3 1
2
1
0
25 50 75
half-equivalence point volume of 0.10 mol dm–3 NaOH/cm3
7
6 equivalence 5
5 3 point < 7
4
3 2
2 1
1
0
25 50 75
volume of 0.10 mol dm–3 NH3/cm3
7 3
6 (difficult to identify)
2
5 1
4
3
2
1
0
25 50 75
volume of 0.10 mol dm–3 NH3/cm3
•
first changes colour in an acid–
determine which combination of acid and base is reacting base titration
• based on the graph of this combination estimate the pH values for the steep
portion of the curve that occurs around the equivalence point
• choose an indicator with a pH range that lies within the steep portion
of the pH curve.
Essential idea:
Energy conversions between electrical and chemical energy lie at the
core of electrochemical cells.
•
electrical energy
has the symbol E (E is used for standard conditions)
•
potential difference the
difference in voltage between the changes with different combinations of half-cells.
anode and cathode in a cell
standard hydrogen electrode a
reference half-cell that is used to
measure the electrode potentials
Understanding: The standard hydrogen electode (SHE)
of other half-cells consists of an inert platinum electrode in contact with
electrode potential the EMF
that is generated by a half-cell
1 mol dm–3 hydrogen ion and hydrogen gas at 100 kPa
when it is connected to the
standard hydrogen electrode
and 298 K. The standard electrode potential (E ) is the
standard conditions a set of potential (voltage) of the reduction half-equation under
consistent reaction conditions
that is used when measuring cell standard conditions measured relative to the SHE. Solute
potentials
concentration is 1 mol dm–3 or 100 kPa for gases. The E of
the SHE is 0 V.
The standard hydrogen electrode (SHE):
• is a half-cell that can be combined with other half-cells to make a voltaic cell
• uses an inert
platinum electrode
• uses standard
conditions.
platinum
electrode
glass tube with
holes in to allow
bubbles of H2(g)
to escape
acid solution
Figure 19.1 The standard hydrogen containing
electrode (SHE). 1.0 mol dm–3
H+(aq)
• a temperature of 298 K
standard
hydrogen
electrode
cotton
anode
wool Cu2+ copper
1.0 mol dm–3 sulfate
H+(aq) solution
1.0 mol dm–3 Figure 19.2 Measuring the
standard electrode potential for
H2(g) ➝ 2H+(aq) + 2e– Cu2+(aq) + 2e– ➝ Cu(s) Cu2+(aq)/Cu(s).
• are positive when the half-cell is more easily reduced than the SHE Harder to reduce
• are negative when the half-cell is harder to reduce than the SHE. – (easier to oxidize)
Hints for success: E values are given for a large number of half-cells in
section 24 of the IB data booklet.
When two half-cells are combined:
• reduction will occur at the half-cell with the more positive E value
• oxidation will occur at the half-cell with the more negative E value.
Worked example
Calculate the EMF for a voltaic cell constructed from a zinc half-cell and a
copper half-cell, and identify the anode and cathode. Write the equation for the
overall cell reaction.
Solution
Hints for success: When using Ecell = Ered – Eox use the reduction E values
in section 24 of the IB data booklet exactly as they are given: do not
change the signs.
•
electrolysis
oxidized at the anode:
2H2O(l) ➝ 4H+(aq) + O2(g) + 4e– –E = –1.23 V
• Cathode
• Anode
Electrolysis of NaCl(aq)
The possible reactions at the electrodes are:
This means that the cell potential, Ecell, also indicates if the redox reaction occurring
is spontaneous or non-spontaneous.
Worked example
Calculate the standard free-energy change at 298 K for the zinc–copper voltaic
cell, which has a standard cell potential of +1.10 V.
Solution
First write the equation for the overall cell reaction, as we need to know the
number of electrons transferred. We deduced this on page 212.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
ΔG = –nFE
= –2 (mol e–) × 96 500 (C mol–1) × 1.10 (V)
= 212 000 J
∴ ΔG = –212 kJ
Worked example
How many grams of copper are formed on the cathode of an electrolytic cell containing CuCl2(aq) if a
current of 2.00 A is run for 15 minutes?
How many grams of copper would be formed if CuCl(aq) was used instead of CuCl2(aq)?
Solution
CuCl2(aq) CuCl(aq)
Step 1: Determine the Q = It Q = It
total charge (Q) produced
= 2.00 A × 900 s = 2.00 A × 900 s
by the current flow:
= 1800 C = 1800 C
Q = It
Step 2: Determine the Q Q
n(e–) = n(e–) =
number of moles of F F
electrons contained in the 1800 C 1800 C
= =
total charge: 96 500 C mol–1 96 500 C mol–1
Q = 0.01865 mol = 0.01865 mol
n(e–) =
F
Step 3: Use the balanced Cu2+(aq) + 2e– ➝ Cu(s) Cu+(aq) + e– ➝ Cu(s)
oxidation (or reduction) n(e–)
reaction to determine the n(Cu) = n(Cu) = n(e–)
2
moles of product formed 0.01865 mol = 0.01865 mol
=
2
= 9.33 × 10–3 mol
Step 4: Determine the m(Cu) = nM m(Cu) = nM
mass of product formed: –3 –1
= 9.33 × 10 mol × 63.55 g mol = 0.01865 mol × 63.55 g mol–1
m = nM
= 0.593 g = 1.19 g
•
a layer of metal on another
a salt solution containing the cations of the metal to be deposited conducting object
onto the object
• a cathode made of the conducting object that will be electroplated General vocabulary
+ – e–
e–
Ag(s) ➝ Ag+(aq) Ag+(aq) ➝ Ag(s)
Replenishes the Ag+ ions anode cathode Layer of silver forms
reduced at cathode on the fork
cathode
anode metal object to be plated
bar of silver
solution of silver salt Figure 19.3 Apparatus for
e.g. silver cyanide electroplating silver.
Essential idea:
Key organic reaction types include nucleophilic substitution,
electrophilic addition, electrophilic substitution, and redox reactions.
Reaction mechanisms vary and help in understanding the different
types of reactions taking place.
•
aprotic unable to donate an H+
or form hydrogen bonds polar protic solvents are polar solvents that are able to form hydrogen bonds.
protic able to donate an H+ or Examples: water (H2O), methanol (CH3OH), ammonia (NH3).
form hydrogen bonds
A curly arrow ( ) shows the movement of a pair of electrons in a
heterolytic fission the breaking reaction mechanism.
of a covalent bond to form a
cation and an anion Heterolytic fission is the breaking of a covalent bond with one atom getting both
nucleophilic substitution electrons. This forms a cation and an anion:
reactions a reaction in which a
leaving group is substituted by a
nucleophile A B A+ + B–
leaving group the atom or
group that is substituted by a (The curly arrow shows the bond breaking and the pair of electrons moving onto B.)
nucleophile in a nucleophilic
substitution reaction Nucleophilic substitution reactions happen when a leaving group is substituted
by a nucleophile. The reaction mechanism involves heterolytic fission of the bond
between the leaving group and a carbon atom.
The reaction of halogenoalkanes with hydroxide ions to make alcohols is a
nucleophilic substitution reaction:
– –
HO + C X C OH + X
X = Cl, Br, I
X is the leaving group
There are two nucleophilic substitution reaction mechanisms: SN1 and SN2.
• is a bimolecular reaction (has two species involved in the rate-determining step) stereospecific produces a
•
particular stereoisomer
has the rate equation: rate = k[halogenoalkane][nucleophile].
inverted of the opposite
The reaction between OH– and CH3CH2Cl occurs by an SN2 mechanism. orientation
• Although the mechanism shows a fast and slow part this is a concerted reaction
and there is only one step.
General vocabulary
•
concerted occurs in one step
The one step that occurs is the rate-determining step and it involves two
reactants, OH– and CH3CH2Cl:
rate = k[OH–][CH3CH2Cl]
• The transition state must be drawn inside square brackets. If the transition state
has a charge this must be written at the top right of the brackets.
• The reaction works best with aprotic polar solvents because they cannot form
hydrogen bonds to the nucleophile, which would slow down the reaction.
H3C H3C
carbocation intermediate
H3C C H3 H3C
+ fast
Step 2: C C H3 + OH – HO C C H2 + H3C C OH
H3C C H3 H3C
inversion retention
• The first step is the rate-determining step and it involves one reactant, (CH3)3Cl:
rate = k[(CH3)3Cl]
• The carbocation is sp2 hybridized so the carbon centre is trigonal planar. This
means the nucleophile can attack from above or below the plane and there will
be a 50:50 mix of inversion and retention in the alcohol product.
H C H3 C H3
primary secondary tertiary
carbocation carbocation carbocation
Skill: Explanation of how the rate depends on the identity of the halogen
(the leaving group), whether the halogenoalkane is primary, secondary,
or tertiary, and the choice of solvent.
20.1 Types of organic reactions 221
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Factors affecting the rate of nucleophilic substitution reactions of halogenoalkanes.
C C + H X H3C C C H
H H
H H
Interhalogens: the less electronegative halogen will attach to the alkene carbon that is
bonded to the smaller number of carbons.
Br I
H3C H
C C + I Br H3C C C H
H H (I is less
H H
electronegative
than Br)
+
H H
Br
H H CH3
–
fast
Step 2: H C C H + Br H C C H
+
Br Br Br
• Br adds to the alkene carbon that creates the most stable carbocation. This is the
carbocation that has the positive charge on the carbon atom that is bonded to
the most carbons. (It will be stabilized by a larger inductive effect.)
Step 2
• The reactant adds to the carbon that gives the most stable carbocation.
Example: Addition of HCl to propene
H CH3
H CH3
δ+ δ– slow
Step 1: C C + H Cl H C C H + Cl
–
+
H H
H
H CH3 H CH3
–
fast
Step 2: H C C H + Cl H C C H
+
H H Cl
+
H CH3
I
H CH3 H CH3
–
fast
Step 2: H C C CH3 + Br H C C CH3
+
I I Br
Benzene:
H C H H C H
C C C C
C C C C
H C H H C H
H H
(Because benzene exists as a hybrid of the two resonance structures the C–C
bond order is 1.5.)
• is aromatic (the three π bonds in the ring are delocalized) Subject vocabulary
• attracts electrophiles because the benzene ring is electron rich due to the
delocalized pi electrons.
aromatic having a ring structure
with delocalized π bonds
nitration an electrophilic
Skill: Deduction of the mechanism of the nitration (electrophilic substitution reaction in which an
substitution) reaction of benzene (using a mixture of concentrated nitric NO2 group replaces an H atom
acid and sulfuric acid).
The nitration of benzene occurs by an electrophilic substitution reaction. The
electrophile NO2+ is produced from a mixture of HNO3 and H2SO4.
H H NO2 NO2
+NO
2 + + H+
conc. H2SO4
C6H6 + HNO3 ⎯⎯⎯⎯→ C6H5NO2 + H2O
50 °C
Comments on the electrophilic substitution mechanism:
reflux
nitrobenzene phenylammonium ion
2 NH3+ NH2
+ OH– + H2O
• The acidic conditions in the first stage means that the product is the
phenylammonium ion.
• The addition of OH– in the second stage results in an acid–base reaction that
converts the phenylammonium ion to phenylamine.
Essential idea:
Organic synthesis is the systematic preparation of a compound from a
widely available starting material or the synthesis of a compound via a
synthetic route that can often involve a series of different steps.
alcohol
ester primary secondary
= mechanism required
= oxidation aldehyde ketone
= reduction/addition of H2
= substitution
= addition carboxylic
= condensation acid
Sn, conc.HCI
NH3+
NaOH
NH2
Worked example
Explain how propyl propanoate can be synthesized from a single carboxylic acid.
Give equations and conditions for all reactions, and state the type of reaction
occurring at the functional group at each step.
Solution
Start by analysing the target molecule: the ester is the result of condensation
between propanoic acid, C2H5COOH, and propanol, C3H7OH.
Worked example
You are required to make butanone starting from any alkene of your choice. Show
the steps involved in retro-synthesis to determine the identity of the alkene.
Solution
Essential idea:
Stereoisomerism involves isomers which have different arrangements
of atoms in space but do not differ in connectivity or bond
multiplicity (whether single, double, or triple) between the
isomers themselves.
Isomerism
compounds with same molecular formula
but different arrangements of the atoms
Subject vocabulary When there are different substituent groups on each end of the alkene, cis and trans
isomers are possible. A reference plane drawn through the double bond is used to
cis isomers isomers where the
substituents are on the same side define whether the isomer is cis or trans.
of a double bond in an alkene
• cis isomers: similar substituents are on the same side of the reference plane
•
or on the same side in a ring
compound trans isomers: similar substituents are on opposite sides of the reference plane.
trans isomers isomers where the Example: but-2-ene has cis and trans isomers:
substituents are on the opposite
sides of a double bond in an H H H 3C H
alkene or on the opposite sides in
a ring compound
C C C C
cycloalkanes alkanes in which
the carbon atoms are linked to
form a ring structure
H3C CH3 H CH3
heteroanalogues cycloalkanes in cis-but-2-ene trans-but-2-ene
which a carbon atom is replaced
by another atom such as O or N
(a heteroatom) Hints for success: trans means on opposite sides, cis means on the same side.
Cycloalkanes are ring structures made up of carbons that are linked together by
single bonds. Cycloalkanes have the general formula CnH2n, e.g.
H
H H
C
C C
H H
H H
H
H H
C C
C C
H H
H H
cyclopropane cyclobutane
H H
H H
cis and trans isomers also occur for cycloalkanes and their heteroanalogues.
C3 C3
H3C CH3 H3C H
H H
C1 C2 C1 C2
H H H CH3
cis-1,2-dimethylcyclopropane trans-1,2-dimethylcyclopropane
Cl H Cl H
C3 C2 C3 C2
H Cl H H
H H H H
C4 C1 C4 C1
H H H Cl
cis-1,3-dichlorocyclobutane trans-1,3-dichlorocyclobutane
• Z isomers: the highest priority groups are on the same side of the
reference plane.
C C C C
H3C H H3C H2C CH3
2 2 1 2
(Z)-3-methylpent-2-ene (E)-3,4-dimethylhex-3-ene
Subject vocabulary
Understanding: A chiral carbon is a carbon joined to four chiral carbon a carbon atom
that has four different groups
different atoms or groups. attached to it
non-superimposable cannot
A chiral carbon (or asymmetric carbon) is bonded to four different groups or be placed on top of each other to
give the same arrangement
atoms. This arrangement has two isomers that are non-superimposable mirror
images of each other.
isomer II
isomer I
•
opposite directions
are also called enantiomers
•
chiral molecules a molecule
that contains chiral carbons and are non-superimposable mirror images of each other
rotates plane-polarized light
• rotate plane-polarized light in opposite directions (are optically active)
enantiomers optical isomers
• have identical physical properties
plane-polarized light light that
only oscillates in one plane
optically active rotating plane-
• have the same chemical properties except for their reactions with other
chiral molecules.
polarized light
C C
H5C2 OH HO C2H5
H H
= chiral C atom
(b) mirror
H H
C C
Cl CH3 H3C Cl
C2H5 C2H5
• do not have opposite configurations at all chiral carbons contain more than one chiral
•
carbon and are not mirror images
are not mirror images of each other of each other
H C* OH HO C* H HO C* H
H C* OH HO C* H H C* OH
Figure 20.5 Examples of four-
carbon sugars showing the
CH2OH CH2OH CH2OH difference between enantiomers
and diastereomers. Red asterisks
mark the position of chiral carbon
enantiomers diastereomers atoms. Note the diagram is
opposite configuration at opposite configuration at simplified to two-dimensional
both chiral centres only one chiral centre representation for this purpose.
θ
ordinary
light
polarizer plane
polarized chiral compound
light in solution emerging
analyser light beam
Features of a polarimeter:
• an analyser measures the angle (θ) through which the plane of the polarized
light has been rotated by the solution.
The two optical isomers of a chiral compound are identified by the direction in
which they rotate polarized light in a polarimeter, e.g.
Essential idea:
Although spectroscopic characterization techniques form the
backbone of structural identification of compounds, typically no one
technique results in a full structural identification of a molecule.
•
environment so only one signal
is recorded. In low-resolution 1H NMR a single peak is observed for each hydrogen
chemical environment.
CH3
absorption
CHO
The total integrated area for all of the peaks in the splitting pattern is proportional
to the number of hydrogen atoms in the chemical environment.
The splitting pattern observed for each chemical environment depends on the
number of hydrogen atoms on the neighbouring carbon atoms. (Hydrogen atoms
on N and H atoms do not cause splitting.)
In the table below:
• the blue H are the hydrogen atoms generating the 1H NMR signal shown in
the final column
• the red H are the neighbouring hydrogen atoms that cause the splitting pattern.
O 1
0 singlet
H3 C C O H
OH
1 1
1 H3 C C C H3 doublet
2
1 1
2 H3C C H2 OH triplet
3 3
Worked example
The 1H NMR spectrum of a
compound with the empirical
formula C2H4O is shown.
(a) Deduce the molecular 2 3 3
H
formula of the compound.
(b) Use section 27 of the IB data
booklet to identify a structure
that is consistent with the H
1
H NMR spectrum and account
for the number of peaks and
H
the splitting patterns in
the spectrum.
5 4 3 2 1 0
chemical shift (ppm)
Solution
(a) The 1H NMR spectrum shows the presence of eight hydrogen atoms so the
molecular formula is C4H8O2. The hydrogens are in three different chemical
environments, in a 2 : 3 : 3 ratio.
From the formula, IHD = 1, so there is either a ring or one double bond.
H O
H C C H H
H O C C H
H H
Worked example
(a) An unknown compound is found to have the following composition:
% composition by mass
C 85.6
H 14.4
Deduce the empirical formula of the compound.
(b) The mass spectrum of the compound is shown below. Deduce the molecular
formula and the IHD of the compound.
100
56
90
80
70
relative abundance
60
84
50
40
30 42
20 28
10
0
0 10 20 30 40 50 60 70 80 90
mass/charge
absorption
TMS
2 1 0
chemical shift, δ/ppm
Solution
(a) To find the empirical formula calculate the relative number of moles:
C H
mass / g 85.6 14.4
moles 85.6 14.4
= =
12.01 1.01
= 7.13 = 14.3
simple 7.13 14.4
= =
ratio 7.13 7.13
= 1.00 = 2.02
The empirical formula is CH2.
(b) The mass spectrum shows a parent ion at 84. The molecular formula is CnH2n.
n(12.01) + 2n(1.01) = 84
14.03n = 84
84
n= = 5.99
14.03
The molecular formula is C6H12.
The saturated non-cyclic compound with six carbons is C6H14, so the IHD = 1.
The molecule contains a C=C or a ring.
(c) The absence of peaks at 15 or 69 (84 – 15) suggests that the molecule
probably does not contain a methyl group.
(d) The absorption close to 2900 cm–1 is due to the C–H bond.
The absence of an absorbance at 1600 cm–1 suggests that the molecule does
not contain a C=C bond. It must, therefore, have a ring structure.
(e) The NMR spectra shows only one peak, so all H H
the hydrogen atoms are in the same chemical H H
C
environment. This confirms that the molecule has a C C
ring structure. It is cyclohexane. H H
H H
C C
C
H H
H H
resultant
wave 2
Figure 21.5 Destructive interference occurs when the waves are out of phase. Waves cancel each other out to
produce a region of low-intensity X-rays on the screen.
• the angle that the X-rays hit the surface of the crystal (θ), which is known as
the incident angle Subject vocabulary
• the distance between the atoms in the crystal (d). incident angle that angle at
which X-rays hit the surface of a
X-rays are in scattered X-rays have travelled different crystal
phase as they distances so may be out of phase
enter the crystal θ electron density map a
representation (or map) of where
at an angle θ
electrons are in a compound
path difference
Figure 21.6 The angle of diffraction at which constructive interference occurs (θ) depends on the wavelength of
the incident radiation (λ) and the interatomic distance (d).
• An electron density map can be determined from the diffraction pattern and
this shows where the atoms are located. The positions of hydrogen atoms
cannot be determined directly using X-ray crystallography because they have a
low electron density and do not diffract X-rays.
0 1 2 3
A.
scale
Figure 21.7 The electron density map of anthracene and its molecular structure. The carbon–carbon bond
lengths and angles can be obtained directly from the map. The high electron density between the atoms shows
the presence of covalent bonding.
The investigation
During your two-year IB Chemistry course, you are expected to carry out an
individual scientific investigation, sometimes known as an exploration. This must be
written up as a full report, and contributes to your final assessment on the course.
The investigation will be based on a topic of your own interest, and have a
purposeful research question and scientific rationale. Your approach and
methodology may rely on the collection of primary data through experimental
work, or it may involve analysis of secondary data. Possibilities include the use of
spreadsheets for analysis and modelling, extraction and analysis of data from a
database, or the use of open-ended simulations. The investigation may also use
a mix of approaches and data sources. In all cases, the investigation is marked
according to the same five criteria, which are summarized below. Note the
following general points:
• The reports of a sample of students from your class will be re-marked by the
IB, a process known as moderation, which ensures that the same standards are
applied across all candidates.
• The mark awarded for your investigation contributes 20% towards your
final IB result.
• The investigation can be based on a topic within the course content, or it can be
on extension material beyond the topic specifications in the IB Chemistry guide.
The investigation should provide clear evidence of the knowledge and skills that you
have acquired with respect to the Nature of Science, and an awareness of the aims
of the course. You also have the opportunity here to demonstrate the attributes of
the IB learner profile.
The marking descriptors for each criterion are shown below, with some additional
notes for guidance.
Mark Descriptor
0 The student’s report does not reach a standard described by the
descriptors below.
1 The evidence of personal engagement with the exploration is limited
with little independent thinking, initiative, or creativity.
The justification given for choosing the research question and/or the
topic under investigation does not demonstrate personal significance,
interest, or curiosity.
There is little evidence of personal input and initiative in the designing,
implementation, or presentation of the investigation.
Your report should include some background to the choice of your investigation
and what inspired you in the planning. Maybe it was something that you thought
about in class, or during an experiment, or something you read? You will be
credited for an investigation that is innovative and somewhat unique, that
shows evidence of independent thinking and creativity in any or all stages. Try to
demonstrate personal interest and genuine enthusiasm in your report.
Exploration
Mark Descriptor
0 The student’s report does not reach a standard described by the
descriptors below.
1–2 The topic of the investigation is identified and a research question of
some relevance is stated but it is not focused.
The background information provided for the investigation is superficial
or of limited relevance and does not aid the understanding of the
context of the investigation.
The methodology of the investigation is only appropriate to address
the research question to a very limited extent since it takes into
consideration few of the significant factors that may influence the
relevance, reliability, and sufficiency of the collected data.
The report shows evidence of limited awareness of the significant safety,
ethical, or environmental issues that are relevant to the methodology of
the investigation*.
3–4 The topic of the investigation is identified and a relevant but not fully
focused research question is described.
The background information provided for the investigation is mainly
appropriate and relevant and aids the understanding of the context of
the investigation.
The methodology of the investigation is mainly appropriate to
address the research question but has limitations since it takes into
consideration only some of the significant factors that may influence the
relevance, reliability, and sufficiency of the collected data.
The report shows evidence of some awareness of the significant safety,
ethical, or environmental issues that are relevant to the methodology of
the investigation*.
* This indicator should only be applied when appropriate to the investigation.
In this part of the report you must record data, both qualitative and quantitative,
and demonstrate how the data are processed. This includes showing accuracy
in calculations, the correct use of units, and consideration of the impact of error
propagation. Data should be put into a table where possible. Graphs must be clearly
titled, have appropriate scales and labelled axes with units, and show accurately
plotted data points. Suitable best-fit lines or curves need to be marked distinctly.
Your processed data need to be interpreted carefully in the context of the research
question to develop a valid conclusion.
Mark Descriptor
0 The student’s report does not reach a standard described by the
descriptors below.
1–2 A conclusion is outlined which is not relevant to the research question
or is not supported by the data presented.
The conclusion makes superficial comparison to the accepted scientific
context.
Strengths and weaknesses of the investigation, such as limitations of the
data and sources of error, are outlined but are restricted to an account
of the practical or procedural issues faced.
The student has outlined very few realistic and relevant suggestions for
the improvement and extension of the investigation.
3–4 A conclusion is described which is relevant to the research question and
supported by the data presented.
A conclusion is described which makes some relevant comparison to
the accepted scientific context.
Strengths and weaknesses of the investigation, such as limitations of the
data and sources of error, are described and provide evidence of some
awareness of the methodological issues* involved in establishing the
conclusion.
The student has described some realistic and relevant suggestions for
the improvement and extension of the investigation.
You must give a conclusion that is both justified by the results presented, and puts
the findings in a broader context. This involves comparison with literature values
and accepted data sources. This is a good place to show off your relevant scientific
knowledge, and you should aim to include molecular level explanations for your
findings where possible. Full consideration of errors is essential, and should lead
you to evaluate the methodology chosen and suggest relevant modifications. Make
sure that the modifications are specific to the weaknesses identified. A table with
three columns can help you to keep this part of the report focused, as shown in
the example below:
Mark Descriptor
0 The student’s report does not reach a standard described by the
descriptors below.
1–2 The presentation of the investigation is unclear, making it difficult to
understand the focus, process, and outcomes.
The report is not well structured and is unclear: the necessary
information on focus, process, and outcomes is missing or is presented
in an incoherent or disorganized way.
The understanding of the focus, process, and outcomes of the
investigation is obscured by the presence of inappropriate or irrelevant
information.
There are many errors in the use of subject-specific terminology and
conventions*.
* For example, incorrect/missing labelling of graphs, tables, images; use of units,
decimal places.
A good scientific report communicates in clear and concise language with no
unnecessary comments. There is no fixed narrative style or outline of sub-headings
expected, but your report should follow a logical sequence and make correct use
of the subject-specific terms. Scientific journal articles provide a good source of
ideas for how to structure your report. This criterion will be judged based on the
communicative quality of the report as a whole.
4 Be enthusiastic!
This is your chance to put the Nature of Science into action. You are in charge of
your choice of topic, and have the flexibility here to be creative, so enjoy it fully.
Some advice
Before you start
Read a copy of the subject-specific details of an Extended Essay in chemistry,
including the assessment criteria. Read some previous essays and try to identify
their strengths and weaknesses. Draw up a list of possible research questions,
including the techniques you would use to address these questions. Many of the
best essays are written by students investigating relatively simple phenomena using
apparatus and materials that can be found in most school laboratories.
You may find it useful to consider some of the following techniques when planning
your research; it is often appropriate to use a combination of two or more of
these approaches:
• electrophoresis
• spectrophotometry
• calorimetry
• Use a range of resources to find out what others have done in the area.
Textbooks should never be the only source of information.
• Keep written records of everything that you do and make a note of all
references, including the date when internet sites were accessed, so that you can
build up your footnotes and bibliography as you go along.
Chemistry Extended Essay 247
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• Record all experimental data, including the dates when the experiments
were performed and any uncertainties in your measurements. In your
preliminary investigations, write down any problems and challenges you
encountered and record any modifications. Use your imagination to design new
equipment if necessary.
• Use the appropriate chemical language and make sure that all chemical
equations are balanced.
• Check any calculations and make sure that all experimental data are
presented correctly.
• Discuss the limitations of the experimental method and any systematic errors.
• Consider any questions which are unresolved at the end of your research and
suggest new questions and areas for possible further investigation.
• Let your enthusiasm and interest for the topic show and emphasize clearly your
own personal contribution.
• Ensure that your word count is close to 4000. You will often find you can cut
quite a number of words as you polish your essay at the end.
• Use the assessment criteria to grade your essay. Are you satisfied with the grade
you award yourself?
• Topic
• Context
• Research
• Structure
• Process
• Research question
• Subject-specific
• Analysis
• Layout
• Research focus
• Is it clear that the sources you are using are relevant and appropriate to your
research question?
• Do you have a range of sources or have you relied on one particular type, for
example internet sources?
• Is there a reason why you might not have a range? Is this justified?
• Layout • Are the title and research question mentioned on the cover page?
• Does your bibliography contain only the sources cited in the text?
• Did you use the same reference system throughout the essay?
• Have you proofread the text for spelling and grammar errors?
Correct chemical should be used consistently. Relevant chemical formulas
(including structural formulas), balanced equations (including state symbols), and
mechanisms should be included. The correct units for physical quantities must
always be given. For experiments where numerical results are calculated from
data obtained by changing one of the variables, try to show one example of the
calculation. The remainder can be displayed in tabular or graphical form.
The conclusion must be consistent with the argument presented. It should not
repeat material in the introduction or introduce new points to the argument. It
is a good idea to consider unresolved questions and to suggest areas for further
investigation.
• Research focus
Have you highlighted challenges you faced and how you overcame them?
Will the examiner get a sense of your intellectual and skills development?
Will the examiner get a sense of your creativity and intellectual initiative?
Will the examiner get a sense of how you responded to actions and ideas in the
research process?
You should also demonstrate that you understand the theory behind any
techniques or apparatus used and also identify possible weaknesses.
Viva voce
After you have handed in the final version you may be given a short interview or
viva voce by your supervisor, who is required to write a report on your project. This
is an opportunity to discuss the successes and challenges of the project, and for you
to reflect on what you have learned from the experience.