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Describe properties of particles and thermochemical
principles
ATOMIC STRUCTURE AND BONDING
Properties of particles include:
 electron configuration of atoms and ions of the first 36 elements (using s,p,d notation)
 special characteristics of transition metals (variable oxidation state, colour) related to
electron configuration. Transition metals will be limited to iron, vanadium,
chromium, manganese, copper and zinc
 periodic trends in atomic radius, ionisation energy, and electronegativity, and
comparison of atomic and ionic radii
 Lewis structures and shapes (up to six electron pairs about the central atom for
molecules and polyatomic ions, including those with multiple bonds)
 polarity of molecules
 attractive forces between atoms, ions, and molecules. These will include ionic
bonds, covalent bonds, and intermolecular attractions due to temporary dipoles
and permanent dipoles (including hydrogen bonding).
Bohr-Rutherford Model of the Atom
Experiments carried out by Ernest Rutherford showed that the atom was composed of a very
small, massive, positively charged nucleus surrounded by negative electrons. Later the
Danish scientist Niels Bohr modified the model and showed that electrons can only occupy
orbits with certain allowable (“quantised”) energy levels. Atomic orbitals define regions of
space in which there is a high probability of finding an electron. Each orbital has a particular
shape and associated energy
values.
In the normal ground state of an atom, the electrons occupy orbitals with the lowest possible
energies. On heating, the electrons can be excited to orbitals with higher energy - the
‘excited state’. As the electrons fall back to lower energy levels (orbitals) they will emit
electromagnetic radiation, which may be in the region of visible light ie. it appears coloured.
Each element has its own characteristic emission spectrum that can be used to identify that
element - its “chemical fingerprint”.
These spectra are used to identify the elemental composition of distant stars.
Energy levels, Sub-levels and Orbitals
The position and energy of any electron in an atom is identified using a set of four quantum
numbers.
1) The first quantum number identifies which main energy level (“shell”) the electron is in.
The first 2 electrons occupy the 1st energy level (1st row of the periodic table, H and He),
then the next 8 electrons fill the 2nd energy level (2nd row of the periodic table Li to Ne).
Following this electrons go into level 3 and then level 4 etc.
2) The second quantum number identifies the sub-level which are classified as either s, p, d
(or f).
The first level has only 1 sub-level - 1s
The second level (or row) has 2 sublevels - 2s and 2p.
The 3rd level has 3 sub-levels - 3s, 3p and 3d
3) The third quantum number identifies the specific orbital (region of space) that the electron
occupies. Each orbital can accommodate a maximum of 2 electrons. The number of orbitals
in any particular sublevel is always the same
 s sublevels only have one orbital and therefore have a maximum of 2 electrons
 p sublevels contain 3 orbitals and therefore contain a maximum of 3 x 2 = 6 electrons
 d sublevels contain 5 orbitals and therefore contain a maximum of 5 x 2 = 10 electrons
 f sublevels contain 7 orbitals - a maximum of 14 electrons. (Available from the 4th
energy level)
4) The fourth quantum number uniquely identifies each of the two electrons in a specific
orbital by defining its spin as either clockwise or anticlockwise.
Aufbau Principle
When placing electrons in orbitals they always fill up the sub-levels of lowest energy first.
The order of energies of sublevels is 1s 2s 2p 3s 3p 4s 3d 4p and this is related to the
periodic
table as follows:
Groups
1
2
3 4 5 6 7 8 9 10 11 12
13
14
15
16
1s1
17
18
1s2
2s1
2s2
2p1
2p2
2p3
2p4
2p5
2p6
3s1
3s2
3p1
3p2
3p3
3p4
3p5
3p6
4s1
4s2
4p1
4p2
4p3
4p4
4p5
4p6
“s” block
3d1
to
3d10
“d” block
“p” block
Thus atoms which are in the first 2 groups (columns) of the periodic table have either 1 or 2
valence (outer shell) electrons in an s sub-shell. Atoms found in groups 3 to 12 will have 1
to 10 electrons in the d sublevel. These elements are called transition metals. Atoms in
groups 13 to 18 will have from 1 to 6 valence electrons in a p sub-shell.
NOTE:
The third level of the periodic table “should” contain a total of 18 elements (3s23p63d10) but
actually only contains 8 elements. The “missing” 10 elements appear slightly later as the
transition elements because electrons placed in the 4s orbital are at a lower energy level
than those in the 3d sublevel and the 4s orbital is therefore filled before any electrons go into
the 3d orbitals.
Electron configuration
The electron configuration for any atom is a statement giving the number of electrons at
each sublevel. Each sub-level is filled, starting with the one of lowest energy, before
proceeding to the
next sublevel.
eg
Li (atomic number 3) 1s2 2s1 (total 3 electrons)
F (at no 9)
1s2 2s2 2p5 (total 9 electrons)
K (at no 19)
1s2 2s2 2p6 3s2 3p6 4s1 (19 electrons)
Fe (at no 26)
1s2 2s2 2p6 3s2 3p6 3d6 4s2 (26 electrons)
NOTE:
In Fe (like all of the transition metals) the 4s orbital (which is filled first) is full but the 3d
sublevel is only partially filled.
Exercise - Write electron configurations for each of the following atoms:
a)
C
b)
Br
c)
S
d)
Ca
e)
Ti
f)
Ni
NOTE:
The electron configuration can be written in an abbreviated form in which the inner core of
electrons (those not in the outer valence shell) are represented by the corresponding inert
gas configuration.
eg
Mn is written [Ar] 3d5 4s2
P is written [Ne] 3s2 3p3
Stability of half-filled sub-levels
Because all orbitals within the same sub-level have the same energy, they are filled so
thatas many electrons remain unpaired as possible as this reduces the repulsion resulting
from electrons occupying the same region of space.
In the same way that atoms react to try and achieve completely full energy levels, a halffilled sub-level is also a particularly stable (or low energy) state. This explains the unusual
electron configurations for the transition metals Cr and Cu, where in each case an electron is
promoted from
the slightly lower 4s sub-level to the 3d sub-level to achieve half-filled and full sub-levels
respectively.
Cr
1s2 2s2 2p6 3s2 3p63d54s1 OR [Ar] 3d54s1
Cu
1s2 2s2 2p6 3s2 3p63d104s1 OR [Ar]3d104s1
This stability also explains the small “kinks” in the graph of ionisation energy vs atomic
number.
e.g. the 1st IE of sulphur is lower than the 1st IE of phosphorus despite sulfur’s higher nuclear
charge. This is because the resulting S+ ion has a stable half filled p sub-level whereas the
ionistaion of phosphorus requires the loss of an electron from the half filled sub-level. Note
that this concept will not be assessed in this level 3 achievement standard.
Electron Configurations of Monatomic Ions
a) When non-metal atoms (found in Groups 13 – 17) gain electrons and form anions the
electrons are added to the outermost, p sub-level. The atom will form a negative ion having
the same electron configuration as the inert gas at the end of the same row of the periodic
table.
This means P3–, S2–, Cl– and Ar will all have the same electron arrangement of 1s2 2s2 2p6
3s2 3p6. They are all isoelectronic.
b) When metal atoms lose electrons they are removed from the sub-level furthest from the
nucleus. This means Na loses its 3s1 electron to form Na+ which has the same electron
configuration as Ne.
NOTES:
 When forming a cation the metal atoms only lose electrons from their valence shell.
Atoms in Groups 1 and 2 only lose the electrons from their outer s orbital and form
either M+ or M2+ ions. For the atoms in the first transition series, both the 3d and 4s
electrons are part of the valence shell and can be removed to form transition metal
cations.
Because of this they can form species with a variety of oxidation states.

Although the 4s sublevel is filled before the 3d, when forming a cation the transition
metals always lose the 4s electrons before losing electrons from the 3d sublevel.
This is because on average, the 3d sub-level is closer to the nucleus than the 4s sublevel despite it having a slightly higher energy level.
e.g.
Fe 1s2 2s2 2p6 3s2 3p6 3d6 4s2
Fe2+ 1s2 2s2 2p6 3s2 3p6 3d6 NOT [Ar] 3d4
4s2
 Fe2+ can be further oxidised to form the Fe3+ ion because of the xtra stability arising from
its half filled 3d5 sub-level.
Exercise
1.
Give the electron configuration of each of the following ions.
a)
Cu2+
b)
Br
c)
Mn2+
e)
Ga3+
d)
Ca2+
2. Complete the following table.
Element
Atomic
No
11
No. of
Protons
Mass
Number
23
Fe
No. of
neutrons
23
Electron
Configuration
0
30
16
Charge
3+
1s22s22p63s23p6
34
28
[Ar] 3d3
Periodic Trends
There are three major trends in the periodic table that you will need to explain. They are
atomic and ionic radii, ionisation energy, and electronegativity. When attempting to explain
these trends you will need to consider the relative size of the electrostatic attraction between
the protons in the nucleus and the valence electrons. This electrostatic attraction depends
on
 the number of protons in the nucleus (electrostatic attraction between positive nucleus
and valence electrons increases with the size of the nuclear charge)
 the distance of the valence electrons from the nucleus (electrostatic attraction decreases
as the distance between the positive and negative charges increases)
 the amount of electron-electron repulsion. (When the number of electron shells increases
this effect is often described as increased shielding from the inner shells of electrons.
This results in decreased electrostatic attraction between the protons and valence
electrons).
As a general rule when going across a period the effect of the increased nuclear charge
is the most important factor whereas when going down a group the increased number of
electron shells (i.e. increased radius) is the more important factor.
1. Atomic Radii
The atomic radius of a metal is half the distance between the nuclei in two adjacent atoms.
For elements that exist as diatomic molecules the atomic radius is half the distance between
the nuclei of the two atoms in a molecule.
The diagram below shows the relative size of some atoms.
Two trends in atomic size are obvious:
a) Down a group the atomic size increases
This is because as you go down a group the valence electrons are occupying orbitals which
are at higher energy levels and are further from the nucleus. This factor is greater than the
effect of the
increased nuclear charge, particularly as the inner shells of electrons provide additional
shielding.
b) Across a row (or period) the atomic radius decreases
Going across a row there is an increase in the number of protons in the nucleus and
therefore an increase in the nuclear charge. The additional electrons are all added to the
same energy level, and therefore provide no additional shielding. As a result, the increased
electrostatic attraction results in
the electrons being attracted more closely to the nucleus and a decrease in atomic size.
2. Ionic Radii
The relative radii of positive and negative ions
is shown on the right. The radius of a positive
ion is always smaller than the radius of the
atom from which it was formed. Removing
one or more electrons reduces electronelectron repulsion so the same attractive force
from the nuclear charge results in the
electrons being closer to the nucleus.
Also the formation of cations often results in
the loss of all the electrons in the outer
valence shell. For example, Na, 1s2 2s22p6
3s1 forms a Na+ ion by losing the electron from
its 3s orbital. Since the valence electrons in
the sodium ion are only in the 2nd energy level
(2s and 2p orbitals) the radius of the ion is
considerably smaller than its atom.
When an anion forms the increased electronelectron repulsion resulting from the additional
electron increases the size of the electron cloud. The increased radius results in decresased
electrostatic attraction between nucleus and valence electrons.
3. Ionisation Energy
The first ionisation energy is the energy required to remove the least tightly held electron
from each
atom in one mole of gaseous atoms in their ground state.
Na(g)
(The second and third
ionisation energies are the
energies needed to remove
the second and third
electrons respectively.)
Ionisation energies are all
positive values as the
breaking of the electrostatic
attraction between the
valence electron and the
nucleus must be an
endothermic process.

Na+(g)
+
1e
The graph of ionisation energy against atomic number on the right shows the periodic trend
both down and across the periodic table.
The explanations for the trends are essentially the same as the explanations for atomic size.
e.g. although both the nuclear charge and the number of filled energy levels increases down
a group, the shielding from the inner shells of electrons more than compensates for the
increased nuclear charge, so it requires less energy to remove electrons.
a) Across the table there is an increase in ionisation energy.
This is because going across a row there is an increase in nuclear charge, but the electrons
are added into the same energy level with no additional shielding. The increase in
electrostatic attraction between the nucleus and the valence electron results in an increase
in ionisation energy.
b) Down a group the ionisation energy decreases.
Down a group there is an increase in nuclear charge, but the electrons are being added to
energy levels at higher energies further from the nucleus with additional shielding from inner
shells of electrons. The electrostatic attraction between nucleus and valence electron
therefore decreases and this results in a decrease in the ionisation energy.
An obvious consequence of the above trends is that the low ionisation energies of metal
atoms means they readily form positive ions, whereas non-metal atoms do not form positive
ions.
Also the decrease in ionisation energy going down a group (e.g. Li to Fr) can be related to
the increase in reactivity of metals as it requires less energy to form the cation.
NOTE:
The removal of an electron from an inert gas is very difficult, reflecting the very stable
arrangement resulting fom a full energy level.
4. Electronegativity
The electronegativity of an element is a measure of the ability of an atom to attract towards
itself the
electrons shared in a chemical bond.
A chemist named Linus Pauling devised a scale of relative electronegativities. Some values
are shown below. You are not expected to recall these values in the exam, but you should
be able to explain the trend in electronegativity both down a group and across a row.
H
2.1
Li
1.0
Be
1.5
B
2.0
C
2.5
N
3.0
O
3.5
F
4.0
Na
0.9
Mg
1.2
Al
1.5
Si
1.8
P
2.1
S
2.5
Cl
3.0
K
0.8
Ca
1.0
Br
2.8
Electronegativity values show the same trends as 1st ionisation energies, but their
interpreted values are much more convenient to use than the precise measured values of
ionisation energies.
This means that an atom such as fluorine that has a high ionisation energy (does not lose
electrons
easily) also has a high electronegativity. In fact fluorine is the most electronegative
element.
NOTES:
 In contrast to metals, the reactivity of non-metals increases up each group of the
periodic table as the greater electronegativity values reflect the greater attraction for
electrons required to gain a complete energy level.
 The elements of Group 18 were not given electronegativity values as they do not tend to
form bonds with other elements.
Exercise
a) Electronegativity increases going across a row. Explain this trend.
b) Electronegativity decreases going down a group. Explain this trend.
NOTE: Atoms of elements with widely different electronegativities tend to form ionic bonds
with each other since the atom of the less electronegative element gives up its electron(s) to
the atom of the more electronegative element. Atoms of elements with more similar
electronegativities tend to form covalent bonds. If the atoms have the same, or almost the
same electronegativities the covalent bond is non-polar and the electrons of the covalent
bond are shared equally between the two atoms. The greater the difference in
electronegativity the more polar the covalent bond, and the atom which has the greater
electronegativity will form the slightly negative end of the polar bond. The chlorides across
any row of the periodic table show a continuum from ionic bonding between a metal and a
non-metal with a big electronegativity difference (e.g. NaCl) through to polar covalent
bonding between non-metal atoms of slightly different electronegativities (e.g. SiCl4) to nonpolar covalent bonding between atoms of identical electronegativity (e.g. Cl2).
Remember from Year 12, that just because a molecule contains polar bonds does not
necessarily mean it will be polar overall. This will be discussed further on page 35.
TRANSITION METALS AND THEIR COMPOUNDS
Electronic Structure of d-block elements.
Transition metals (d-block elements) lie in groups 3 - 12 of the periodic table. In the first
transition series (Sc - Zn) the 3d shell is being filled (as has been covered on page 26-27).
Exercise - Write the electron configuration of the following transition metal atoms.
a) Cu
b) Fe
c) Zn
Almost all the properties of the transition elements are related to their electronic structures
and the relative energy levels of the orbitals available for their electrons.
Characteristic Properties of Transition Elements
1. Transition Elements are all metals - In the absence of a surface oxide coating they
have a metallic lustre; high m.ps and b.ps, and are good conductors of heat and
electricity. Most are grey/silver in colour, although copper is a pink metal. Like other
metals they are usually malleable, ductile and sonorous.
2. Size, Electronegativity and Ionisation Energy.
Because the electrons are being added to the inner 3d-subshell, there is only a very
small decrease in size across the first transition row. Similarly there is only a marginal
increase in electronegativity and ionisation energy.
3. Variety of Oxidation State
In the first row of d-block elements, both the 3d and 4s electrons can be considered as
part of the valence shell. In forming ions these elements lose the 4s electrons and a
variable number of 3d electrons. This means the elements show variable valence. It is
this ability of the transition elements to use at least some of the underlying d electrons in
bonding which makes much of transition element chemistry distinctive.
This ability to have a variety of oxidation states applies to all the transition elements of
the first row with the exception of Sc and Zn. The scandium atom has electron
configuration [Ar] 3d14s2 so only has one electron to lose from the 3d level. This means
the maximum charge on a scandium ion is +3 (Sc3+). The zinc atom does not behave
as a typical transition metal atom as its electron structure [Ar] 3d104s2 means it only
forms the Zn2+ ion by losing the 2 electrons from the 4s level, leaving a stable full 3d
sublevel.
The number of oxidation states generally increases with the number of unpaired
electrons, with elements in the middle of the row (eg. Mn) having the widest range of
oxidation states. The maximum oxidation state for transition elements up to Mn is equal
to the total number of valence electrons available.
Element
V
Cr
Mn
Fe
Cu
Electrons
3d34s2
3d54s1
3d54s2
3d64s2
3d104s1
Ox. States +2 to +5
+2 to +6
+2 to +7
+2, +3
+1, +2
Because of the variety of oxidation states available, transition elements are involved in
many redox reactions. Species in which an element is in a high oxidation state (MnO4,
ox state +7, Cr2O72, ox state +6) tend to be good oxidising agents. Conversely, some
species in which the element has a low oxidation state eg. Fe2+, or the metals
themselves, are good reductants.
4. Bonding - Ionic to Covalent
When the transition elements are in low oxidation states (+2, +3) they exist as monatomic
ions in ionic compounds e.g. manganese(II) chloride (MnCl2) and copper(II) nitrate
(Cu(NO3)2). Compounds with transition elements in oxidation states higher than +3
generally have the transition elements covalently bound in polyatomic ions such as
dichromate Cr2O72 and permanganate MnO4.
5. Coloured Compounds
The compounds of transition elements are usually coloured due to the partially filled d
orbitals and the ability to excite electrons into a higher energy level by absorption of
visible light. The magnitude of the energy absorbed and the resulting colour depends on
the nature of any ligands attached to the metal ion eg most Cu2+ salts are blue but CuCO3
is green and both CuS and CuO are black. The colouring in most ceramics and precious
stones are the result of transition metal oxides.
Zn2+ compounds, with the electron arrangement [Ar]3d10, have no vacant 3d orbitals and
these compounds are therefore white. Similarly Sc3+ compounds are white because they
do not have any d electrons.
Chemistry of Individual Transition metals
Manganese
Mn is an important component of alloys. In steel it increases the hardness, toughness and
resistance to abrasion. Mn exhibits all oxidation states from II to VII. The most stable
oxidation states are +2, +4, and +7. The most common compound is manganese dioxide,
MnO2. The permanganate ion, MnO4, is purple and is a strong oxidising agent (especially in
acid solution).
The reduction product of MnO4- depends on the pH of the solution.
MnO4
acid
neutral or
strongly
mildly alkaline
alkaline
Mn2+
MnO2
pale pink or
colourless
brown solid
MnO42
green ion
Chromium
Chromium is a bright, lustrous, corrosion resistant metal. Emeralds and rubies owe their
colour to traces of chromium compounds. Cr is used in stainless steel (about 15% Cr) and
for chrome plating.
The oxides of Cr vary from basic through amphoteric to acidic.
Basic
CrO (low ox state, +2) – dissolves in acid to form Cr2+ ions in solution.
Amphoteric
Cr2O3 (ox state +3)-like Cr(OH)3 dissolves in both acid (to form Cr3+ ions
in solution) and base (to form Cr(OH)4).
CrO3 (high ox state +6) – dissolves in water to form acidic solution of
chromic acid, H2CrO4. This chromic acid in turn gives rise two types of
salt - the yellow chromates (containing the CrO42 ion in basic solution)
and orange dichromates (containing the Cr2O72 ion in acidic solution).
Both these oxyanions contain Cr in the +6 oxidation state. The
equilibrium between these two anions is NOT a redox reaction but an
example of an acid-base equilibrium. The ion present in an aqueous
solution depends on the pH.
Acidic
2CrO42 + 2H+ 
yellow –present in alkaline soln
Cr2O72 + H2O
orange – present in acid soln
Vanadium, V
Vanadium is a soft, grey metal produced by reducing vanadium(V) oxide, V2O5, or
vanadium(II)
chloride, VCl2.
Vanadium pentoxide V2O5 is an orange-yellow compound used as an oxidising agent and
catalyst in the contact process for the manufacture of sulfuric acid. Many vanadium
compounds form coloured solutions eg. the blue of the vanadyl ion, VO2+, has led to the use
of vanadium compounds as glazes
in the ceramics industry. A solution made by dissolving NH4VO3 in NaOH and then adding
sulfuric acid is yellow due to the VO2+ ion. If this solution is shaken with zinc the colour
gradually changes as the vanadium species is reduced:
VO2+(aq) → VO2+(aq) →
V3+(aq)
→
V2+ (aq)
Yellow
blue
green
violet
Iron, Fe
Fe is the second most abundant metal in the earth's crust (after Al). It is mainly found in
oxide ores e.g. magnetite Fe3O4 or in N.Z. in iron sand, titanomagnetite. It is also found in
the bright, shiny yellow ore called "fool's gold", FeS. The strength of iron metal is improved
by the addition of carbon to form steel, and also by alloying with other metals.
Fe is quite reactive and corrodes in moist air. It reacts with acids to form hydrogen gas and
iron II (ferrous) or iron III (ferric) salts. Fe2+ is a mild reductant and Fe3+ is a mild oxidant.
Eo(Fe3+/Fe2+ ) = + 0.77 V
Iron(II) sulfate, FeSO4, is a green soluble salt. Iron(II) sulfide, FeS, is black and insoluble.
If NaOH is added to a solution of Fe2+ a green gelatinous precipitate of Fe(OH)2 forms. This
is readily oxidised by air to Fe2O3 - a red-brown iron(III) oxide (ferric oxide).
Iron(II) hydroxide (ferrous hydroxide) is a basic hydroxide i.e. it dissolves in acid not base.
Fe(OH)2
2H+
+

Fe2+
+ 2H2O
Iron(III) oxide - Fe2O3 is also a basic oxide.
Fe2O3
+
6H+

2Fe3+
+ 3H2O
Aqueous solutions of iron(III) ions are yellow due to the presence of the [Fe(H2O)5OH]2+ ion.
The proton tranfer to form this ion means a solution of Fe3+ is acidic, pKa(Fe3+,aq) = 2.17.
The presence of Fe3+ ions in solution is detected by the formation of the dark red complex
ion FeSCN2+ when a solution of potassium thiocyanate is added.
Fe3+ + SCN
 [FeSCN]2+
Copper
Copper is an unreactive metal found mostly in sulfide ores. Cu is an excellent conductor of
heat and electricity. Mixed with zinc it forms the alloy brass, and with tin it forms bronze. Cu
is too unreactive
a metal to displace hydrogen gas from dilute acids.
Cu slowly corrodes in moist air (H2O, O2 and CO2) to form a pale green layer of basic copper
carbonate, Cu2(OH)2CO3(s). This is what gives copper and bronze objects their
characteristic green
colour called patina. This compound adheres to the copper surface and protects the metal.
Free cuprous ions (Cu+) are unstable in water and disproportionate to form metallic Cu and
blue Cu2+ ions. Copper(I) oxide (Cu2O ) is the red precipitate that forms when the blue Cu2+
ions in Fehling’s solution or Benedicts solution are reduced by aldehydes.
When KI(aq) is added to CuSO4(aq) a white precipitate of CuI is formed.
2Cu2+(aq)
+
4I(aq) 
2CuI(s)
+
I2(aq)
This is used in the volumetric analysis of copper in fungicides or alloys.
Aqueous solutions of [Cu(H2O)6]2+ are blue. When dilute ammonia is added the insoluble
light blue precipitate of Cu(OH)2 forms. As it is a basic hydroxide it dissolves in acid to give
a solution of Cu2+. With excess ammonia, the precipitate dissolves as the royal blue complex
ion [Cu(NH3)4]2+ forms.
Zinc
Zinc is a silver reactive metal, used mainly for galvanising iron where it is protected by a
hard film of basic carbonate. Typical of metal carbonates, this substance neutralises acid,
eg.
ZnCO3 + 2HCl

ZnCl2 + H2O + CO2
The amphoteric nature of Zn, ZnO and Zn(OH)2 are covered on page 81. Zn(OH)2 dissolves
in excess ammonia due to the formation of the colourless zinc tetrammine ion [Zn(NH3)4]2+.
Formation of the Zn2+ ion occurs by the loss of the 4s2 electrons in the same way as the
formation of the Ca2+ ion. However, the first ionisation energy of Zn is much higher because
of the larger nuclear charge of the Zn compared to Ca.
LEWIS STRUCTURES

Lewis structures and shapes (up to 6 electron pairs about the central atom for
molecules and polyatomic ions, including those with multiple bonds)
A Lewis diagram (or dot diagram) shows how the atoms (with their valence electrons) are
linked in a molecule e.g. the Cl2 molecule is formed by linking the two Cl atoms, each with
their 7 valence electrons. Each Cl atom will need to share 1 extra electron to form a
covalent bond.
Cl
non-bonding electrons (6 pairs)
Cl
bonding electrons of the covalent bond
Before starting to draw Lewis diagrams, it is essential to count the total number of valence
electrons.
Hydrogen and halogen atoms must always be peripheral atoms because they can only from
one bond. The central atom is identified from symmetry or because it is the least
electronegative atom.
Then start by linking each atom into the correct structure using a single covalent bond. This
bond is equivalent to 2 electrons i.e. a bonding electron pair. The remaining valence
electrons are placed around the outer atoms and then the central atom, until all valence
electrons are used.
When drawing Lewis diagrams many molecules obey the octet rule i.e. in drawing the Lewis
diagram there will be 8 electrons (4 pairs) of electrons around each atom, except for
hydrogen which only needs 2 electrons to complete its valence shell. However this octet
rule only holds for the first two periods of the periodic table. In the 3rd row and beyond, the
central atom in a molecule is large enough and has vacant orbitals (eg 3d) that may
accommodate 10 or 12 electrons eg PCl5 or SiCl62.
Common examples of Lewis diagrams are
H2O (a total of 8 valence electrons)
H O
H
Cl
CCl4 (a total of 34 electrons)
Cl
C
Cl
Cl
In some molecules the sharing of one pair of electrons will not result in a complete valence
shell for all atoms. It may be necessary to share two pairs of electrons (4 electrons)
between two atoms. This results in a double covalent bond.
Atoms such as oxygen and sulfur may form two single bonds or a double covalent bond
when linked to only one other atom as this allows the atom to share two extra electrons.
oxygen gas, O2
O
O
or
O
O
Note that if the O atom is linked to two other atoms (as in H2O) then each bond will be a
single bond. Similarly, a nitrogen atom with 5 electrons in the valence shell needs 3 more
electrons. If joined to 3 other atoms each bond will be a single covalent bond but if the atom
is linked to only 1 other atom it will form a triple covalent bond (sharing of 3 electron pairs).
nitrogen gas, N2
N
N
or
N
N
NOTE: To draw any Lewis diagram always start by linking the atoms so that they are all
connected by a single bond (in the appropriate arrangement). If there are not enough
electrons to complete the octet for each atom (except H) then non-bonding pairs will need to
be shifted to form a double covalent bond (or triple bond).
Exercise: Draw Lewis diagrams of the following molecules.
CH4, NH3, H2S, H2CO, PCl3, CS2, CH3OH, PCl5, SF6, H2SO4, XeF2, XeF4
Lewis Structures of Ions
The Lewis structures of ions will have more or less electrons depending on the charge on
the ion. For example a PO43– ion has 5 electrons from the P atom, 24 electrons from the four
O atoms PLUS 3 electrons from the negative charge on the ion, giving a total of 32
electrons. The NH4+ ion will have 5 electrons from the N atom and 4 electrons from the 4 H
atoms, but one electron is removed with the +1 charge giving a total of 8 electrons. The
Lewis structure of the ion is then enclosed in square brackets and the charge is shown on
the outside.
3–
O
O P
O
O
Exercise: Draw Lewis diagrams for each of the following ions.
NO2–,
SF3+,
SF5–,
SO32–,
CO32–,
NH4+,
SCN–
Shapes of Molecules
Lewis diagrams show the sharing of electrons but do not adequately represent the threedimensional shape of the molecule. In any molecule the electron pairs mutually repel each
other, and to reduce this repulsion the molecule adopts a shape which allows the electron
pairs (bonding and non-bonding) to be as far apart as possible. (This is sometimes referred
to as VSEPR - valence shell electron pair repulsion).
The shape of the molecule depends on both
a) the number of atoms linked to the central atom
and b) the number of regions of electron density (bonded and non-bonded electron pairs)
around the central atom.
The following are the shapes you will need to be able to identify.
1.
Tetrahedral - occurs when there are 4 atoms linked to a central atom by 4 single
covalent bonds (no additional non-bonding pairs on the central atom) eg. CH4
H atom in the plane of the paper
H
C
H
H atom in behind plane of paper
H
H
H atom in front of plane of paper
The bond angles are all 109.5o, the tetrahedral angle
2.
Trigonal pyramid (or triangular pyramid) - occurs when the central atom is linked to
only three other atoms and also has a non-bonding pair of electrons (i.e. a total of 4
electron pairs around the central atom but only 3 involving bonding
with another
atom).
P
Common for compounds of Group 15 atoms such as N and P eg. PH3
H
H
H
The bond angles are also 109.5o, the tetrahedral angle.
3.
Trigonal planar (or triangular planar) - occurs when the central atom is linked to
three other atoms and has no additional non-bonding electron pairs. Commonly found
with compounds of carbon in which the carbon is linked by a double covalent bond to
one atom, and 2 single covalent bonds to two other atoms eg. H2CO
H
C O
H
The bond angles are
120o.
4.
Bent (or V-shaped)- occurs when the central atom is linked to only two other atoms
but also contains at least one other non-bonding pair of electrons. It is commonly
found in molecules of Group 16 atoms such as O and S eg. H2S and SO2
S
H
5.
H
Bond angles 109.5o.
Linear - occurs when the central atom is linked to two other atoms but there are no
additional non-bonding electrons. Commonly found with molecules having two double
bonds (eg. CO2) and is the only option for molecules containing only two atoms eg.
HCl, O2.
O
C
O
O
O
H Cl
In a linear molecule the bond angle is 180o
6.
Other shapes are less common, but occur when the central atom has more than 8
electrons.
These shapes are summarised in the table on the next page.
Note that where there are six regions of charge the angles are all 90o. The shape of the
molecule
with six bonded pairs is described as octahedral because the resulting 3D shape has 8
faces.
Exercise
Give the shape of all the Lewis structures drawn in the exercise on page 32.
CH4,
NH3,
H2S,
H2CO,
PCl3,
CS2,
CH3OH,
PCl5,
SF6,
H2SO4,
XeF2
XeF4
NO2–,
SF3+,
SF5–,
SO32–,
CO32–,
NH4+,
SCN–
POLAR AND NON-POLAR MOLECULES

polarity of molecules
Molecules can be classified as being polar or non-polar.
a) Polar molecules must have at least one polar bond (covalent bond between two atoms of
different electronegativity) and these polar bonds cannot be arranged symmetrically. This
means the dipole moments of the polar bonds do not cancel out eg. NH3 is a polar molecule
because it contains three polar N-H bonds and a non-bonding pair of electrons on the central
atom and because of this asymmetric arrangement of polar bonds the molecule is polar.
Another example is HCN which is a linear molecule H-CN and the two bonds are of
different polarity so the dipole moments do not cancel out (the arrangement is not
symmetric). Similarly CHCl3 is polar because although the tetrahedral shape can be called
symmetrical, the molecule does not have a symmetric arrangement of polar bonds (so the
dipole moments do not cancel out).
b) Non-polar molecules are those that either
(i) have no polar bonds eg. O2
or (ii) have polar bonds which are arranged symmetrically so that the dipole moments of
the polar bonds cancel out. eg. CF4 has four polar C-F bonds but these are all
equivalent and in a symmetric (tetrahedral) shape.
Some common examples of polar and non-polar molecules are
polar substances
water, ethanol, methanol, ethanoic
acid, ammonia, hydrogen chloride
non-polar substances
cyclohexane, hydrocarbons (eg. petrol, wax,
turps), tetrachloromethane
Exercise
State whether each of the following molecules (whose shapes you gave on p 40) are polar or
non-polar .
CH4,
NH3,
H2S,
H2CO,
PCl3,
CS2,
CH3OH,
PCl5,
SF6,
H2SO4
XeF2
XeF4
Physical properties of polar substances
 If an electrically charged object is placed near a stream of a polar liquid, the polar
molecules will align themselves so that the oppositely charged end of the molecule is
attracted to the rod. Non-polar molecules are not charged and there is no deflection of
the liquid stream. This deflection of the liquid can be used to test whether the liquid is
polar or not.
 Polar molecular substances dissolve in polar solvents (eg. ethanol dissolves in water)
whereas non-polar substances dissolve in non-polar solvents (eg. wax, a hydrocarbon,
and solid iodine both dissolve readily in cyclohexane but not in water). This solubility
behaviour is referred to as “like dissolves like”. Note that water is sometimes referred to
as the universal solvent because as a polar solvent it not only dissolves many polar (and
ionic) substances but is also able to dissolve some non-polar substances (eg I2) to a
limited extent.
 Molecular substances do not conduct electricity as they do not have electrically charged
particles (electrons or ions) that are free to move under the influence of an electric field.
However, many solutions of molecular substances conduct electricity because the
molecule undergoes a hydrolysis reaction with the water to produce ions. Examples
include weak acids such as CH3COOH partially dissociating in water and the hydrolysis of
non metal chlorides such as PCl3.
PCl3 + 3H2O  H3PO4 + 3H+ + 3Cl-
THERMOCHEMISTRY
Thermochemical principles include:
 transfer of heat between the system and the surroundings
 calculations involving the use of specific heat capacity
  cH,  fH,  rH,  vapH,  subH, and  fusH
 Hess’s Law including application of
 rH(=  fH((products) –  fH((reactants)
 bond enthalpies.
Solid structures
If the temperature is low enough all substances form solids. The temperature at which this
occurs depends on the strength of the attractive forces between the particles. H2 is a very
small non-polar molecule that solid also depends on the types of forces that hold the atoms,
ions or molecules tightly packed together in an array. When the particles are packed in an
orderly array the solid is crystalline. Metallic elements such as copper and iron, non-metallic
elements such as sulfur and iodine, and ionic solids such as sodium chloride all form crystal
structures.
Classification of solids
Crystalline solids are classified according to the bonds that hold their atoms, ions or
molecules in place. There are four classes of solids:
 metals (metallic solids)
 ionic solids
 covalent network solids
 molecular solids
Metallic crystals
All the atoms in a crystal of a metallic element have the same electronic arrangement. In the
crystal the atoms are stacked neatly in layers, and are held together by metallic bonds. To
explain this type of bonding it is necessary to recall that metal atoms have loosely held
valence electrons, and low ionisation energies. The metallic bond is the electrostatic
attraction between the “sea” of valence electrons and the positively charged atomic cores.
These metallic bonds have very little directional character and this means that the atom can
easily be pushed past its neighbours without too much effort. Metals are therefore malleable
and ductile.
Metals are also good conductors of electricity and heat as the mobile (or ‘delocalised’)
electrons are free to move. The mobility of the electrons also accounts for the lustre of
metals.
Ionic Crystals
Ionic solids are composed of an extended three dimensional network of oppositely charged
ions (anions and cations) packed together. The ions are held together by the electrostatic
attraction between the positive and negative ions, ionic bonds. Melting or vaporising an
ionic substance requires that the strong ionic bonds that extend throughout the crystal are
broken. Because ionic bonds are strong (compared to van der Waals forces) this requires a
large amount of energy and the melting points of ionic solids are moderate to high.
Ionic solids are brittle because the ions cannot be pushed past each other without setting up
repulsions between ions of the same charge. When we strike an ionic solid, it tends to
shatter into fragments, although it is possible to cleave a crystal along the packing line.
In the solid state, ionic solids do not conduct electricity as the charged particles (cations and
anions) are not free to move past each other. Once molten the ions are free to move towards
the charged electrodes and hence can conduct electricity. An alternate way of separating
ions without heating the ionic solid, is to dissolve it in water. Many ionic solids are soluble in
water, and when dissolved, the ions become separated and form an electrolyte solution.
Some ionic solids, such as MgO and AgCl, are insoluble in water as the attraction between
their cations and anions is too strong to allow them to separate.
In solution, each of the ions is hydrated i.e. they are surrounded by water molecules which
are attracted to the ion by ion-dipole attractions.
Covalent network solids
The atoms in covalent network solids are joined to their neighbours by strong covalent bonds
to form a network that extends in three dimensions throughout the crystal. Covalent network
solids are usually very hard and rigid materials. They have high boiling and melting points
as strong covalent bonds need to be broken for the solid to melt.
The two allotropes of carbon, diamond and graphite, are both covalent network solids, but
differ in the way that the atoms are linked. Each C atom in diamond is covalently bonded to
four other C atoms and the tetrahedral framework extends throughout the entire solid,
making diamond one of the hardest substances known. Graphite, the “lead” of pencils, is a
black, shiny, slippery solid that is a good conductor of electricity. It consists of flat sheets of
C atoms, each one covalently bonded to three other atoms but with only weak van der
Waals forces between the sheets. Graphite is used as a high temperature lubricant because
the weak attractive force between sheets is easily overcome so they can slide past each
other. Because only three electrons per atom are used in covalent bonding, the fourth
valence electron on each C atom is able to move between the layers making graphite a good
conductor of electricity. It is therefore commonly used to make electrodes.
Other important covalent network solids are silicon and silicon dioxide, SiO2, (silica). In SiO2,
each silicon is attached to four oxygen atoms and each O is attached to two Si atoms. The
structure is similar to diamond and the covalent bonds are very difficult to break. Silica is the
substance that makes up sand, which clearly has a very high melting point as we can have
bonfires and barbeques on the beach without the sand melting. Sand can be mixed with
other substances such as limestone, melted at very high temperatures and cooled slowly to
form glass.
Molecular solids
A molecular substance is made up of discrete particles -atoms (eg. Ne), or molecules (eg.
CO2 and H2O) held together in the solid and liquid states by weak intermolecular forces
(often called van der Waals forces). These intermolecular forces are the electrostatic
attraction between the slightly positive end of one molecule and the slightly negative end of
another.
+ -
slightly positive
end
H - Cl slightly negative
end
Within any molecule the atoms are linked by strong covalent bonds, but the process of
melting or boiling does not alter the structure of the molecule or break any covalent bonds.
solid iodine
I-I
I-I
I-I
I-I
I-I
I-I
I-I
I-I
gaseous iodine
small amount of
heat energy
I-I
I-I
I-I
Because the intermolecular forces are only weak, molecular substances have low melting
and boiling points.
There are three types of intermolecular forces –their names and strengths are summarised
in the table below, and can be compared to the much higher force of interaction between a
positive and negative ion (ion-ion interaction) which occurs in an ionic solid.
Intermolecular (and inter-ionic) forces
Type of interaction
Permanent dipole-dipole
Typical energy
/kJ mol-1
About 0.3 - 2
Interacting species
polar molecules
instantaneous dipoledipole (dispersion
forces)
hydrogen bonding
2
all types of molecules
20
c.f. ion-ion
250
H atom attached to O, N or F on 1 molecule
and non-bonding electron pair on O, N or F
of an adjacent molecule
ions only
NOTES:
 Covalent bonds are of similar strength to ionic bonds and both are about 100x stronger
than intermolecular forces (except for very large molecules such as polymers).
The total intermolecular force acting between two molecules is the sum of all the separate
types of forces. i.e. Instantaneous dipoles are found in both polar and non-polar, molecules.
1. Instantaneous dipole attractions (London Forces or Dispersion Forces)
All molecules (and individual atoms such as the inert gases) are attracted by electrostatic
attractions between instantaneous dipoles. These are often called London forces or
dispersion forces. For non-polar molecules, it is the only interacting force between
molecules. These attractive forces result from instantaneous (‘temporary’) dipoles that
occur as a result of the rapid, random motion of the orbiting electrons ie at any given instant
of time electrons may be slightly more concentrated at one end of the molecule forming a
partial negative charge for an instant. At the same time the other end will similarly develop a
slightly positive charge. These instantaneous partial charges on different molecules attract
each other and the molecules are temporarily attracted to each other.
+
-
+
-
The instantaneous dipoles on two molecules are continually changing direction, but over a
period of time remain in step long enough for the molecules to attract each other.
Even particles with very few electrons such as helium atoms and hydrogen molecules have
intermolecular attractions which are sufficiently large to hold themselves together as a solid,
provided they are at extremely low temperatures where the kinetic energy is very low.
The strength of the instantaneous dipole-dipole attractions between molecules increases
with the size of the electron cloud (molar mass). Heavier molecules have more electrons,
and so there is a greater probability of electrons being unequally distributed at any particular
instant. The larger the electron cloud the stronger the instantaneous dipole-dipole attractions
and therefore the higher the m.p. and b.p. Methane, CH4, and propane, C3H8, for example,
are both non-polar molecules but propane has a higher b.p. due to the larger electron cloud
(higher molar mass). The variation in strength of instantaneous dipole-dipole attractions with
molar mass also explains why the halogens F2 and Cl2 are gases, whereas Br2 is a liquid,
and I2 is a solid at room temperature.
The strength of instantaneous dipole attractions is also determined in part by the shape of
the molecules. For example, both pentane and 2,2-dimethylpropane have the molecular
formula C5H12, so they have the same number of electrons (or molar mass). Their very
different boiling points,
36 oC and 10 oC respectively, can be explained by looking at the structure and shape of the
molecules below.
+
+
+
pentane molecules
+
2,2-dimethylpropane molecules
The molecules of pentane are relatively long and rod-shaped, so that the instantaneous
partial charges can interact strongly. In contrast, the 2,2-dimethylpropane molecules are
more spherical and cannot get so close to one another because only a small region of the
molecule can come into contact. The instantaneous dipole attractions between pentane
molecules occur over a larger surface area of contact and are therefore stronger and the
substance has a higher boiling point.
2. Permanent dipole-dipole attractions
In addition to the attractions between temporary dipoles, polar molecules also have a
permanent dipole-dipole interaction arising from the dipole on one polar molecule
interacting with the dipole of another. Therefore, a polar molecule will have a higher melting
and boiling point than a non-polar molecule of similar molar mass. e.g. Br2 and ICl both
have a molar mass of 70 g mol-1 but the boiling point of the polar ICl molecule is 38oC higher
than the non-polar Br2 molecule.
The strength of these permanent dipole-dipole interactions depends on the magnitude of the
bond dipoles ie how polar the covalent bonds are. This is related to the difference in
electronegativity of the atoms involved in the covalent bond. It also depends on the shape of
the molecule, since in a polyatomic molecule with a symmetric shape, the individual bond
dipoles may cancel, so that overall the molecule has a zero dipole moment.
The more polar the molecule the stronger the permanent dipole-dipole attractions and the
higher the m.p. and b.p. Thus ethanal, CH3CHO will have a higher boiling point than ethane
CH3CH3 because, although they are similar in molar mass, the ethanal is more polar than
ethane because of the presence of the polar C=O bond.
Exercise
By considering the shape of the molecule, and therefore the overall polarity, decide whether
cis-1,2- dichloroethene or trans-1,2-dichloroethene would have the higher boiling point.
Cl
cis-1,2- dichloroethene
C
H
Cl
Cl
C
trans-1,2-dichloroethene
C
H
H
H
C
Cl
3. Hydrogen Bonding
Water has an unusually high boiling point and melting point for its molar mass. It is also less
dense in the solid form (ice) than in the liquid form, and as a result ice floats on water.
These unique properties are due to a third type of intermolecular attraction called hydrogen
bonding. It is a special type of permanent dipole-dipole attraction that occurs between a
hydrogen atom bonded to a small, highly electronegative atom (O, N or F) and the nonbonding pairs of electrons on the O, N or F of another molecule. This strong attraction is
only possible due to the large difference in electronegativity between H and the atoms O, N
and F. The small size of these atoms also allows the molecules to get much closer together
resulting in a stronger intermolecular attraction.
Boiling Point / oC
At temperatures just above freezing these attractive forces hold the water molecules
together in an
open, cage-like, solid framework. We can see the effect of this attractive force known as
hydrogen bonding when we plot the boiling points of the hydrides of the elements in Groups
14 to 17.
Period (or row)
The trend in Group 14 is what would be expected for similar compounds that differ in their
number of electrons - the boiling points increase with molar mass because their
instantaneous dipole-dipole attractions are increasing. In contrast, the hydrides of nitrogen,
oxygen and fluorine show unusual behaviour. Water boils at a much higher temperature
(100 oC) than hydrogen sulfide (-60 oC), even though H2S has far more electrons and
stronger instantaneous dipole-dipole attractions.
Molecules such as ethanol (CH3CH2OH) and aminomethane (CH3NH2) will hydrogen bond to
each other, and can also hydrogen bond with a water molecule making these compounds
particularly soluble in water. Molecules that undergo hydrogen bonding have higher m.p.s
and b.p.s than other polar molecules of similar molar mass. In fact, hydrogen bonding is so
strong it can, in some cases, survive in a vapour state. An example is ethanoic acid vapour
which contains dimers, or pairs of molecules, linked together.
Trees stand upright on account of hydrogen bonds. Cellulose molecules, which have many
OH groups, can form many hydrogen bonds with one another and the strength of wood is
due in large part to the strength of hydrogen bonds between neighbouring ribbon-like
cellulose molecules. The viscosity of a substance like glycerol (propane-1,2,3-triol)
compared to propan-1-ol occurs because glycerol can form three hydrogen bonds per
molecule and this extra attraction makes it harder for the molecules to move past each other
in the liquid state.
Exercise
Which of the following molecules can form hydrogen bonds with water? Explain why.
CH4
CH3OH
CH3COOH
CH3COCH3
The diagram below shows the relationship between all three types of intermolecular forces.
temporary dipole-dipole attrations
(all molecules e.g. Br2, HBr and HF)
permanent dipole-dipole attrations
(polar molecules only e.g. HBr)
hydrogen bonding
(H bonded to O, N or F e.g. HF)
The physical properties of molecular solids depend on the strength of their intermolecular
forces. Some such as paraffin, a long-chain hydrocarbon, are soft and waxy. In its solid
state the molecules are only held together by weak attractions between instantaneous
dipoles. Other molecules are brittle and hard eg. sucrose molecules are held together by
hydrogen bonds between their many OH groups. The hydrogen bonds are so strong that by
the time the melting point has been reached (at 184 oC), the molecules themselves have
started to decompose.
Molecular solids do not conduct electricity as there are no mobile charged particles (ions or
electrons). Some molecular solids will conduct in aqueous solution because the molecules
undergo a reaction to produce ions. For example:
SO2 + H2O
H+
+
HSO3
ENTHALPY CHANGES
Thermochemical principles includes:
 transfer of heat between the system and the surroundings
 the enthalpy change for any process is the sum of the enthalpy changes for the
steps into which the process can be divided
 definition of the following terms:  cH,  fH,  rH,  vapH and  fusH.
Heat of Reaction
Heat is the transfer of energy from regions of high temperature to regions of low
temperature. As temperature is a measure of the movement of molecules (average kinetic
energy), heat moves from regions of high thermal motion to a region where there is less
thermal motion.
Enthalpy
When considering the heat change in chemical reactions we use a quantity called enthalpy,
H, which is a measure of the chemical potential energy stored in the bonds of the
substances involved.
A change in enthalpy of a system is equal to the heat released or absorbed at constant
pressure.
It is measured in kilojoules, kJ (or joules, J). The change in enthalpy is given by rH, where
rH = Hfinal - Hinitial
= Hproducts - Hreactants
Exothermic and Endothermic processes
For an exothermic reaction, heat is released and rH < 0. In this case the products are more
stable because they contain less enthalpy than they started with. The chemical potential
energy is not lost but is converted into increased kinetic energy of all the particles (both
reactants and products) and also transferred to the surroundings. This means the
temperature of all species and the surroundings increases. All combustion reactions are
exothermic.
Reactants
Enthalpy, H
rH < 0
Products
An endothermic reaction absorbs heat from the surroundings because the enthalpy of the
products is more than reactants and rH > 0. This means the temperature of the system
gets colder. Spontaneous endothermic reactions tend to be relatively rare and usually only
absorb a small amount of heat. Dissolving may be endothermic or exothermic depending on
the nature of the solute and solvent e.g. dissolving NH4NO3(s) in water is endothermic while
dissolving conc H2SO4 is exothermic.
Exercise:
In the space below draw a labeled diagram for an endothermic reaction, showing reactants,
products and rH.
Enthalpy of physical changes
A heating curve of a substance shows the variation of temperature of a sample as it is
heated.
gas
b.p.
Temperature
/ oC
liquid
m.p.
solid
Time
At both its melting point and boiling point, even though the sample is being heated,
temperature remains constant. All heat being added is used to break the forces holding the
molecules in their solid or liquid states, rather than being converted into kinetic energy.
Enthalpy of vaporisation, vapH, is the difference in enthalpy per mole of molecules between
the vapour and liquid states of a substance.
vapH = Hvapour - Hliquid
An example would be the vaporisation of water.
vapH (H2O, 100 oC) = +40.7 kJ mol-1
The reaction equation which would have this value of vapH (H2O, 100 oC) is
H2O(l) → H2O(g)
Vaporisation is always endothermic as molecules need energy to overcome the attractive
forces holding the molecules in the liquid state. Cooling a gas to a liquid will always be
exothermic as heat is released as the bonding forms between molecules in the liquid state.
The enthalpy of fusion, fusH, is defined as the enthalpy change that occurs when one mole
of molecules changes from solid to liquid state. All enthalpies of fusion (melting) are
positive as energy must be added to cause the molecules to vibrate more vigorously until
they can move past each other and flow as a liquid. The equation for the reaction that has
an enthalpy change value given by fusH(CH3OH) is:
CH3OH(s) → CH3OH(l)
The enthalpy change for freezing is the negative value of the equivalent enthalpy change of
fusion.
The enthalpy of sublimation, subH is the enthalpy change that occurs when one mole of
solid is turned into one mole of gas eg C(s) → C(g)
Enthalpy of Chemical Change
Like all hydrocarbons methane undergoes complete combustion to form carbon dioxide and
water.
CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)
rH0 = -890 kJ mol-1
The enthalpy value given is for the reaction carried out at standard state (one atmosphere
pressure). This means that the water produced is in the liquid rather than gaseous state. If
the water produced were in the gaseous state, then the value of rH0 would not be the same.
CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)
rH0 = -802 kJ mol-1
CH4(g) + 2O2(g)
802 kJ
CO2(g) + 2 H2O(g)
890 kJ
88 kJ
CO2(g) + 2 H2O(l)
The diagram above shows that the conversion of two moles of H2O(g) to H2O(l) results in the
release of 88 kJ of heat.
Exercise 1
Based on the information above, what is the enthalpy of vaporisation of water, vapH0?
Note that the unit for rH0 is kJ mol-1. This refers to the number of moles of reactants and
products in the specific balanced equation for which the value of rH0 is given. It may or
may not be 1 mole of each reactant.
If the combustion reaction were reversed and 1 mole of CH4 were produced, then the value
of rH0 is +890 kJ mol-1. This is the same numerical value but with the opposite sign.
For combustion of two moles of CH4(g) then rH0 = - 1780 kJ mol-1.
The process of photosynthesis is an endothermic process in which energy from the sun is
trapped and stored in the bonds of glucose.
6CO2(g) + 6H2O(l)

C6H12O6(aq) + 6O2(g)
rH0 = +2808 kJ mol-1
It is however easier to measure the enthalpy change for the reverse reaction, the
combustion of glucose (i.e. the process of respiration).
C6H12O6(aq) + 6O2(g) 
6CO2(g) + 6H2O(l)
rH0 = -2808 kJ mol-1
NOTES:
 While most thermochemical data are reported at 25 oC, we can have a standard state at
any temperature.
 Because the state of the reactants and products is highly significant in thermochemical
equations, it is important that they are always included.
Standard enthalpy of combustion, cH0
The standard enthalpy of combustion is the change in enthalpy per mole of a substance
when it reacts completely with oxygen under standard conditions. This means the enthalpy
of combustion refers only to the equation involving the combustion of one mole of reactant,
and the units of cH0
are kJ mol-1.
In the complete combustion of an organic compound the carbon atoms are converted to
CO2(g) and the hydrogen atoms are converted to H2O(l).
Exercise 2 Write equations for the reactions which have the following enthalpies of
combustion:
cH0 (H2, 25 oC) = -286 kJ mol-1
cH0 (CH3OH, 25 oC) = -726 kJ mol-1
Standard enthalpy of formation, fH0
fH0 is the standard reaction enthalpy for the formation of one mole of substance from its
elements in their most stable form at standard state. The units of fH0 are kJ mol-1 and the
equation it refers
to must have only one mole of product.
NOTE: If 4C(s) + 6H2(g) + O2(g) 
Then fH0 (C2H5OH, l) = -555  2
2C2H5OH (l)
rH0 = -555 kJ mol-1
= - 277.5 kJ mol-1
The standard enthalpy of formation of any element in its standard state is arbitrarily
defined to be zero kJ mol-1. This is because we can only measure changes in energy and
therefore a reference point is needed to compare against. This is the same principle as
used in electrochemical cells where Eo(H+/H2) is assigned a zero value.
Exercise 1 Write balanced equations for the formation reactions of
(a) C6H12O6(s)
(b) CO(g)
(c) HCl(g)
ENERGY CALCULATIONS
Calculations involving thermochemical principles may include:
 relating enthalpy changes to heat and mass, and use of specific heat capacity of
water
 application of  rH =  fH (products) -  fH (reactants)
 application of Hess’s Law
 use of average bond energies in enthalpy change calculations.
Measuring enthalpies of reaction – calorimetry
To measure enthalpy changes, the reaction is carried out in an insulated container or
calorimeter (such as a polystyrene cup) and the temperature change is measured. Using
this temperature change, ΔT, and the value of the specific heat capacity, c, the amount of
energy transferred to the mass m of
substance can be calculated using the expression
E = m c ΔT
If the reaction takes place in aqueous solution, and the solutions are placed in an insulated
container, the specific heat capacity of the solutions is 4.18 J oC-1 g-1. If the densities of
these aqueous solutions is taken as 1.0 g mL-1 then 100 mL would have a mass of 100 g.
If the solution in the calorimeter gets hotter the reaction itself is exothermic. Assuming all
the heat has been transferred to the solution then the measured E = –  rH. In reality some
heat is often lost to the surroundings.
Exercise 2
25.0 mL of 1.0 mol L-1 HCl is placed in a polystyrene cup and its temperature is measured as
21.0 oC. Following the addition of 25.0 mL of 1.0 mol L-1 NaOH solution, the mixture is
stirred and the final temperature recorded as 27.8 oC. Calculate the molar enthalpy change,
ΔrH in kJ mol-1, for this reaction. (Note: If the temperature increases then the reaction is
exothermic and ΔrH is negative)
Hess’s Law
If an overall reaction can be broken down into a series of two or more steps, then the
corresponding overall enthalpy of reaction is the sum of the enthalpies of the individual
reaction steps. None of the steps need to be a reaction that can be carried out in the
laboratory. e.g. it is impossible to experimentally measure the enthalpy of formation for
CO(g) as some CO2 is always formed as a by-product. However, using the energy cycle
below, it is clear that as ΔH1 = ΔH2 + ΔH3 then the required ΔH2 can be calculated from the
known values of ΔH2 and ΔH3.
ΔH2
C(s) + ½ O2(g)
CO(g)
ΔH3 + ½ O2(g)
ΔH1
CO2(g)
Another way of saying this is that the energy difference depends only on the difference in
energy between the reactants and products, not on the reaction path taken.
reactants
Reaction 1, H1
Enthalpy,H
Reaction 2, H2
Reaction 3, H3
Htotal
products
rHtotal = rH1 - rH2 +
rH3
Example
The octane in petrol may burn to carbon monoxide, CO, if the air supply is limited. Given the
standard enthalpies of combustion of C8H18(l) and CO(g), calculate the standard reaction
enthalpy for the incomplete combustion of one mole of octane.
C8H18(l) + 8.5O2(g)  8CO(g) + 9H2O(l)
Eqn 1
cHo (C8H18, l) = -5471 kJ mol-1
cHo (CO, g) = -283 kJ mol-1
Solution:
Start by writing equations for the two combustion reactions .
C8H18(l) + 12.5O2(g) 
CO(g) + 0.5O2(g) 
8CO2(g) + 9H2O(l)
CO2(g)
Eqn 2
Eqn 3
Equation 1 shows that 8 moles of CO are needed as a product, so equation 3 is reversed
and multiplied by 8.
8CO2(g)  8CO(g) + 4O2(g)
rH0 =(+283 x 8) kJ mol-1
Equation 1 also shows that 1 mole of C8H18 is required as a reactant, so this can be added
to the equation above to obtain the required overall equation.
8CO2(g)

8CO(g) + 4O2(g)
rH0 = + 2264 kJ mol-1
C8H18(l) + 12.5O2(g) 
8CO2(g) + 9H2O(l)
rH0 = - 5471 kJ mol-1
C8H18(l) + 8.5O2(g) 
8CO(g) + 9H2O(l)
rH0 = -3207 kJ mol-1
Exercises
1. Calculate the enthalpy change for the formation of ethanol from CH4 according to the
equation:
2CH4(g)
+
O2(g) 
2CH3OH(l)
Use the following equations and heats of reaction.
CH4(g) + H2O(g) 
2.
rHo = +206 kJ mol-1
CO(g) + 3H2(g)
2H2(g) + CO(g) 
CH3OH
rHo = -128 kJ mol-1
2H2(g) + O2(g) 
2H2O(g)
rHo = -484 kJ mol-1
P4(s)
PCl3(l)
+
+
6Cl2(g) 
4PCl3(l)
rHo = -1279 kJ mol-1
Cl2(g) 
PCl5(s)
rHo = -124 kJ mol-1
Using the information above calculate the enthalpy for the following reaction
P4(s)
+
10Cl2(g) 
4PCl5(s)
3. Calculate the reaction enthalpy for the synthesis of hydrogen chloride gas.
H2(g) + Cl2(g)  2HCl(g)
Use the following data.
NH3(g) + HCl(g) 
NH4Cl(s)
rHo = -176 kJ mol-1
N2(g) + 3 H2(g) 
2 NH3(g)
rHo = -92 kJ mol-1
N2(g) + 4 H2(g) + Cl2(g) 
2 NH4Cl(s)
rHo = -629 kJ mol-1
Calculating rHo given the standard heats of formation of reactants and products.
Another consequence of the Law of Conservation of Energy is that the standard enthalpy of
any reaction can be obtained by subtraction of the standard enthalpies of formation of
reactants from those of the products.
rHo =  n fHoproducts -  n fHoreactantss
where n is the stoichiometric coefficient of each substance in the reaction equation.
Example
Using the standard heats of formation of CO2(g), H2O(l), and C6H12O6(s), calculate the
standard enthalpy of combustion of glucose.
fHo(CO2, g) = -394 kJ mol-1
fHo(H2O, l) = -286 kJ mol-1
fHo(C6H12O6, s) = -1268 kJ mol-1
Hint - Start by writing an equation for the combustion of 1 mole of glucose.
C6H12O6(s) + 6O2(g)

6CO2(g) + 6H2O(l)
rHo
=  nfHoproducts -  nfHoreactants
rHo
= ( 6 x -394 + 6 x -286) - ( -1268 + 0) = - 2812 kJ mol-1
Exercises
1. a) Using the standard heats of formation given calculate the enthalpy change for the
following reaction.
2SO2(g) + O2(g)
 2SO3(g)
fHo(SO2, g)
= -297 kJ mol-1
fHo(SO3, g)
= -396 kJ mol-1
b) What is the enthalpy change when 100 g of SO2 burn to form SO3?
2. Calculate fHo(PCl5, s) from the following information.
fHo(PCl3, l)
= -320 kJ mol-1
PCl3(l) + Cl2(g)

PCl5(s)
 rH = -124 kJ mol-1
3. Ethanoic acid can be formed by the oxidation of ethanol.
C2H5OH(l) + O2(g) 
CH3COOH(l) + H2O(l)
This reaction occurs when wine goes sour. By calculating the standard reaction enthalpy
for this oxidation reaction, decide whether the reaction is exothermic or endothermic.
fHo(H2O, l) = -286 kJ mol-1
fHo(CH3COOH, l) = -485 kJ mol-1
fHo(C2H5OH, l) = -278 kJ mol-1
Using Bond Enthalpies to calculate rHo
Bond enthalpy is the change in enthalpy when the covalent bond, in a gaseous molecule,
is broken. It is always a positive value because bond breaking always requires an input of
energy. Conversely, making bonds releases energy and this means that, in general, the
more bonds a substance can form the more stable it will be.
The strength of a covalent bond depends on the electrostatic attraction between the positive
nuclei and the shared electron pair. Obviously this means that trends in bond enthalpy down
a group, (eg decreasing bond enthalpy from H-Cl to H-I), can be explained by the increasing
atomic radius of the halogen atom and therefore decreasing electrostatic attraction. The
stronger a covalent bond, the higher the value of the bond enthalpy. The units are kJ mol-1.
Exercise:
Explain the relative values of the following bond enthalpies:
E (C-F) = +452 kJ mol-1
E (C-Cl) = +293 kJ mol-1
The high value for the C-F bond explains why polymers of fluorocarbons such as teflon are
very resistant to chemical attack and can therefore be used to line the interior of chemical
reaction vessels.
In a polyatomic molecule, the bond strength between a given pair of atoms varies slightly
from one compound to another. Thus in CH4 the value of E(C-H) is the average value for all
four C-H bonds.
A multiple bond is always stronger than a single bond because more electrons bind the
multiply bonded atoms together. These trends can be clearly identified in the table below
showing average bond enthalpies.
Bond enthalpy /kJ mol-1
Bond enthalpy /kJ mol-1
Bond enthalpy /kJ mol-1
H-H
436
C-H
412
C=C
612
H-O
463
C - Cl
338
C≡C
837
H-N
388
C-F
484
C=O
743
H - Cl
431
C-O
360
O=O
496
H-F
565
C-C
348
NN
944
F-F
158
O–O
146
Cl - Cl
242
It is possible to calculate an approximate rHo for any reaction (provided all products and
reactants are in the gas phase) by calculating the difference between the total bond
energies for reactants and products in the equation for the reaction.
rH0
=  Ereactants -  Eproducts
Note that in this particular calculation we subtract the total bond energies of the products
from the reactants. This is because the bond enthalpy is defined as the energy required to
break the bond. It is always an endothermic reaction with a positive bond enthalpy.
Example
Use the average bond enthalpies given in the previous table to calculate the enthalpy
change for the following reaction
CH4(g)
+ 2F2(g)

CH2F2(g) + 2HF(g)
HINTS:
(i) Draw the molecules out as expanded structural formulae and clearly identify those bonds
that
are broken and made.
(ii) Remember that some bonds maybe muliple bonds eg O=O and N≡N
(iii) Remember to allow for the number of moles present eg in 2CO2 there are 4 C=O bonds.
H
H
H C H
+ 2F-F

H
H
rHo
C
F
+ 2H-F
F
=  Ereactants -  Eproducts
= (4 x EC-H + 2 x EF-F ) - (2 x EC-F + 2 x EC-H + 2 x EH-F)
= 1964 – 2922
= - 958 kJ mol-1
Exercises
1. Use the bond enthalpies given on the previous page to calculate the standard enthalpy
of combustion of the following molecules in the gas phase.
a) H2(g)
b) CH3CH3(g)
2. Calculate the reaction enthalpy for the following reaction:
CH2=CH2(g)
+
HCl(g) 
CH3CH2Cl(g)
3. Show that the reaction enthalpy for the conversion of solid carbon, C(s) to gaseous CO2
is
–274 kJ mol–1.
Note that in this case you will need to use bond enthalpies for the conversion of C(g) to
CO2(g) and combine this with the value of subH(C,s) = 716 kJ mol–1 (remember Hess’s
law)