Chapter 0.10 - Chemical Bonding
It is an observable fact that matter is usually found in nature aggregated into
masses of some kind. These aggregations may take the form of solids, liquids or
gases. Even gases condense to liquids and solids as the temperature is
decreased. The only explanation for such behaviour is the existence of forces that
pull the particles together.
Measurement and observation of the particles in matter leads to the conclusion
that the very atoms themselves are bonded together in some way to form the
underlying structure of matter. In fact, atoms are almost never to be found on
their own in nature (the exception is the noble gases). It seems that it is more
energetically favourable for atoms to be bonded to other atoms than for them to
exist on their own.
The study of bonding presents theories to demonstrate the forces and conditions
to which atoms are subject in order to explain the observable nature of matter
and materials.
In Chapter 0.10
0.11 - The forces of nature
0.12 - Covalent bonding
0.13 - Covalent bond characteristics
0.14 - Macromolecules
0.15 - Lewis structures
Section 0.11 - The forces of nature
IB Syllabus ref: 4.2.1
4.2.1 Describe the covalent bond as the electrostatic attraction between a pair of
electrons and positively charged nuclei.
There are four fundamental forces in nature that can act over a distance. We are
very familiar with gravity, for example, and accept that it acts on any object in a
gravitational field. Two of the other forces are applicable only to the nuclei of
atoms and, as such, do not concern us here. The fourth force is the
electromagnetic force. This has various manifestations, such as a magnetic field
acting on magnetic materials. One other manifestation of this force is the
electrostatic field that is associated with charged particles. This is the force
responsible for the structure of all matter.
•Electrical charge
•Electrostatic field
•Electrostatic force
•Forces in atoms
•Forces between atoms
Electrical charge
Some of the fundamental sub-atomic particles carry electrical charges. These are
static electrical charges; they are a characteristic of the particle itself and cannot
be lost.
•Electrons carry a negative charge, represented by a minus sign:
•Protons carry a positive charge, represented by a plus sign:
The words 'positive' and 'negative' are just conventions to describe the opposite
nature of the two types of charge.
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Electrostatic field
An electrically charged particle has an associated electrostatic 'field'. This means
that any other charged particle entering that region of space experiences a force.
If the approaching particle has the same electrical charge, then it feels a
repulsive force. If it has an opposite electrical charge it feels an attraction.
An electrostatic field cannot be seen (unlike the demo), but it is there
nevertheless and any other charged particle entering this region of space feels a
corresponding force due to this field.
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Electrostatic force
The forces created by electrostatic fields interacting are very straightforward.
Positively charged particles attract negatively charged particles, and vice versa.
This force of attraction acts along an imaginary line joining the centres of the two
charges.
Particles with like charges, i.e. both negative, or both positive, repel one another.
This effect is the fundamental force behind the formation of all matter. Atoms are
held together by electrostatic forces, atoms are bonded to one another by
electrostatic forces, molecules are attracted to one another by electrostatic
forces, metals are held together by electrostatic forces and finally ionic
compounds are held together by electrostatic forces.
You could say that without this force there would be no universe.
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Forces in atoms
Atoms have protons in the nucleus and electrons in regions of space around the
outside of the atom. The nucleus of atoms is positively charged and the individual
electrons each have a single negative charge. Atoms are always neutral, meaning
that the number of protons, and the magnitude of the positive charge, is always
balanced by the same number of electrons and negative charge.
Example: The hydrogen atom.
1 proton, 1 electron= 1 positive charge and 1 negative charge
Overall charge 1 + 1- = 0
Why do the electrons not fall into the nucleus?
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Forces between atoms
When there is a force of attraction between positive and negative charges from
different atoms, it draws the atoms together and holds them in position. This is
called a chemical bond. The bonded atoms have lower overall energy than the
individual component atoms, so the bonded situation is more stable and
preferred.
Even though the positive nucleus has its charge balanced by the electron cloud, a
single electron approaching from one side will experience the repulsion from the
electrons that are nearest, but will be more powerfully attracted by the nucleus.
This also applies to electrons in different atoms; they are attracted to
neighbouring atoms.
It is this force of attraction between the nucleus of one atom and the the
electrons of another that is the basis of chemical bonding. These forces have the
effect of holding the nucleus of one atom to the nucleus of another atom, i.e.
bonding the two atoms together.
In reality, these bonding forces take slightly different forms in different situation,
but it is true that we are always dealing with some kind of electrostatic force
between atoms.
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Section 0.12 - Covalent Bonding
Syllabus ref: 4.2
4.2.1 Describe the covalent bond as the electrostatic attraction between a pair of
electrons and the positively charged nuclei.
Covalent bonding usually occurs between non-metal atoms. It is the simplest kind
of chemical bonding.
•The covalent bond
•Bonding theory
•Linear combination of atomic orbitals
•Hydrogen molecule
•Full outer energy shells
•Summary of covalent bonding
The covalent bond
Atoms are held together in covalent bonding by means of shared pairs of
electrons. This constitutes a single covalent bond. The various theories as to how
and why these bonds are formed are discussed below. It is important to recognise
that formation of covalent bonds (or any other type of bond) is an exothermic
process, one that releases energy. Similarly, to break a bond always requires
energy. Any breakage of covalent bonds involves a chemical reaction and new
substances are necessarily formed.
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Bonding theory
Atoms are unstable unless they have fully occupied outer shells of electrons. We
know from observation and inference that atoms bond together to make larger
structures. In order to explain how this happens, different theories are proposed
that explain these observations. All of these theories revolve around the accepted
idea that opposite charges attract (electrostatic attraction).
The scientific method
A theory helps to explain all of the observable facts. If it fits all of the evidence,
then it is adopted as a useful explanation of how the universe works. However, if
further observations throw up anomalies, or observations that cannot be
explained by the current theory, then it must be either adapted, or even
discarded completely.
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Linear combination of atomic orbitals
This theory of bonding proposes that molecules are formed by overlapping
regions of space, allowing atoms to mutually share pairs of electrons. This pair of
electrons is then held in a position between the two nuclei of the atoms holding
them together. We can consider this theory by looking at the hydrogen molecule.
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The hydrogen molecule
The hydrogen molecule is the simplest structure formed between atoms. In this
case two hydrogen atoms share one pair of electrons between them, forming a
diatomic molecule.
The negative charge clouds (atomic orbitals) overlap, placing a region of negative
charge between the two hydrogen nuclei.
The electrons in each atom is also attracted to the nucleus of the neighbouring
atom. The atoms are pulled together. This view of chemical bonding is known as
covalent bonding as the valence electrons of both the hydrogen atoms are
shared. ('co' = shared and 'valent' = valence electrons)
The regions of space in which electrons are found are called orbitals and we say
that this bond arises from direct overlap of atomic orbitals. This is sometimes
called the Linear Combination of Atomic Orbital Theory (LCAO).
This model works very well for a simple picture of bonding between atoms. The
nuclei of the atoms are held to one another by the positive - negative - positive
attractions caused by electron pairs between the nuclei.
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Full outer energy shells
The only atoms that occur in nature not combined into some kind of structure are
the inert gases. These all have full outer (valence) shells. It seems to be that this
particular electronic configuration is stable for some reason.
Atom configuration
helium 2
neon 2,8
argon 2,8,8
krypton 2,8,18,8
xenon 2,8,18,18,8
Covalent bonding occurs between atoms so that they can attain a full outer shell
by SHARING electrons.
In the case of hydrogen molecules, above, the atom would have a full outer shell
if it were to attain two electrons. By an arrangement in which two atoms share
one pair of electrons, each atom achieves the requirement. The system is now
stable, it does not change further.
Covalent bonding usually occurs between non-metal atoms; they attain a full
outer shell of electrons by sharing electrons. However, as we shall see in the
following section there are exceptions.
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Summary of covalent bonding
•Covalent bonding happens when non-metal atoms combine
•Electron pairs are shared to give a full outer shell of valence electrons to each
atom
•The final units are molecules of atoms joined together by shared pairs of
electrons
The covalent bond (shared pair) may be represented by a line to show typical
stick structures such as:
ethyne propanal propyne
A single line is drawn to represent one pair of electrons. A double bond containing
two pairs of electrons between atoms is shown as a double line, and three pairs,
a triple bond, as a three lines. It is not possible for two atoms to share more than
three pairs of electrons.
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Section 0.13 - Covalent bond characteristics
Syllabus ref: 4.2.4
4.2.4 State and explain the relationship between the number of bonds, bond
length and bond strength. The comparison should include bond lengths and bond
strengths of: two carbon atoms joined by single, double and triple bonds, the
carbon atom and the two oxygen atoms in the carboxyl group of a carboxylic
acid.
The characteristics, or properties of bonds formed by sharing electron pairs is
examined in this section.
•Single bonds and the octet rule
•Dative covalent bonds
•Double bonds
•Triple bonds
•Electron deficient molecules
•Bond length
•Bond strength
•Worked examples
Single bonds and the octet rule
As stated in the previous section, single bonds consist of a shared pair of
electrons. Each bonded atom usually provides one electron for the shared pair.
The result is a full outer shell of electrons. This is known as the octet rule. This
rule suggests that covalent bonds form so as to provide each atom in the
molecule with a full outer shell of electrons, usually eight, therefore 'octet' (group
of eight).
Each fluorine atom provides one electron to the shared pair.
The fluorine atoms both now have a full outer shell (octet) of electrons
Fluorine molecule
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Dative covalent bonds
Usually, the two atoms involved in a covalent bond provide one electron each to
make the pair. However, occasionally, both of the electrons come from one of the
atoms. In this case, the bond is said to be dative (= giving) covalent. The fact
that an electron pair is dative has no influence on the final structure.
The central oxygen atom is bonded to one of the other oxygen atoms by a dative
covalent bond (blue pair).
All of the oxygen atoms have a full outer shell (octet) of electrons.
Ozone molecule
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Double bonds
Double bonds consist of two shared pairs of electrons between the bonded atoms.
The two single bonds are actually different in character and are given names to
differentiate them, sigma and pi bonds. A double bond always consists of 1 sigma
bond and 1 pi bond.
Oxygen molecule
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Triple bonds
Triple bonds consist of three shared pairs of electrons between the bonded atoms.
These bonds are not all the same and consist of one 'sigma' bond and two 'pi'
bonds.
In this case one of the 'pi' bonds is formed by donation of one pair of electrons
from the oxygen to the bond. It is a dative bond.
Carbon monoxide molecule
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Electron deficient molecules
There are exceptions to the octet rule, such as NO and NO2, in which the nitrogen
atom does not have a full outer shell. The molecule is said to be electron
deficient, as it is missing an electron.
Nitrogen has only seven electrons in its outer shell. It is electron deficient.
Molecules such as this are usually very reactive.
nitrogen(II) oxide molecule
Electron deficient molecules can be identified by counting up all of the available
valence electrons. If they add up to an odd number then there MUST be one
electron left over somewhere, as an unshared single electron.
In the case of the nitrogen(II) oxide molecule (diagram above), the formula is
NO. Nitrogen is from group V and has five valence electrons. Oxygen is from
group VI and has six valence electrons.
Total valence electrons = 5 + 6 = 11
This is an odd number and so there has to be an unpaired electron somewhere.
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Bond strength
The strength of the bond depends very much on the atoms being bonded.
However, carbon forms single, double and triple bonds with other carbon atoms
and we can use this to determine the relative strength of these bonds.
Bond type Bond strength
C - C 348 kJ mol-1
C = C 612 kJ mol-1
C C 837 kJ mol-1
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Bond length
Single bonds use one pair of electrons to hold atoms together. Greater force of
attraction between the electron pairs and the two nuclei draw the atoms closer
together reducing the bond length. The values in the table below are given in nm
(nanometres) = 1 x 10-9 m
Bond type Bond length
C - C 0.154 nm
C = C 0.134 nm
C C 0.120 nm
This is not just the case for carbon carbon bonds, the principle applies to all
single and multiple bond systems; i.e. a carbon - oxygen single bond is longer
than a carbon - oxygen double bond.
Bond type Bond length
C - O 0.133 nm
C = O 0.122 nm
C O 0.111 nm
We can make use of these facts to make logical deductions about the nature of
bonding in certain molecules.
For example, studies show that the carbon - oxygen bond lengths in carboxylic
acids is different for the two bonds. This can be explained by the fact that one of
the bonds is a double bond whereas the other is a single bond.
However, bond lengths measured in the carboxylate ion, indicate that both C-O
bonds are the same length. The reason for this is dealt with in the resonance
section.
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Worked examples
Q013-01 The most likely molecule to be formed by the reaction of boron and
chlorine is:
A.BCl
B.BCl3
C.BCl4
D.BCl6
Answer
-------------------------------------------------------------------------------Boron is in group 3 and has three valence electrons and a valency of 3. Each
chlorine can share one pair to get a full outer shell. The final formula is BCl3
This is an electron deficient compound, as there are only six electrons in the outer
shell of the boron atom.
-------------------------------------------------------------------------------Q013-02 In which molecule would you find a triple covalent bond?
A.O2
B.N2
C.O3
D.FeCl3
Answer
-------------------------------------------------------------------------------Nitrogen is from group 5 of the periodic table. It needs to share three pairs of
electrons to attain a full outer shell. Therefore the nitrogen molecule has three
bonds per atom.
-------------------------------------------------------------------------------Q013-03 Which diatomic species should have the greatest bond dissociation
energy?
A.H2
B.F2
C.NO
D.CO
Answer
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The bond dissociation energy is the energy required to break the bond. It is a
measure of the strength of the bond. The question is really asking "which has the
strongest bond?". Triple bonds are stronger than double bonds which are stronger
than single bonds.
This time the triple bond is found in the carbon monoxide, CO molecule:
-------------------------------------------------------------------------------Q013-04 When C2H4, C2H2 and C2H6 are arranged in order of increasing C-C
bond length what is the correct order?
A.C2H6, C2H2, C2H4
B.C2H4, C2H2, C2H6,
C.C2H2, C2H4, C2H6
D.C2H4, C2H6, C2H2
Answer
-------------------------------------------------------------------------------The stronger the bond the shorter it is, therefore the order we are looking for is
triple. double, single bonds.
This is the order of C2H2, C2H4, C2H6
-------------------------------------------------------------------------------Q013-05 Which statement is correct about two elements, whose atoms form a
covalent bond with each other.
A.The elements are metals
B.The elements are non-metals
C.The elements have very low electronegative values
D.The elements have very different electronegativity values
Answer
-------------------------------------------------------------------------------Covalent bonds are normally formed between two non-metals (although not
exclusively)
-------------------------------------------------------------------------------Q013-06 Which of the following increases for the bonding between carbon atoms
in the sequence of molecules C2H6, C2H4, C2H2?
I - Number of bonds
II - Length of bonds
III - Strength of bonds
A.I only
B.I and III only
C.III only
D.I, II and III
Answer
-------------------------------------------------------------------------------The three molecules, ethane, ethene and ethyne have C-C bonds of single, double
and triple in that order.
•I - Number of bonds is correct, it increases.•II - Length of bonds is incorrect, it
decreases•III - Strength of bonds is correct, it increasesTherefore the correct
answer is I and III only
-------------------------------------------------------------------------------Q013-07 How do bond length and bond strength change as the number of bonds
between two atoms increase?
bond length bond strength
A increases increases
B increases decreases
C decreases decreases
D decreases increases
Answer
-------------------------------------------------------------------------------Increasing bond number decreases bond length and increases bond strength =
option D
-------------------------------------------------------------------------------Q013-08 The length of the bond between carbon and oxygen is shortest in:
A.CO
B.CO2
C.CH3CH2OH
D.CH3CHO
Answer
-------------------------------------------------------------------------------In CO the two atoms are bonded by a triple bond. It is the shortest.
-------------------------------------------------------------------------------Q013-09 Which molecule has the longest nitrogen - nitrogen bond length?
A.N2
B.N2F2
C.N2H4
D.N2H2
Answer
-------------------------------------------------------------------------------The longest bond length applies to a single bond. Only N2H4 contains an N-N
single bond.
-------------------------------------------------------------------------------Q013-10 In which of the following is there at least one double bond?
I - O2
II - CO2
III - C2H4
A.I only
B.III only
C.II and III only
D.I,II and III
Answer
-------------------------------------------------------------------------------I - O2 yes II - CO2 yes III - C2H4 yes
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Section 0.14 - Macromolecules
IB Syllabus ref: 4.2.9; 4.2.10
4.2.9 Describe and compare the structure and bondg in the three allotropes of
carbon (diamond, graphite and C-60 fullerene)
4.2.10 Describe the structure and bonding in silicon and silicon dioxide
A giant molecular structure, or network solid, has a virtually infinite arrangement
of atoms, all of which are bonded into position using strong covalent bonds.
Effectively the whole unit is one molecule.
•Allotropes
•Giant molecular structures
•Diamond
•Graphite
•Fullerenes
•Silicon
•Silicon dioxide
•Worked examples
Allotropes
Allotropes are different physical forms of the same element. They have different
structures on a molecular scale.
Allotropic forms can arise in two fundamental ways.
•The atoms themselves can be bonded together in a different way to make
different molecules.
•The molecules can pack together in different arrangements
Most elements have allotropic forms. Oxygen can exist in the gaseous state as
simple molecules of O2, or ozone, O3. sulfur can exist in several physical forms,
each of which has a different crystal structure. Phosphorus exists as white
phosphorus, red phosphorus and purple, or black phosphorus.
However, it is probably carbon that has the most famous allotropes, diamond,
graphite and the more recently discovered fullerene. Two of these allotropes are
macromolecular, or giant covalent structures, and the other is simple molecular.
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Giant molecular structures (macromolecules)
These may be either elements or compounds.
Note: Students often ask "but how does the structure end?"
The answer is that the final bonding electrons are used up bonding to oxygen, or
another suitable element. However, this is unimportant compared to the vast bulk
of the atoms on the 'inside' of the structure
A giant molecular structure, or network solid, has a virtually infinite arrangement
of atoms, all of which are bonded into position using strong covalent bonds.
Effectively the whole unit is one molecule. Diamond is an example of a three
dimensional network solid element.
It clearly is not possible to draw a diagram showing the whole molecular structure
of a macromolecular solid, so normally only a representative part is shown.
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Diamond
Diamond is a three dimensional lattice (network) of tetrahedral carbon atoms.
The structure is horribly difficult to draw well, but a small part may be shown
fairly effectively.
All of the carbon atoms are bonded to four other carbon atoms in a tetrahedral
arrangement. All of the valence electrons are used for bonding. There are no
valence electrons free for conduction of electricity. This rigid lattice gives diamond
its unique properties. It has high melting and boiling points and is the hardest
naturally occurring substance.
Each carbon atom has sp3 hybridisation with bond angles of 109.5º (HL only).
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Graphite
Graphite is another naturally occurring allotrope of carbon. In this case the
structure is made up of giant 2-dimensional layers, each layer held together in
place by Van der Waal's attractions.
The layers themselves consist of hexagons of carbon atoms in which each carbon
is bonded to three others. This leaves one free electron per carbon atom that can
be delocalised throughout a layer and conduct electricity.
Graphite is a black slippery solid with a very high melting and boiling point. It's
slippery nature comes from the ability of the layers to slide over one another.
This is because the Van der Waals' forces that hold the layers to one another are
weak.
Click on the diagram at the right to see the layers.
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Fullerenes
The graphite layer structure can be folded and bent into balls and tubes providing
the number of carbon atoms in the sheet is large enough. These balls and tubes
(bucky balls and nano tubes) have all been developed from the discovery by
Kroto of spherical molecules consisting of 60 carbon atoms (and greater). These
molecules were called fullerenes after the architect who popularised the
geodhesic dome (Richard Buckminster Fuller), as the sphere is effectively two
hemispherical geodhesic domes stuck together to make a sphere.
Fullerene is an allotrope of carbon with 60 carbon atoms in the molecule. It is not
a giant structure (C60 is just not that large when compared to 'real' giant
structures, such as graphite and diamond). It was first isolated from soot and can
be crystallised as black, or red crystals, with a melting point of about 280ºC.
Fullerene is soluble in many non-polar solvents.
The structure consists of five and six membered carbon rings in which each
carbon atoms is attached to three other carbon atoms. The hybridisation at each
carbon is sp2 (HL only). It has delocalised electrons spread out over the whole
structure in one large molecular orbital.
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Silicon
Silicon also forms giant macromolecular structures similar to diamond, in which
all of the valence electrons are used to link each of the silicon atoms into a giant
array of tetrahedral atoms.
The properties of silicon is consistent with its macromolecular structure. It is a
hard solid with a very high melting and boiling point.
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Silicon dioxide
Silicon dioxide is a macromolecular compound with a structure which can be
thought of as having oxygen atoms interposed between each pair of silicon atoms
within the diamond-type lattice.
Silicon showing only one part of the giant macromolecular structure Silicon
dioxide showing how the oxygen atoms are bonded between silicon atoms
Silicon dioxide has the characteristic properties of a macromolecular compound; a
high melting and boiling point and very hard. All of its valence electrons are used
in bonding so it is a non-conductor. The silicon atoms (and the oxygen atoms)
use sp3 hybridisation.
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Worked examples
Q014-01 Which statements are correct about diamond, graphite and a C60
fullerene?
I - The poorest electrical conductor of the three is diamond
II - The atoms in graphite and C60 fullerene are attached to three other carbon
atoms
II - The atoms in diamond and C60 fullerene are arranged in hexagons
A.I and II only
B.I and III only
C.II and III only
D.I, II and III
Answer
-------------------------------------------------------------------------------•I - The poorest electrical conductor of the three is diamond. This is true, all of its
valence electrons are used up in bonding.•II - The atoms in graphite and C60
fullerene are attached to three other carbon atoms. This is true.•II - The atoms in
diamond and C60 fullerene are arranged in hexagons. They are not arranged in
true hexagons in either structure.Correct response - A options I and II only
-------------------------------------------------------------------------------Q014-02 State the structures of, and the bonding, in diamond and graphite.
Answer
-------------------------------------------------------------------------------Diamond and graphite are both giant molecular structures. In diamond, each
carbon is bonded to four others, and in graphite each carbon atoms is bonded to
only three others in layers. Each layer in graphite has an associated delocalised
electron layer and is held to other layers by weak van der Waals' forces.
-------------------------------------------------------------------------------Q014-03 Compare and explain the hardness and electrical conductivity of
diamond and graphite.
Answer
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Diamond is a hard, rigid structure, as each carbon atom is held in place by four
strong covalent bonds. It is a non-conductor, as all of its valence electrons are
used in bonding.
Graphite is a soft slippery solid, as the layers can slide easily over one another,
because they are held only by weak van der Waals' forces. The carbon atoms use
only three out of four electrons to make the framework structure of a layer and
the remaining electron is delocalised over the whole layer, allowing graphite to
conduct well.
-------------------------------------------------------------------------------Q014-04 Predict and explain how the hardness and electrical conductivity of C60
fullerene would compare with that of diamond and graphite.
Answer
-------------------------------------------------------------------------------Although there is delocalisation within a C60 fullerene molecule, it is expected to
be a non conductor as the simple molecules are not connected to one another by
any formal bonds. It is similarly expected to be soft.
Diamond is a hard, rigid structure, as each carbon atom is held in place by four
strong covalent bonds. It is a non-conductor, as all of its valence electrons are
used in bonding.
Graphite is a soft slippery solid, as the layers can slide easily over one another,
because they are held only by weak van der Waals' forces. The carbon atoms use
only three out of four electrons to make the framework structure of a layer and
the remaining electron is delocalised over the whole layer, allowing graphite to
conduct well.
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Section 0.15 - Lewis structures -SL
IB Syllabus ref: 4.2.3
4.2.3 Deduce the Lewis (electron dot cross) structure of molecules and ions for
up to four electron pairs on each atom. A pair of electrons can be represented by
dots and crosses or by a line.
Lewis structures are diagrams of molecules that represent electron pairs by dots,
crosses or lines. They show all of the valence electrons, including lone, or nonbonding pairs.
•Dot-cross representation•Lewis structures
•Multiple bonds
•Electron deficient molecules
•Odd electron species
•Simple ions
•Multi-centre ions
•Worked examples
Dot cross representation
Otherwise known as Lewis structures, dot-cross representations show the
electrons in the outer shell (valence electrons) of all of the atoms involved in a
molecule. Covalent bonds consist of electron pairs and there are also lone, or
non-bonding pairs.
Conventionally, either dots or crosses can be used to represent the electrons in
each outer shell:
In the image at the left, the four hydrogen atoms are bonded to the central
carbon atom by means of covalent bonds. Each bond consists of a pair of
electrons, conveniently coloured to distinguish those electrons that 'originally'
came from carbon from those that 'belong' to the hydrogen atoms.
Atoms (with the exception of the noble gases) are not stable on their own. They
MUST bond to produces structures in which the electronic arrangement is stable.
In 99% of cases this means that the outer shell has a full 'octet', or eight
electrons.
Methane Lewis Structure
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Lewis structures
The number of outer shell electrons may be found from the position of the atom
in the periodic table. For instance, fluorine is in group VII and so it has seven
outer (valence) electrons. All single covalent bonds are represented by a pair of
electrons, each bonding atom providing one for the pair. There are two fluorine
atoms in the fluorine molecule and so there are 15 electrons altogether. Two
electrons are used in the single F-F bond leaving 12 electrons to share over the
two atoms. These are arranged as three non-bonding pairs per atom.
Example: Draw the Lewis structure of the ammonia molecule.
Ammonia has the formula NH3, therefore there is one N atom and three H atoms.
•Nitrogen is in group 5, hence it has 5 electrons in the outer shell
•Hydrogen has one outer electron only; 3 x hydrogen = 3 electrons
Total outer electrons = 3 + 5 = 8
Each N-H bond requires 2 electrons, therefore 3 x N-H = 6 electrons used in the
three bonding pairs. Remaining electrons = 8 - 6 = 2 electrons, therefore one
non-bonding pair.
Animate
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Double and triple bonds
A double bond is shown as two shared pairs of electrons, each of the bonded
atoms provide two electrons for the bond. In the oxygen molecule at the left,
their are two shared pairs of electrons giving a stable octet (eight) of electrons
around each oxygen atom.
In triple bonds there are three pairs of electrons holding the two atoms together.
The molecule shown at the right is the ethyne (acetylene) molecule. Each of the
carbon atoms (group IV) originally started with 4 outer electrons. In the ethyne
molecule at the right, both of the carbon atoms now have a full octet (eight) of
electrons and the two hydrogen atoms have a full first shell, with two electrons.
There are no non-bonding pairs.
Animate
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Electron deficient molecules
These are strucures in which the outer (valence) shells do not have a full octet of
electrons. This may be because the molecule has been formed from
electropositive atoms that have few electrons in the outer shell, or possibly the
molecule is just an exception. Such molecules are by definition unstable and
there are few exceptions that need to concern us here.
Boron trifluoride is electron deficient, as the boron atom is relatively
electropositive and only has three outer electrons. The fluorine atoms themselves
do have a full outer shell, but the boron atom in the centre has only six electrons
in its valence shell.
Another example of electron deficiency is the beryllium dichloride molecule,
BeCl2.
To construct the Lewis structure of beryllim dichloride we have to first consider
the number of outer electrons in the central atom Beryllium is from group II and
has two electrons in the outer shell. It bonds to two chlorine atoms. Chlorine can
only use a single bond when bonding as it has 7 electrons in its outer shell. There
are two chlorine atoms therefore the beryllium shares two pairs of electrons, one
with each chlorine. However this then leaves beryllium with only four electrons in
its outer (valence) shell. It is electron deficient.
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Odd electron species
Occasionally molecules may manage to become stable even though they do not
have a full outer shell. This is not common, but is the case for nitrogen(IV) oxide,
NO2, at elevated temperatures. As you can see from the Lewis structure, there
are only seven electrons around the central nitrogen.
It is impossible to draw a Lewis structure in which all of the atoms are surrounded
by an octet of electrons. However, the molecule does exist. It is reactive and at
lowish temperatures soon pairs up to form a more stable structure, N2O4, in
which all of the atoms do have a full octet.
2NO2 N2O4
Nitrogen(IV) oxide
-------------------------------------------------------------------------------top
Simple ions
Ions are atoms or groups of atoms that carry a formal electrical charge. This
charge may be positive or negative. Negative charges mean that the atoms, or
groups of atoms have gained extra electrons to form a stable structure. Positive
ions, conversely, have lost electrons in order to attain a stable electronic
structure.
For example, a chlorine atom has the electronic configuration 2, 8, 7. It does not
have a full outer shell and when on its own is extremely reactive - it is called a
free radical. It's extreme reactivity is due to the instability of an incomplete outer
shell.
[Note that a chlorine atom and a chlorine free radical are two different ways of
describing exactly the same thing]
A chloride ion has one extra electron needed to complete the outer shell.
Consequently it carries a negative charge.
All simple ions can be thought of as the result of the atom needing to gain or lose
electrons to make up a full outer shell octet. Ions that consist of two or more
atoms can be treated in exactly the same way.
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Multi-centre ions
A multi-centre ion consists of two or more atoms that have gained or lost
electrons leaving a stable electronic structure. For example, the hydroxide ion has
the formula OH-. It is made up of one oxygen atom and one hydrogen atom,
which together have 6 + 1 = 7 valence electrons. There is a negative charge on
the ion, meaning that the two atoms have gained one more electron, making a
total of 8 electrons. The O-H bond requires one pair of electrons, leaving 6 to be
distributed over the oxygen and hydrogen atoms. The Hydrogen atom does not
need any more electrons as it already has a share of 2 in the outer shell. The
remaining 6 electrons therefore can go onto the oxygen atom as three nonbonding pairs.
In the diagram, the extra electron is show in green, although in reality there is no
way of distinguishing it from the others.
Hydroxide ion
Example: Show the Lewis structure of the carbonate ion.
The carbonate ion has the formula CO32-. It consists of one central carbon atom
and three oxygen atoms. The total number of valence electrons = 4 + (3 x 6) =
22. It also has a 2- charge making a total of 22 + 2 = 24 electrons.
The carbon shares two pairs of electrons with one of the oxygen atoms and one
pair of electrons with the other 2 oxygen atoms. This makes 8 electrons around
the central carbon atom, leaving 16 electrons to be distributed over the three
oxygen atoms. One of the atoms already shares four electrons, so it needs
another two non-bonding pairs (4 electrons). The other two oxygen atoms need
three non-bonding pairs (6 electrons).
Once again, the extra electrons are shown in green, but are indistinguishable
from the others.
Carbonate ion Lewis Structure
Test yourself
-------------------------------------------------------------------------------See if you draw the Lewis structure of the following molecules and ions. Check
your answer by clicking on the name.
Molecules
Ions
Boron trichloride
Boron trifluoride
Beryllium chloride
Aluminium chloride
Ammonia
Carbon monoxide
Sulfur dioxide
Image Ammonium ion
Hydronium ion
Nitrate ion
Hydroxide ion
Carbonate ion
Sulfate ion
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Worked examples
Q015-01 When the Lewis structure for HCOOCH3 is drawn how many bonding
and how many lone pairs of electrons are present?
Bonding pairs Lone pairs
A. 8 4
B. 7 5
C. 7 4
D. 5 5
Answer
-------------------------------------------------------------------------------The best way is to draw out the structure:
Counting up there are 4 lone pairs and 8 bonding pairs.
-------------------------------------------------------------------------------Q015-02 What is the Lewis (electron dot) structure of sulfur dioxide?
Answer
-------------------------------------------------------------------------------Sulfur dioxide has the formula SO2. Sulfur has six valence (outer shell) electrons
and each oxygen also has six valence (outer shell) electrons, giving a total of 18.
By double bonding one of the oxygens to the sulfur, the oxygen and sulfur now
have a full octets.
The sulfur can then donate a pair of electrons into the second oxygen's outer
shell to leave all the atoms with full octets
-------------------------------------------------------------------------------Q015-03 Butane, C4H10, propanal, C3H6O and propan-1-ol, C3H8O have very
similar molar masses (59±1). Draw the Lewis structures of each of these
molecules.
Answer
-------------------------------------------------------------------------------Butane propanal propan-1-ol
Click on the compound name for the Lewis structure
-------------------------------------------------------------------------------Q015-04 Hydrazoic acid can be represented by two possible Lewis structures, in
which the atoms can be arranged as NNNH. Draw the two possible Lewis
structures of N3H
Answer
-------------------------------------------------------------------------------Lewis structure 1 Lewis structure 2
-------------------------------------------------------------------------------Q015-05 Draw Lewis structures of each of the following species, NO2- and NO2+
Answer
-------------------------------------------------------------------------------With the negative ion, NO2-, you have to add one electron to the structure. So N
(5 valence electrons ) and 2 x oxygen (six valence electrons) = 17 electrons + 1
for the negative charge = 18 electrons.
With the positive ion, NO2+, you have to subtract one electron from the
structure. So N (5 valence electrons ) and 2 x oxygen (six valence electrons) =
17 electrons - 1 for the negative charge = 16 electrons.
NO2- NO2+
-------------------------------------------------------------------------------Q015-06 Draw Lewis structures to represent BF3 and NF3
Answer
-------------------------------------------------------------------------------BF3 NF3
-------------------------------------------------------------------------------Q015-07 Write two Lewis electron dot structures for the methanoate ion HCOOAnswer
-------------------------------------------------------------------------------Structure 1 Structure 2
-------------------------------------------------------------------------------Q015-08 Write Lewis electron dot structures for H2NNH2 and HNNH. What bond
angle is expected for the H-N-N atoms is each molecule?
Answer
-------------------------------------------------------------------------------Structure 1 Structure 2
Both nitrogen atoms are electronically tetrahedral. The tetrahedral bond angle is
109º 28', but there is a lone pair squeezing the angle together. The expected
bond angle is 107º. Both nitrogen atoms are electronically trigonal planar (three
regions of electrical charge). The trigonal bond angle is 120º', but there is a lone
pair squeezing the angle together. The expected bond angle is a little less than
120º.
-------------------------------------------------------------------------------Q015-09 Draw Lewis (electron dot) structures for CO2 showing all valence
electrons.
Answer
-------------------------------------------------------------------------------Count up the available valence electrons:
•Carbon = 4, 2 x oxygen = 12•Total valence electrons = 16 (8 pairs)Connect
both of the oxygen atoms to the central carbon by a double bond. O=C=O
This uses up four pairs of electrons, leaving four pairs to place on the structure,
so that all atoms have a full octet.
-------------------------------------------------------------------------------Q015-10 Draw the Lewis structure of NCl3. Predict giving a reason the Cl-N-Cl
bond angle in NCl3
Answer
-------------------------------------------------------------------------------Count up the valence electrons.
•Nitrogen from group V = 5 valence electrons•Each chlorine provides 1 electron =
3 electrons•Total = 8 electrons = 4 pairsThe molecule is electronically
tetrahedral, but only 3 pairs of electrons are used in bonding. Thus it has a lone
pair.
The expected bond angle for a tetrahedral electronic arrangement is 109º 28'.
However, the repulsion expected between the lone pair and the bonding pairs
squeezes the Cl-N-Cl bonds together by about 2.5º.
Therefore, the expected bond angle is 107º
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Chapter 0.20 - Shapes of molecules
Syllabus ref: 4.3.7; 14.1.1; 14.2.1,2,3; 14.3.1
4.3.7 Predict the shape and bond angles for species wth four, three and two
negative charge centres on the central atoms using the valence shell electron pair
repulsion theory (VSEPR)
Examples should include CH4, NH3, H2O, NH4+, H3O+, BF3, C2H4, SO2,C2H2,
CO2.
14.1.1 Predict the shape and bond angles for for species with 5- and 6-negative
charge centres using the VSEPR theory. Examples should include: PCl5, SF6,
XeF4 and PF5-.
14.2.1 Describe sigma and pi bonds.
Treatment should include:
sigma bonds resulting from the axial overlap of orbitals
pi bonds resulting from sideways overlap of parallel p orbitals
double bonds formed by one sigma and one pi bond
triple bonds formed by one sigma and two pi bonds.
14.2.2 Explain Hybridisation in terms of the mixing of atomic orbitals to form new
orbitals for bonding. Students should consider sp, sp2 and sp3 hybridization, and
the shapes and orientation of these orbitals.
14.2.3 Identify and explain the relationships between Lewis structures, molecular
shapes and types of hybridization (sp, sp2 and sp3). Students should consider
examples from inorganic as well as organic chemistry.
14.3.1 Describe the delocalization of pi orbital electrons and explain how this can
account for the structures of some species.
Examples should include NO3-, NO2-, CO32-, O3, RCOO- and benzene.
Make sure that you understand what is meant by the term "shape of molecule". It
is the shape adopted by the atoms in a molecule with respect to one another.
Confusion arises occasionally, because the shape adopted by the regions of
electronic charge around an atom has to be known before determining the
location of the atoms. Students confuse this electronic organisation with the
molecular shape. In fact, they may be the same, but often they are not.
In Chapter 0.2
0.21 - Valence Shell Electron Pair Repulsion theory
0.22 - 2, 3 and 4 charge centre molecules and ions
0.23 - 5 and 6 charge centre molecules and ions
0.24 - Hybridisation
0.25 - Delocalisation
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Section 0.21 - Valence shell electron pair repulsion theory
Syllabus ref: 4.2.7
4.2.7 Predict the shape and bond angles for species wth four, three and two
negative charge centres on the central atoms using the valence shell electron pair
repulsion theory (VSEPR)
Examples should include CH4, NH3, H2O, NH4+, H3O+, BF3, C2H4, SO2,C2H2,
CO2.
The valence shell is the outer shell of electrons. These are the electrons that are
involved in bonding.
•The octet rule
•Electron pair repulsion
•Two regions of electron density (charge)
•Three regions of electronic charge
•Four regions of electronic charge
•Electronic shape and molecular shape
•Unequal repulsion - VSEPRT
•Dealing with ions
•Other molecules and ions
•Worked examples
The octet rule
In the section on Lewis structures we discussed the idea that electrons tend to
exist in pairs around the outer shells of atoms. The stable situation appears to be
for the electrons to adopt a situation where four pairs are in the outer (valence)
shell (with the exception of hydrogen, which is 'full' with one pair of electrons).
This arrangement of eight electrons is called the 'octet rule'.
Methane, CH4, molecule showing the four pairs of electrons around the central
carbon atom.
The methane molecule
-------------------------------------------------------------------------------top
Electron pair repulsion
Like charges repel, negative charges repel other negative charges. The electrons
are constrained around the nucleus in pairs. Each pair exerts a repulsion on other
electron pairs. To ascertain the actual shape adopted by molecules we must
consider these repulsions.
The electrons in a charge region are also attracted to the nucleus. Thus any
repulsion will cause the electron pairs to move as far apart as possible, while still
remaining attached to the central nucleus.
It is rather like having two magnetic north poles attached to two strings with the
strings attached to a central pivot. The magnets will push as far away as possible
while staying tied to the central pivot. The final electronic shape adopted depends
on the number of regions of electronic charge around the nucleus. shapes
-------------------------------------------------------------------------------top
Two regions of electron density (charge)
The two regions move as far apart as possible while remaining fixed to the
nucleus. They adopt an arrangement in which they are at 180º to one another.
Here we can see the beryllium dichloride molecule that has only two pairs of
electrons on the central atom. They move as far apart as possible and adopt an
arrangement as shown.
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Three regions of electronic charge
Three regions move as far apart as possible while remaining attached to the
central atom. They adopt a trigonal planar arrangement in which the angle
formed between the electron pairs is 120º.
Here we can see the boron trifluoride molecule. It has three pairs of electrons on
the central atom. They move as far apart as possible and adopt an arrangement
as shown.
The bond angle subtended by the F-B-F atoms is 120º in all cases.
The boron trifluoride molecule
-------------------------------------------------------------------------------top
Four regions of electronic charge
Four regions of charge adopt a tetrahedral arrangement. This is a three
dimensional shape that can be considered formed from the peaks of a regular
pyramid (tetrahedron)
with the central atom in the very centre of the pyramid. The molecule methane
has four regions of electronic charge around the central atom, which adopt a
tetrahedral arrangement.
Methane in two-dimensions Methane showing the tetrahedral 3-D arrangement
-------------------------------------------------------------------------------top
Electronic shape and molecular shape
The shape of a molecule is given by the relative positions of its constituent
atoms. This may, or may not, be the same as the electronic shape of the
molecule. If all of the electrons around the central atom are involved in bonding,
then the molecular shape equals the electronic orientation. However, if there is
one, or more, lone pairs (non-bonding pairs) of electrons, then the molecular
shape will be different from the electronic orientation.
If, for example, the central atom has three regions of electron density (three
charge centres), but is attached to only two atoms then the electronic shape is
trigonal planar, but the molecular shape is angular (bent).
Trigonal planar electronic shape - three charge centres Angular molecular shape 2 out of 3 charge centres used for bonding.
The shape adopted by electron pairs around a central atom depends on the
number of regions of electrical charge. Each region will move so as to minimise
the repulsive forces experienced. The molecular shape is not necessarily the same
as the electronic shape, as only the positions of the atoms themselves are used to
describe the molecular shape.
To determine the molecular shape it is necessary to first determine the shape
adopted by the electrons, and only then can the positions of the atoms be known.
The electronic shape is tetrahedral, but the shape of the water molecule considers
only the H-O-H. It is angular, or bent.
-------------------------------------------------------------------------------top
Unequal repulsion - VSEPR
One of the successes of the Valence Shell Electron Pair Repulsion theory lies in its
ability to predict, or explain, the bond angles of molecules. To do this, it considers
that electron pairs that are shared by two atoms (bonding pairs) experience less
repulsion than lone, or non-bonding pairs, of electrons. The logic of this
assumption comes from the idea that shared pairs have their electron density
relatively displaced away from the central atom by the bonded atom, when
compared to a lone (non-bonding) pair.
This idea gives rise to an order or repulsive force experienced by electron pairs, in
which lone pair - lone pair repulsion is greater than lone pair - bonding pair
repulsion, which in turn is greater than bonding pair - bonding pair repulsion.
These different forces of repulsion distort the perfectly symmetrical shapes that
would be adopted by the electron pairs under ideal conditions. We can appreciate
this by studying examples such as the ammonia molecule, NH3.
Ammonia
The ammonia molecule has a central nitrogen atom with four pairs of electrons.
These electrons tend to adopt a tetrahedral orientation. A perfect tetrahedron has
a bond angle of 109º 28'. However, there is a greater repulsion between the lone
pair (non-bonding pair) of electrons and the bonding pairs, than there is between
the bonding pairs themselves. This greater repulsive force has the effect of
'squeezing' the hydrogen atoms closer together, creating smaller H-N-H bond
angles of 107º.
The ammonia molecule
The ammonia molecule - VSEPR
1. The four electron pairs adopt a tetrahedral arrangement.
2. Only three pairs of electrons are actually used for bonding.
3. The lone pair of electrons, shown in red, repel the bonding pairs more than the
bonding pairs repel each other.
4. The final bond angle is 107º.
The ammonia molecule
-------------------------------------------------------------------------------Water
The water molecule has four pairs of electrons on the central oxygen atom. These
electron pairs, or regions of electron density, adopt a tetrahedral arrangement.
The theoretical bond angle of the electron pairs would be 109º 28'. However, only
two of the electron pairs are actually used in bonding and the other electrons are
lone (non-bonding) pairs. The lone pairs of electrons repel the bonding pairs more
than the bonding pairs repel each other. This has the effect of closing the 109º
28' bond angle down to 104º 30'.
The water molecule
The water molecule - VSEPR
1. The oxygen atom has four pairs of electrons. They are arranged in a
tetrahedral arrangement.
2. Only two of the pairs of electrons are used in bonding
3. The lone pairs repel the bonding pairs more than the bonding pairs repel one
another.
4. The final H-O-H bond angle is 104.5º
The water molecule
-------------------------------------------------------------------------------top
Dealing with ions
Ions are atoms or grops of atoms that have lost or gained electrons to attain a
stable electronic configuration. In the case of ions involving more than one atom
the atoms always have an octet of electrons.
When counting up the total number of electrons available, it is important to
remember that a positive ion has lost an electron and a negative ion has gained
one or more electrons with respect to the constituent atoms.
For example, the ammonium ion has the formula NH4+. It is comprised of one N
atom (group V) and four hydrogen atoms. The total valence electrons from the
atoms = 5 + 4 = 9. However, the ion has a positive charge, so it has lost a
valence electron. It therefore has only 8 valence electrons.
The central nitrogen atom is attached to four hydrogen atoms by for shared pairs
of electrons = 8 electrons. Thus there are no electrons left over.
The ammonium ion - VSEPR - NH4+
The ammonium ion - VSEPR - NH4+
1. The ammonium ion has four pairs of electrons. They are arranged in a
tetrahedral arrangement.
2. All four of the pairs of electrons are used in bonding
3. There are no lone pairs to influence the position of the bonding pairs so the ion
is in the form of a perfect tetrahedron.
4. The H-N-H bond angle is 109.5º
The ammonium ion
Negative ions have electrons added, for example the NH2- ion. In this case the
central nitrogen atoms provide 5 electrons and each hydrogen provides 1
electron. This makes a total of 7 valence electrons. However, the ion carries a
negative chrge that must be added into the valence shell. So the overall total of
valence electrons is 7 + 1 = 8.
The amide ion - VSEPR - NH2-
1. The amide ion has four pairs of electrons. They are arranged in a tetrahedral
arrangement.
2. Only two of the pairs of electrons are used in bonding
3. There are two lone pairs which influence the position of the bonding pairs, so
the angular shape of the H-N-H atoms is squeezed into a smaller angle.
4. The H-N-H bond angle is 104.5º
The amide ion
In all cases, ions have full octets in the valence shells.
-------------------------------------------------------------------------------top
Other molecules and ions
-------------------------------------------------------------------------------top
Worked examples
Q021-01 What is the valence shell electron pair repulsion theory (VSEPRT) used
to predict?
A.The energy levels in an atom
B.The shapes of molecules and ions
C.The electronegativities of elements
D.The type of bonding in compounds
Answer
-------------------------------------------------------------------------------The VSEPR theory is used to predict the shapes of molecules and ions
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Section 0.22 - 2, 3 and 4 charge centre molecules
Syllabus ref: 4.2.7
4.2.7 Predict the shape and bond angles for species with four, three and two
negative charge centres on the central atom using the valence shell electron pair
repulsion theory (VSEPR).
Examples should include CH4, NH3, H2O, NH4+, H3O+, BF3, C2H4, SO2, C2H2
and CO2.
In the previous section we looked at how the electrons organise themselves
around an atom to minimise repulsions. Now, by using the Lewis structures, we
can predict the shapes of molecules and ions.
•Two charge centre systems
•Three charge centre systems
•Four charge centre systems
•Worked examples
•Now test yourself
Two charge centre systems
Two regions of electron density (charge centres) always adopt a linear orientation
to minimise repulsion between the negative charges.
In beryllium chloride the beryllium atom has only two regions of electron density.
It uses both of them to bond the two chlorine atoms.
•Electronic shape = linear
•Molecular shape = linear
•Bond angle = 180º
-------------------------------------------------------------------------------top
Three charge centre systems
When the central atom has three charge centres the electronic shape adopted is
trigonal planar. The actual shape of the molecule depends on the number of
attached atoms.
•Three attached atoms - trigonal planar
•Two attached atoms - angular
The boron trifluoride molecule has the central boron atom with three regions of
electron density. These adopt a trigonal planar orientation.
•All three electron regions are used to bond fluorine atoms.
•The molecular shape is trigonal planar.
•F-B-F bond angle = 120º.
Three regions of electron density but only two attached atoms produces an
angular molecule
The sulfur dioxide molecule has the central sulfur atom with three regions of
electron density. These adopt a trigonal planar orientation.
Only two out of the three electron regions are used to bond oxygen atoms.
The molecular shape is angular.
O-S-O bond angle = 119º. The deviation from the expected bond angle of 120º is
due to slightly different degrees of repulsion between the sulfur-oxygen bonding
electrons and the lone pair on the sulfur atom.
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Four charge centre systems
Molecules with four charge centres are very common, as four pairs of electrons
make up a full octet of electrons. The charge centres adopt a tetrahedral
orientation, but the molecular shape depends on the number of atoms or groups
attached to the central atom.
Number of attached atoms or groups Molecular shape
4 tetrahedral
3 trigonal pyramidal
2 angular (bent)
The bond angles that are finally produced depends on the inter-electron repulsion
between the bonding pairs of electrons and the non-bonding, or lone pairs. This is
best understood using some examples.
The methane molecule - CH4 The ammonia molecule - NH3 The water molecule H2O
The methane molecule - VSEPR
•The carbon atom has four pairs of electrons. They are arranged in a tetrahedral
arrangement.•All four of the pairs of electrons are used in bonding•The molecular
shape is tetrahedral.•All of the H-C-H bond angles are 109.5º
The ammonia molecule - VSEPR
•The four electron pairs adopt a tetrahedral arrangement. •Only three pairs of
electrons are actually used for bonding.•The lone pair of electrons, shown in red,
repel the bonding pairs more than the bonding pairs repel each other. This has
the effect of 'squeezing' the N-H bonds closer together.•The final bond angle is
107º.
The water molecule - VSEPR
•The oxygen atom has four pairs of electrons. They are arranged in a tetrahedral
arrangement.•Only two of the pairs of electrons are used in bonding•The lone
pairs repel the bonding pairs more than the bonding pairs repel each other.•The
final H-O-H bond angle is 104.5º
-------------------------------------------------------------------------------top
Worked examples
Q022-01 The geometry and bond angle of the sulfite ion (SO32-) are best
described as:
A.Pyramidal, 107º
B.Tetrahedral, 109º
C.Bent, 104º
D.Trigonal planar, 120º
Answer
-------------------------------------------------------------------------------sulfur is in group VI = 6 valence electrons. There are three oxygens (group VI)
each with 6 valence electrons. The ion carries a 2- charge. Therefore total
number of valence electrons = 26.
The best structure gives all the bonded oxygen atoms a full octet and leaves the
sulfur with a lone pair. This makes 26 electrons. The sulfur then has four regions
of electron density (electronically tetrahedral), but only three regions are used for
bonding. It is therefore a trigonal pyramid.
-------------------------------------------------------------------------------Q022-02 Which shape correctly describes the stated molecule?
A.PCl3; trigonal pyramidal
B.P2H4; planar
C.N2O; bent
D.NO2; linear
Answer
-------------------------------------------------------------------------------The first option PCl3; trigonal pyramidal is correct. Phosphorus is from group 5
and each chlorine shares one pair of electrons. Therefore the central phosphorus
atom has 8 electrons = 4 pairs. It is tetrahedral electronically, but only three
pairs are used in bonding.
-------------------------------------------------------------------------------Q022-03 Which molecule has the smallest bond angle?
A.CO2
B.NH3
C.CH4
D.C2H4
Answer
-------------------------------------------------------------------------------Carbon dioxide is linear (bond angle 180º), Ammonia is trigonal pyrimidal (bond
angle 107º), methane is tetrahedral (bond angle 109º 28') and ethene has
carbon atoms with a HCH bond angle of 120º. The smallest is therefore ammonia.
-------------------------------------------------------------------------------Q022-04 The geometries of the molecules BF3 and NF3 are trigonal planar and
trigonal pyramidal respectively. Which statement best accounts for the
difference?
A.N is more electronegative than B.
B.BF3 is ionic, which NF3 is covalent.
C.B is a metalloid, whereas N is a non-metal.
D.N has a non-bonding pair of valence electrons, while B does not.
Answer
-------------------------------------------------------------------------------Consideration of the Lewis structures shows us that NF3 has a lone pair of
electrons whereas BF3 does not.
nitrogen trifluoride boron trifluoride
-------------------------------------------------------------------------------Q022-05 In which compound would the molecules be expected to be linear?
A.O3
B.SO2
C.H2O
D.CO2
Answer
-------------------------------------------------------------------------------Carbon dioxide is the linear molecule:
-------------------------------------------------------------------------------Q022-06 Which one of the following is not the expected geometry of the indicated
molecule or ion?
A.CCl4 - regular tetrahedron
B.BeCl2 - linear
C.SO3 - triangular pyramid
D.CH2O - trigonal planar
Answer
-------------------------------------------------------------------------------CCl4 - regular tetrahedron - correct; BeCl2 - linear - correct; SO3 - triangular
pyramid -this is incorrect. SO3 is a trigonal planar molecule.
-------------------------------------------------------------------------------Q022-07 According to VSEPR theory, repulsion between electron pairs in a
valence shell decreases in the order:
A.lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
B.bond pair - bond pair > lone pair - bond pair > lone pair - lone pair
C.lone pair - lone pair > bond pair - bond pair >lone pair - bond pair
D.bond pair - bond pair > lone pair - lone pair > lone pair - bond pair
Answer
-------------------------------------------------------------------------------The greater density of electrons has the greatest repulsive effect. The correct
order is lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
-------------------------------------------------------------------------------Q022-08 The arrangement of the atoms in the ammonia molecule at its lowest
energy is described as
A.trigonal pyramid
B.trigonal planar
C.trigonal bipyramid
D.square planar
Answer
-------------------------------------------------------------------------------Nitrogen has 5 electrons in the valence shell. Each hydrogen provides one
elecrtron = total 8 electrons = 4 pairs (electronically tetrahedral) . Only 3 pairs
are used in bonding to three hydrogens leaving one lone pair, therefore molecular
shape trigonal pyramidal.
-------------------------------------------------------------------------------Q022-09 Which of the following describes the shape of the ethene molecule,
C2H4?
A.linear.
B.planar.
C.pyramidal.
D.shaped like two tetahedra joined at the points.
Answer
-------------------------------------------------------------------------------Each of the carbon atoms in ethene is a trigonal plane and both of these planes
themselves lie in the same plane. Therefore the whole molecule is planar.
-------------------------------------------------------------------------------Q022-10 Which species has a trigonal planar shape?
A.CO32B.SO32C.NF3
D.PCl3
Answer
-------------------------------------------------------------------------------The shapes of ions are obtained by examining the Lewis structure of the ion, not
forgetting to add in the number of electrons from the ionic charge. Carbonate ion:
•Carbon has 4 electrons in the valence shell. •Each oxygen has 6 electrons in the
valence shell. •The ionic charge is 2- = 2 electrons to be added. •Total = 24
electrons.Carbon can double bond to one of the oxygen atoms to satisfy the
valence shell requirements of this oxygen.
Carbon then single bonds to the other two oxygen atoms, fulfilling the
requirements of the carbon atom's valence shell. This leaves both of two oxygens
lacking one electron each, which can then be added in from the ionic charge.
The carbon has three regions of electron density and therefore adopts a trigonal
planar geometry.
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Section 0.23 - 5 and 6 charge centre systems
Syllabus ref: 14.1.1
14.1.1 Predict the shape and bond angles for for species with 5- and 6-negative
charge centres using the VSEPR theory. Examples should include: PCl5, SF6,
XeF4 and PF5-.
Some molecules cannot be easiliy described using the octet rule, as they have
more than four electron pairs around the central atom. This section explores the
phenomenon of octet expansion.
•Octet expansion
•Electronic shapes of five and six charge centres
•Five charge centre systems
•Six charge centre systems
•Summary
•Worked examples
•Now test yourself
Octet expansion
The elements in period 3 of the periodic table can use available 'd' orbitals to
increase the number of electrons that can be accomodated in the outer valence
shell. This only occurs when covalent bonding takes place involving relatively
electronegative elements, such as oxygen, fluorine and chlorine.
This increase in the valence shell electrons is known as 'octet expansion' and it
gives rise to molecules that have more than four charge centres on the central
atom.
Octet expansion can also occur in ions.
Molecules with octet expansion Ions with octet expansion
PCl5, SF6, XeF4, XeF6 PCl6-, SF5+,
Although the actual theory is not needed at IB level, it is useful to understand
how orbital systems can exist with more than four charge centres. The central
atom 'borrows' orbitals from the 3d level (see) and hybridises them along with
the 3s and 3p orbitals (see section 0.24 - hybridisation), to produce the number
of orbitals required to share electrons with covalently bonded atoms. Thus, sp3d2
(six orbitals) can be used to make an octahedral system.
-------------------------------------------------------------------------------top
Electronic shapes of five and six charge centres
There is no perfect shape that allows five regions of electron density to be equally
separated from one another. The shape adopted is trigonal bipyramidal. This
involves a trigonal plane with two axial regions at 90º (perpendicular) to the
plane.
Trigonal bipyramidal shape The electron regions repel as far as possible
Six regions of electronic charge can be accomodated into a perfect octahedral
shape.
The name 'octahedral' seems strange at first sight, as the 'octa' suggests that
there are eight regions. The name 'octahedral' actually means 'eight faces, or
sides'. Each side is produced by joining the ends of three axes.
Octahedral shape The electron regions repel as far as possible
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Five charge centre systems
Five charge centres must involve a minimum of 10 electrons. Group V elements
have 5 outer (valence) electrons. When bonded to five atoms this makes a total
of 10 electrons = five shared pair charge centres.
phosphorus(V) fluoride - PF5
Phosphorus (group V) has 5 valence electrons
Each fluorine provides one electron, there are 5 fluorines therefore 5 electrons.
Total electrons = 10 = 5 pairs
All pairs used for bonding therefore molecular shape = electronic orientation.
Trigonal bipyramidal
Phosphorus(V) fluoride
For molecules, or ions, with five charge centres, but only four attached atoms,
there are two possible positions for the lone pair of electrons. It could either be in
the axial (top or bottom) position of the trigonal pyramid, or it could be on one of
the equatorial positions. The disadvantage of an axial position is that this leaves
the lone pair at 90º to three bonding pairs, whereas if an equatorial position is
adopted, it leaves the lone pair at 90º to only two other bonding pairs. This
second alternative is the lowest energy in terms of repulsion and therefore the
one adopted.
sulfur(IV) fluoride - SF4
sulfur (group VI) has 6 valence electrons
Each fluorine provides one electron, there are 4 fluorines therefore 4 electrons.
Total electrons = 10 = 5 pairs, therefore electronically the pairs adopt a trigonal
bipyramidal orientation.
However, only four pairs are used for bonding, therefore molecular shape:
sawhorse
Sulphur(IV) fluoride
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Six charge centre systems
Six charge centres adopt an octahedral orientation and when all six electron pairs
are used for bonding it gives rise to an octahedral molecule, for example, SF6.
sulfur(VI) fluoride - SF6
sulfur (group VI) has 6 valence electrons
Each fluorine provides one electron, there are 6 fluorines, therefore 6 electrons.
Total electrons = 12 = 6 pairs, therefore electronically the pairs adopt a trigonal
bipyramidal orientation.
All six pairs are used for bonding, therefore molecular shape:
octahedral
Sulphur(VI) fluoride
When one of the electrons pairs is not used for bonding (lone or non-bonding
pair) the remaining shape is a five coordinated, square pyramidal structure. One
of the attached atoms is the top of the pyramid and the other four make up the
base. An example is Iodine (V) fluoride.
Iodine(V) fluoride - IF5
Iodine (group VII) has 7 valence electrons
Each fluorine atom provides 1 electron, there are 5 fluorines, therefore 5 more
electrons.
Total electrons = 12 = 6 pairs, therefore electronically, the pairs adopt a
octahedral arrangement. One of the electron pairs is not used for bonding.
umbrella shape
When five bonding pairs are involved, there is no specific advantage, or
disadvantage, for the lone pair as regards the position it adopts as the octahedral
shape is totally symmetrical, however when only four pairs of electrons are used
for bonding the two remaining lone pairs arrange themselves at 180º to one
another. This leaves the remaining bonding pairs in a square planar shape, for
example XeF4.
Xenon tetrafluoride - XeF4
Xenon (group VIII) has 8 valence electrons
Each fluorine atom provides 1 electron, there are 4 fluorines, therefore 4 more
electrons.
Total electrons = 12 = 6 pairs, therefore electronically, the pairs adopt a
octahedral arrangement. Two of the electron pairs are not used for bonding.
square planar
Xenon tetrafluoride
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Summary - determination of molecular shape
1. Count up the total valence electrons on the central atom. This is done by
consideration of the group number of the central atom and the number of
attached atoms.
2. The total number of electron pairs gives the electronic arrangement adopted:
•two: linear
•three: trigonal planar
•four tetrahedral
•five: trigonal bipyramidal
•six: octahedral
3. Each of the atoms bonds to the central atom by means of a bonding pair of
electrons. If there are no electron pairs remaining, then the molecular shape is
the same as the electronic arrangement. If there are any lone pairs, then only the
bonding pairs are considered in the molecular shape. The lone pairs may distort
the final shape in some cases, due to greater repulsions between lone pairs and
bonding pairs than between bonding pairs and bonding pairs.
Total electron pairs number of lone pairs electronic arrangement molecular shape
2 0 linear linear
3 0 trigonal planar trigonal planar
3 1 trigonal planar angular
4 0 tetrahedral tetrahedral
4 1 tetrahedral pyramidal
4 2 tetrahedral angular
5 0 trigonal bipyramidal trigonal bipyramidal
5 1 trigonal bipyramidal saw-horse
5 2 trigonal bipyramidal distorted T-shaped
5 3 trigonal bipyramidal linear
6 0 octahedral octahedral
6 1 octahedral umbrella
6 2 octahedral square planar
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Worked examples
Q023-01 Draw the Lewis structure, state the shape and predict the bond angles
for the ICl4- species.
Answer
-------------------------------------------------------------------------------•Iodine (group VII) is the central atom with 7 valence electrons•Each chlorine
atom provides one electron = 4 electrons•1 electron for the negative charge
•Total electrons = 12 electrons = 6 pairsThe electronic structure is octahedral but
only four of the sites are occupied. To minimise repulsion the axial sites take the
lone pairs.
The final ionic shape is square planar, with bond angles Cl-Xe-Cl of 90º.
-------------------------------------------------------------------------------Q023-02 Draw and name the shape of the PCl6- ion:
Answer
-------------------------------------------------------------------------------•Phosphorus (group V) is the central atom with 5 valence electrons•Each chlorine
atom provides one electron = 6 electrons•1 electron for the negative charge
(shown in green)•Total electrons = 12 electrons = 6 pairsThe electronic structure
is octahedral with all six sites occupied.
The final ionic shape is octahedral.
-------------------------------------------------------------------------------Q023-03 What is the distribution of electron pairs and the arrangement of atoms
in the triiodide ion, I3-?
Electron pairs Atoms arrangement
A. Tetrahedral bent
B. Square planar linear
C. Trigonal bipyramid linear
D. Trigonal bipyramid bent
Answer
-------------------------------------------------------------------------------•Iodine (group VII) is the central atom with 7 valence electrons•Each other iodine
atom provides one electron = 2 electrons•1 electron for the negative charge
(shown in green)•Total electrons = 10 electrons = 5 pairsThe electronic structure
is trigonal bipyramidal, with only two sites occupied. The occupied sites are both
axial to minimise repulsions between the three pairs of equatorial electron pairs
(the ion is shown on its side).
-------------------------------------------------------------------------------Q023-04 Which of the following contains a bond angle of 90º?
I PCl4+
II PCl5
III PCl6A.I and II only
B.I and III only
C.II and III only
D.I, II and III
Answer
-------------------------------------------------------------------------------PCl4+ has (5 + 4 - 1 = 8 valence electrons) all are used for bonding with no lone
pairs left over. Molecular shape = tetrahedral, bond angle 109º 28'.
PCl5 has (5 + 5 = 10 valence electrons) all are used for bonding with no lone
pairs left over. Molecular shape = trigonal bipyramidal, bond angles 90º and
120º.
PCl6- has (5 + 6 + 1 = 12 valence electrons) all are used for bonding with no
lone pairs left over. Molecular shape = octahedral, bond angles 90º.
Therefore correct response = II and III only
-------------------------------------------------------------------------------Q023-05 Draw a diagram that represents the correct geometry of the BrF5
molecule and state its shape.
Answer
-------------------------------------------------------------------------------•Bromine (group VII) is the central atom with 7 valence electrons•Each fluorine
atom provides one electron = 5 electrons•Total electrons = 12 electrons = 6
pairsThe electronic structure is octahedral, with only five sites occupied and one
lone pair.
This distorts the remaining bonding pairs into an umbrella shape.
-------------------------------------------------------------------------------Q023-06 What is the smallest bond angle found in the PF5 molecule?
A.90º
B.109.5º
C.120º
D.180º
Answer
-------------------------------------------------------------------------------•Phosphorus (group V) is the central atom with 5 valence electrons•Each fluorine
atom provides one electron = 5 electrons•Total electrons = 10 electrons = 5
pairsAll five sites occupied thereforethe electronic shape is the same as the
molecular shape = trigonal bipyramidal, with bond angles of 120º and 90º.
-------------------------------------------------------------------------------Q023-07 Which molecule or ion does not have a tetrahedral shape?
A.XeF4
B.SiCl4
C.BF4D.NH4+
Answer
-------------------------------------------------------------------------------•Xenon (group VIII) is the central atom with 8 valence electrons•Each fluorine
atom provides one electron = 4 electrons•Total electrons = 12 electrons = 6
pairsOnly four out of six sites are used for bonding, therefore the electronic shape
is octahedral and the lone pairs occupy the axial positions to be as far apart as
possible. The molecular shape is square planar, with bond angles of 90º.
All of the other molecules and ions are tetrahedral, therefore the correct response
is XeF4
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Structure and bonding: 0.24 - Hybridisation
Syllabus ref: -14.2.1; 14.2.3; 14.2.3
14.2.1 Describe sigma and pi bonds. Treatment should include:
sigma bonds resulting from the axial overlap of orbitals
pi bonds resulting from sideways overlap of parallel p orbitals
double bonds formed by one sigma and one pi bond
triple bonds formed by one sigma and two pi bonds.
14.2.2 Explain Hybridisation in terms of the mixing of atomic orbitals to form new
orbitals for bonding. Students should consider sp, sp2 and sp3 hybridisation, and
the shapes and orientation of these orbitals.
14.2.3 Identify and explain the relationships between Lewis structures, molecular
shapes and types of hybridisation (sp, sp2 and sp3). Students should consider
examples from inorganic as well as organic chemistry.
Hybridisation is a word that strikes fear into the very heart of students. However,
in reality the concept is relatively straightforward. In this section we explore the
concept of hybridisation and learn to recognise and use it.
•Background
•Linear combination of atomic orbitals
•Sigma bonds
•Pi bonds
•Hybridisation
•Hybridisation in double and triple bonds
•Worked examples
Background
Science goes forward by means of a process of observation, experimentation and
theorisation. The theories and explanations for observations are considered valid
providing there is no evidence to the contrary.
Through the work of Schroedinger and others in the middle of the 20th century,
scientists began to put shapes to the orbitals that hold the electrons around
atoms. However the shape and orientation of the atomic orbitals did not explain
the shapes of molecules. It was clear that some theory was needed to explain
how atomic orbitals could somehow give rise to the shapes of the molecules
formed from them.
There are various approaches to the problem, one of which is the concept of
hybridisation.
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Linear combination of atomic orbitals
Covalent bonds involve shared pairs of electrons. These shared pairs using
consist of one electron from one atom and the second electron being provided by
the other atom. Clearly these two electrons are going to have to be in the same
region of space. As they are held in atomic orbitals these orbitals themselves
must overlap.
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Sigma bonds - direct orbital overlap
When the process of orbital overlapping involves a direct overlap along an axis
the bond is referred to as a sigma bond. Sigma bonds are always single bonds.
One electron is usually provided by each of the overlapping orbitals, although in
certain cases both of the electrons may be donated by the same atom. The
second case is called dative covalent bonding.
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Pi bonds - lateral orbital overlap
Pi bonds can only form AFTER sigma bonds have already formed. A pi-bond
cannot exist on its own. Sigma bond formatin brings suitable orbitals into
sideways (lateral) contact on the adjacent atoms. This causes orbital overlap both
above and below the axis of the sigma bond. This lateral overlap of two parallel
orbitals contributes to the force of attraction between the two atoms - it is called
a pi-bond.
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Hybridisation
This is the idea that atoms involved in bonding rearrange their orbitals shapes
and energies in order to produce orbitals that can overlap successfully with
suitable orbitals on adjacent atoms and be used in bonding.
Carbon, for example, forms four bonds with hydrogen to make the molecule CH4.
Studies show that the methane molecule is symmetrical and the hydrogen atoms
are arranged in a tetrahedral fashion around the central carbon atom.
However, the atomic orbitals in the outer shell of carbon have two electrons in a
2s orbital and 2 electrons in two 2p orbitals. There are three problems here:
•The 2s orbital is spherical and can't overlap with adjacent atoms.
•The 2s orbital is full. It can't share an electron from another overlapping orbital.
•There are only two singly occupied orbitals available for overlap and shaing
electrons suggesting that carbon should only form two bonds and that these
bonds would be at 90º to one another.
Carbon atoms overcome these problems by reforming the orbitals to give four
degenerate orbitals, each of which is singly occupied. This is called hybridisation.
It must be stressed that hybridisation is simply an model to explain the difference
between the shapes and orientation of the orbitals found in atoms and molecules.
It gives us a process that explains how one situation can change into another.
sp3 hybridisation
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Hybridisation in double and triple bonds
Hybridisation can also explain the formation and shape of molecules that contain
double and triple bonds. Using carbon as the example, in the case of a double
bond the central atom needs to only bond to three other atoms. To do this it
needs three degenerate orbitals which it obtains by hybridisation of the three
valence orbitals that have the lowest energy, i.e. the 2s and two of the 2p
orbitals.
Atoms that are triple bonded can only bond to two other atoms. This means that
two degenerate orbitals are needed. These are provided by hybridisation of the
two orbitals from the valence shell that have the lowest energy, i.e. the 2s and
onle of the 2p orbitals. This hybridisation scheme produces two 'sp' orbitals at
180º to one another (linear) that can be used for direct orbital overlap (sigma
bonding) and leaves two unchanged 'p' orbitals that can be used for lateral
overlap with suitable orbitals on adjacent atoms (pi bonding).
It must be stressed that hybridisation occurs in all molecules, not just carboncontaining molecules. It is a simple matter to spot sp3, sp2 and sp hybridisation
by the number and type of bonds on an atom. Carbon, for example, forms four
bonds. If all four bonds are single then the hybridisation is sp3. If there is one
double bond then it is using sp2 hybridisation and if there is a triple bond then
the hybridisation is sp.
Nitrogen forms only three bonds with other atoms and in nitrogen containing
compounds one double bond is indicative of sp2 hybridisation, just like in carbon.
In the ammonia molecule the four electron pairs are arranged in a tetrahedral
orientation using sp3 hybridisation.
In the hydrogen cyanide molecule the nitrogen is attached to the carbon atom by
a triple bond. Both atoms are sp hybridised.
In summary:
•If an atom has only single bonds then it is using sp3 hybridisation.
•if an atom is attached with one double bond then it is sp2 hybridised
•If an atom is attached by a triple bond then it is sp hybridised.
Summary
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Worked examples
Q024-01 Appropriate hybridisation schemes for the C atoms in molecular
CH3CO2H are:
A.sp3 and sp
B.sp3 and sp2
C.sp2 and sp
D.sp3 and sp3
Answer
-------------------------------------------------------------------------------•The first (CH3) carbon is attached to four other atoms and is therefore sp3
hybridised•The second carbon (COOH) is attached to three other atoms, one of
which it has a double bond with, and is therefore sp2 hybridised.The correct
response is sp3 and sp2
-------------------------------------------------------------------------------Q024-02 sp3 hybridisation would not be appropriate for the central atom in:
A.SiF4
B.[PCl4]+
C.XeF4
D.[Me4N]+
Answer
-------------------------------------------------------------------------------The molecule XeF4 has six pairs of electrons around the central atom and cannot
involve sp3 hybridisation.
--------------------------------------------------------------------------------
Q024-03 Suitable hybridisation schemes for Be in BeCl2(g) and BeCl2(s) are,
respectively:
A.sp and sp
B.sp3 and sp3
C.sp3 and sp
D.sp and sp3
Answer
-------------------------------------------------------------------------------BeCl2(g) is simple molecular and has four electrons around the central atom in
two degenerate orbitals. The lowest energy orbitals in the second energy level are
an 's' and a 'p' orbital. Therefore BeCl2(g) used sp hybridisation.
In the solid form the BeCl2 has electrons donated from chlorines of adjacent
molecules into the empty orbitals on the beryllium atom making a giant lattice
structure. Each beryllium atom is now associated with four other atoms, it
therefore uses sp3 hybridisation. The correct answer is sp and sp3
-------------------------------------------------------------------------------Q024-04 Which of the following molecules contain a central atom which is sp2
hybridised?
A.H2SO4
B.H2CO3
C.ICl2
D.H3CCH3
Answer
-------------------------------------------------------------------------------sp2 hybridisation requires an atom to have only three regions of electron charge.
The only structure that fulfills this requirement is the carbon atom in H2CO3.
-------------------------------------------------------------------------------Q024-05 Which molecule utilizes sp3 hybridisation according to Valence Bond
Theory?
A.NH3
B.BF3
C.BeF2
D.XeF4
Answer
-------------------------------------------------------------------------------For sp3 hybridisation there must be four regions of electric charge around the
atom. Only NH3 has four regions, three bonding pairs and one lone pair.
-------------------------------------------------------------------------------Q024-06 The structure for dimethyformamide is shown below: How many C and
N atoms are sp3, sp2, and sp hybridised?
A.4 (sp3), 0 (sp2), 0 (sp)
B.3 (sp3), 0 (sp2), 1 (sp)
C.3 (sp3), 1 (sp2), 0 (sp)
D.2 (sp3), 2 (sp2), 0 (sp)
Answer
-------------------------------------------------------------------------------The carbonyl carbon is sp2 hybridised and the lone pair on the nitrogen atom of
the amide laterally overlaps with the carbonyl 'pi' system, becoming sp2 as well.
The other carbon are sp3 hybridised.
Correct response D (sp3), 2 (sp2), 0 (sp)
-------------------------------------------------------------------------------Q024-07 What is the hybridisation of nitrogen atoms I,II, III and IV in the
following molecules?
H2NINIIH2 HNIIINIVH
I II III IV
A. sp2 sp2 sp3 sp3
B. sp3 sp3 sp2 sp2
C. sp2 sp2 sp sp
D. sp3 sp3 sp sp
Answer
-------------------------------------------------------------------------------The first nitrogen has four regions of electrical charge (three bonds and a lone
pair), the second nitrogen has the same. Nitrogen number III has only three
regions of electrical charge, the same as nitrogen IV.
The correct response is sp3, sp3, sp2, sp2.
-------------------------------------------------------------------------------Q024-08 Identify the types of hybridisation show by the carbon atoms in the
molecule CH2CH2CH2COOH
I sp
II sp2
III sp3
A.I and II only
B.I and III only
C.II and III only
D.I, II and III
Answer
-------------------------------------------------------------------------------The first two carbon atoms are bonded to only three other atoms and have no
lone pairs. They employ sp2 hybridisation. The third carbon atom is sp3
hybridised and the last carbon atom in the COOH group is sp2 hybridised.
Correct response II and III only
-------------------------------------------------------------------------------Q024-09 Which types of hybridisation show by the carbon atoms in the molecule
CH2=CH-CH3
I sp
II sp2
III sp3
A.I and II only
B.I and III only
C.II and III only
D.I, II and III
Answer
-------------------------------------------------------------------------------The first two carbon atoms are bonded to only three other atoms and have no
lone pairs. They employ sp2 hybridisation. The remaining carbon atom is sp3
hybridised.
Correct response II and III only
-------------------------------------------------------------------------------Q024-10 NO3- is trigonal planar and NH3 is trigonal pyramidal. What is the
approximate hybridisation of N in each of these species?
N in NO3- N in NH3
A. sp2 sp3
B. sp2 sp2
C. sp3 sp2
D. sp3 sp3
Answer
-------------------------------------------------------------------------------The NO3- ion is trigonal planar meaning that it must use sp2 hybridisation,
whereas the trigonal pyramidal NH3 must involve sp3 hybridisation.
Correct response sp2 and sp3
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