IB TOPIC 14: Bonding
14. 1 Shapes of molecules.
14.1.1 Predict the shape and bond angles for species with five and six negative charge centres using the VSEPR theory.
You should be able to predict the electron distribution, molecular shape and bond angles of molecules or
molecular ions with 5 or 6 negative centres such as PCl5, SF6, XeF4, and PF6.
14. 2 Hybridisation
14.2.1 Describe σ and π bonds.
Covalent bonds
The sharing of electrons occurs as a result of the interaction or overlapping of a half-filled atomic orbital
(= an orbital containing an unpaired electron) of one atom with a half-filled atomic orbital of another atom.
Both single/unpaired electrons now become paired as they share each others overlapping atomic orbitals
which now form a new single molecular orbital.
A molecular orbital is a region of space around both nuclei with a great probability of finding the shared
electron pair; each molecular orbital can hold two electrons.
Most of the time, the shared electron pair is found in the region between both nuclei as this is where the
atomic orbitals have overlapped; this is also the region where the attraction between the pair of electrons
and the nucleus is the greatest. So in a molecular orbital the electron density is the highest in the
internuclear region, which is the area, surrounding the axis between the nuclei of both atoms.
Covalent bonds and the energy state of the particles involved
Electrons in bond orbitals are attracted by both nuclei. As a result, they experience a larger attraction
than when they were unpaired. Therefore, after the overlapping, these bonding electrons are at a lower
energy (more negative) level then non-bonding valence electrons in separate atoms as the non-bonding
valence electrons are only attracted by one nucleus (see figure below).
After the bond formation, a more stable substance is formed as the molecule is at a lower energy level
than what the separate atoms were at before the bonding. This is because its shared electrons are now
attracted more strongly. This also explains why bond making is an exothermic process and bond
breaking is endothermic because when we want to break a bond, we have to supply the energy the
electrons lost in forming the bond. The greater the difference in energy, the stronger the bond.
The diagram below shows the energy changes that occur in the course of bond formation between 2
hydrogen atoms. The hydrogen molecule is at a lower energy state, and therefore more stable, than the
original atoms.
energy


H(g) + H(g)
electron
in H atomic orbital
436
kJ/mol
electron
in H atomic orbital

H2(g)
Molecular orbital
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(In a way, in ionic bonding the bond orbital has strongly moved towards one atom)
Multiple bonds
Normally, the number of covalent bonds an atom makes = number of its unpaired electrons (provided all
covalent bonds are normal covalent and not dative bonds!). For instance hydrogen has one unpaired
electron so it makes one covalent bond; oxygen has two so….
Sometimes the number of other atoms to bond with are limited and more than one bond (a multiple bond)
is formed with the same atom. The formation of a multiple bond involves a different type of covalent
bond.
The first type of covalent bond formed between two atoms is always a sigma, , bond. Any subsequent
bonds formed are always pi bonds, . This is the case because, as the area along the axis between both
nuclei is already occupied by the  bond/electrons, any further overlap of atomic orbitals has to take
place above and below this axis (=sideways overlap).
So  bonds only form when there is already a  bond. They are only formed by p orbitals!!!
What is the difference between sigma, , and pi, , bonds?
In sigma, , bonds the electron cloud (space of high probability of finding bonding electrons) or
molecular orbital, formed as a result of the end-on overlap or ‘head-on overlap’ of two
atomic orbitals, lies along the axis between the two nuclei. The electron cloud or molecular orbital
encompasses both nuclei and both electrons can move around both nuclei although they are most likely
to be found on the axis between both nucelii.
A  bond can be formed between two s orbitals, 1 s and 1 p orbital, 2 p orbitals and so on.
The diagram on the left shows parallel p orbitals of two separate carbon atoms overlapping sideways in a
ethene molecule. Notice that there is already a sigma bond between both carbon atoms. The diagram on
the right shows the outcome which is a pi, , bond.
(from http://www.chemguide.co.uk/analysis/uvvisible/bonding.html on 15/10/09)
 bonds are symmetrical ‘sausage’ shaped electron clouds lying above and below the axis between the
two nuclei and are formed by the sideways overlap of 2 parallel p orbitals. As p orbitals have two
lobes, the  bond has two regions of overlap, one above and below the axis between both nuclei.
Summary:
a single covalent bond
double bond
triple bond
The diagram below shows the triple bond in ethyne.
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Type of bond
1  bond
1  bond + 1  bond
1  bond + 2  bonds
(from http://www.science.uwaterloo.ca/~cchieh/cact/c120/hybridcarbon.html on 15/10/09)
Comparison of bond strength of  and pi bonds.
As the electron density of the  bond is above and below the axis between the two nuclei of the bonding
atoms, the  electrons are attracted less strongly than the two electrons in a  bond; ‘on average’ they
are always further away than the ‘ electrons’ from both nuclei.
In addition. the area of overlap (sideways overlapping only) is also less although this in not what the
diagram of a pi bond shows; the greater the overlap the stronger the bond.
Both these factors explain why a  bond is weaker than a  bond i.e. less energy is needed to break it
and less energy is released when it is made.
Because they are attracted less strongly, the electrons in a  bond also have more freedom and kinetic
energy.
In a  bond the centre of charge density is above and below the internuclear axis. For any  bond to
form and remain both atoms must lie in the same plane!!! And will need to stay in the same plane after it
has been formed. If these atoms are moved and there is no overlap anymore, the bond is broken.
A triple bond is made up of a  bond and two  bonds, which are at right angles to each other; as there
are 3 areas of overlap this triple bond is the stronger compared to a double or single covalent bond (but
remember that the strongest component is the  bond).
Hybridisation
14.2.2 Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding.
14.2.3 Identify and explain the relationships between Lewis structures, molecular shapes and types of
hybridization (sp, sp2 and sp3).
Use: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf
The theory of hybridisation is a theory which was developed to explain some of the shortcomings of the
valence bond theory.
The valence bond theory states that the chemical behaviour of an element is determined by its valence
electrons, e.g. covalent bonds (sigma or pi) are formed when half-filled atomic orbitals (containing
unpaired electrons) from 2 different atoms overlap to form electron pairs so that both atoms achieve the
noble gas electronic configuration. Therefore the number of unpaired electrons in an atom determines the
number of covalent bonds that the atom makes in a molecule or molecular ion.
But the valence bond theory has not always been able to explain all of the experimental data (on bonding
and shapes of molecules) obtained from the study of the physical properties of molecules or molecular
ions.
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Some examples of some of these shortcomings of the valence bond theory:
 Why does boron make 3 covalent bonds with 3 different atoms although it has only 1 unpaired
valence electron?
 Why does beryllium make two covalent bonds although it does not have any unpaired electrons?
Beryllium although being a metal forms a covalent bond with halogens. This is because of its small
size its 2s electrons are attracted too strongly by the nucleus so that a complete transfer can not
occur.
 Why does carbon make 4 bonds although it has only 2 unpaired electrons? Why do carbon
compounds like methane have a perfectly tetrahedral arrangement around each carbon atom i.e. all
angles are 109? all 4 bonds have the same length and all electron pairs are distributed evenly? This
can only be as a result of all bonds having equal repulsion which can only occur if they are all on the
same energy level.

When applying the VSEPR theory to determine the shape of molecules, we give the same amount of
repulsion to all bonding pairs although they cannot all be on the same energy level, equally we give
the same amount of repulsion to all non-bonding pairs although again they cannot all be on the same
energy levels to justify the same force of repulsion. In water one of the lone pair is on the 2s orbital
which is at a lower energy level than the 2p orbitals but when explaining the shape using the VSEPR
we have considered both lone pairs (one on 2s and one on 2p) to be on a par in terms of energy
(=equivalent) i.e. they cause the same repulsion which should not be the case if one is on a lower
energy level.
New ideas (by Linus Pauling) were added and these include the idea of hybridisation.
Hybridisation is the mixing of a number of non-equivalent (=not the same) but similar atomic
orbitals (e.g. 2s and 2p orbitals) to form the same number of new equivalent atomic orbitals (hybrid
orbtials) for bonding.
These new orbitals are hybrids (= a mixture); they are degenerate/equivalent (= of the same energy level)
and have some s and p orbital character i.e. their energy level is in between the two. Each hybridized
orbital has a different orientation i.e. points in a different direction.
Hybridised orbitals have their own shapes (different from the normal s and p orbitals) i.e. they have 1
large lobe which extends between the two interacting atoms; these large lobes allow them to overlap
better - area of overlap is larger - and form stronger bonds than unhybridised orbitals.
So more energy is released when they are formed as opposed to unhybridised orbitals; that energy is
used to initially promote the electrons during the process.
Hybridisation involves 2 stages:
 the unpairing of an electron and the promotion of an electron e.g. an s electron to a p orbital;
 the actual hybridisation (or mixing) in which all orbitals become degenerate/the same.
Example: hybridisation in boron
promotion
hybridisation
non-
2s
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2p
2s
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2p
sp2
hybridised
Example: hybridisation in carbon
promotion
2s
2p
hybridisation
2s
sp3
2p
Example : hybridisation in beryllium
promotion
2s
2p
hybridisation
2s
2p
sp
nonhybridised
The energy needed for the unpairing of some electrons to a higher energy level is provided when a bond
is made (bond making releases energy) as a result of the hybridisation. The energy released when
making a new bond makes up for the energy needed to promote and hybridise electrons.
Multiple bonds
All sigma covalent bonds in polyatomic molecules are formed as a result of the overlap of two hybridised
orbitals or one hybridised and one unhybridised (e.g. hydrogen).
Pi, , bonds are always formed from unhybridised p orbitals, usually the pz, so they have no impact on
the shape.
Determining the type of hybridization that has occurred in a molecule or molecular ion
Types of hybridization: sp, sp2 and sp3
How can we now identify the type of hybridisation (or which orbitals have hybridised) in a molecule or
molecular ion?
1. draw the Lewis structure
2. determine electron pair geometry or valence electron pair distribution around central atom by counting
the number of electron pairs (lone and bonding pairs) but consider a multiple bond as one;
3. the number you obtained in step 2 = number of hybrid orbitals
4. if 2 hybrid orbitals = sp hybridisation (180)
5. if 3 hybrid orbitals = sp2 hybridisation(120 bond angles between hybrid orbitals)
6. if 4 hybrid orbitals = sp3 hybridisation(109/ bond angles between hybrid orbitals)
The type of hybridization that occurs depends on the number of sigma bonds and lone pairs that have to
be accommodated around the atom.
Hybrid orbitals are responsible for all the σ bonding overlaps and non-bonding pairs in a molecule or
molecular ion. Unhybridised orbitals are responsible for all the  bonding overlaps in a molecule.
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We can also work out the type of hybridization from the shapes of the molecules or the angles between
bonding pairs.
Shapes and orientation of hybridised orbitals
type of
hybridisation
sp
sp2
sp3
how many orbitals
have been
hybridized
one s and one p
orbital
orientation of
hybridized orbitals
1 s and 2 p orbitals
120 bond angles
1 s and 3 p orbital
shape of molecule
180 bond angles
109 bond angles
linear
trigonal planar

all 4 orbitals carry bonding pair:
tetrahedral

1 orbital carries a lone pair:
pyramidal

2 lone pairs: bent
Good website: http://www.chemguide.co.uk/atoms/bonding/covalent.html#top
Link the hybridisation and formation of sigma and pi bonds in this tutorial:
http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/bom5s2_6.swf
14. 3 Delocalisation of electrons
14.3.1 Describe the delocalization of π electrons and explain how this can account for the structures of some
species.
See:


http://www.mikeblaber.org/oldwine/chm1045/notes/Bonding/Resonan/Bond07.htm
http://www.mpcfaculty.net/mark_bishop/resonance.htm
The valence bond theory could also not explain why the bonds in an ozone molecule were of the same
length and strength.
If a Lewis structure is drawn of an ozone molecule there should be 1 double bond and one single bond
which means in an ozone molecule there should be two bonds of different lengths which is not backed up
by experimental evidence. In fact, experimental evidence shows both bonds between the 3 oxygen
atoms to be the same in length and strength.
Resonance or delocalization of electrons occurs when 2 or more equivalent Lewis structures can be
drawn with different electron positions for the same positions of atoms. It can only happen in molecules
or molecular ions in which there are unhybridised p orbitals.
Resonance structures are Lewis structures which do not exist and which differ only in the location of
electron pairs but not in the number of valence electrons or the arrangement of the atoms within a
molecule. Each resonance structure shows a different way in which lone and bonding pairs are
arranged; bonding pairs and lone pairs are interchanged between the 2 or more resonance structures
that can be drawn.
Resonance then is a concept in which the actual structure of a molecule or molecular ion is taken to be
the average of all possible Lewis structures. This average structure is called the resonance hybrid.
When drawing resonance structures it is important that each one is separated by the symbol 
The reason why different arrangements of electrons can be drawn is because sometimes  bonds extend
over more than 2 atoms and this allows some of the electrons to move between more than two atoms,
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they are not localised or fixed between two atoms but they are delocalised over the entire ion or molecule
(or parts of it).
The advantage of this arrangement is that it lowers the energy of the ion/molecule making it more stable
as the negative charge carried by the electrons is more spread out over the molecule or ion; the
delocalised bonding pairs are shared equally between more than two atoms.
The delocalisation or sharing of bonding electrons between three or more atoms occurs as a result of the
overlapping of more than two unhybridised p orbitals, forming the  bond; each orbital overlaps with the
orbitals of the atoms on each side creating a larger continuous molecular orbital.; this only occurs if the p
orbitals are in the same plane(eg trigonal planar)
As the electron in the delocalised  bond move around more - have more freedom - their energy of
attraction is less than the attraction experienced by electrons in a localised sigma and pi bond which is
why the delocalised bond is the weakest bond of all 3 types.
As both a pi bond or a delocalised bond can only occur between two atoms that already have a sigma
bond, the order of bond strength/length is the following:




single bond (weakest ; longest bond length);
single bond and delocalised bond (although a delocalised bond on its own – but this does never occur
– would theoretically be the weakest); a single and delocalised bond together are sometimes referred
to as 1 1/3 bond if the delocalisation is over 3 atoms as e.g. in the ethanoate ion
double bond;
triple bond;
Experimental evidence for resonance:
The theory of resonance has been developed as a result of X-ray measurements indicating equal bond
length and bond strength within molecules which were thought to contain a mixture of double and single
bonds e.g. in sulphur dioxide, ozone, and benzene. When measured the bond length is found to be
intermediate between the single and the double bond. (see data booklet)
The actual bond consists of sigma and part of a pi bond which explains its intermediate strength and
length.
So resonance occurs whenever there is a question as to which atom(s) has/have the double bond(s)
between them. A Lewis structure can then be drawn for each possible arrangement changing the
location of the double bonds; each such a Lewis structure in which the number of electron pairs between
two atoms must differs is referred to as a resonance structure.
It must be realised that any of the possible resonance structures does not actually exist and that the
molecule does not continuously interchanges between the different resonance structures; a more
accurate representation is the resonance hybrid diagram which represents an average of all Lewis
structures. The resonance structures are merely structures showing the delocalisation.
The resonance hybrid which is the most accurate presentation refers to the diagram which summarises in
one structure the various arrangements of valence electrons and which shows the equal bond length; the
resonance hybrid` also shows the only form in which the molecule exists.
The more resonance structures that can be drawn, the greater the delocalisation – and as delocalisation
gives stability - more stable the molecule. So delocalised bonding gives stability to a molecule
Resonance is also used to explain properties such as reactivity (e.g. in the case of benzene and phenol),
stability (benzene) and acidity (ethanoate ion).
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IB past paper questions
PAPER 1
1. (M08) What is the shape of PF5?
A. tetrahedral
2.
B. trigonal pyramidal
C. square planar
D.
trigonal bipyramidal
(M08) The formula of tetrachloromethane is CCl4. Which combination correctly shows the number of
lone pairs of electrons in the Lewis structure, the shape of the molecule and the type of
hybridization of the carbon atom?
3. (N07) What is the shape of the CO32− ion and the approximate O–C–O bond angle?
A. linear, 180
pyramidal, 109
B.trigonal planar, 90
C. trigonal planar, 120
D.
4. (N07) In the molecules N2H4 , N2H2 , and N2, the nitrogen atoms are linked by single, double and
triple bonds, respectively. When these molecules are arranged in increasing order of the
lengths of their nitrogen to nitrogen bonds (shortest bond first) which order is correct?
5. (N07) What is the molecular geometry and the Cl–I–Cl bond angle in the ICl4− ion?
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Topic 14
6. (N07)
7. (M07) Which molecule is square planar in shape?
A. XeO4
B.
XeF4
C.
SF4
D.
SiF4
8. (M07)
9.
(M06) Which is the most volatile substance?
A. Chlorine
B. Fluorine
C. Sodium chloride
D. Sodium fluoride
10. (M06) Which is the smallest bond angle in the PF5 molecule?
A. 90°
B. 109.5°
C. 120°
D. 180°
11. (M06) Which types of hybridization are shown by the carbon atoms in the compound CH2= CH-CH3?
II.sp2
I. sp
A. I and II only
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B. I and III only
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III. sp3
C. II and III only
D. I, II and III
12. (M01) Which species is/are sp2 hybridised?
I. C2H4
A. I only
II. C2H6
B. I and II only
III. C3H6
C. I and III only
D. II and III only
13. (M01) Which species contains no delocalised electrons?
B. NO3-
A. O3
C. CO32-
D. H2SO4
14. (M00) In which of the following is there at least one double bond?
I. O2
II. CO2
A. I only
15.
B. III only
III. C2H4
C. II and III only
D. I, II and III
(M01) Which is the best description of metallic bonding?
A.
B.
C.
D.
The attraction between oppositely charged ions
The attraction between protons and electrons
The attraction between positive ions and delocalised electrons
The attraction between nuclei and electron pairs
16. (N01) The length of bond between carbon and oxygen is shortest in
A. CO
B. CO2
C. CH3CH2OH
D. CH3CHO
17. (N01) In which species can the bonding NOT be described in terms of delocalisation of ρi electrons
A. CH3CH2O-
B. CH3CO2-
D. NO3-
C. O3
18. (M00) According to the VSEPR theory, which molecule would be expected to have the smallest
bond angle?
A. H2O
B. H2CO
C. CH4
D. NH3
19. (M00) In which of the following would hydrogen bonding be expected to occur?
I.
CH4
II. CH3COOH
III. CH3OCH3
A. II only
B. I and III only
C. II and III only
D. I, II and III
20. (M01) In which of the following are the compounds BF3, SiH4, CO2 and SF6 arranged in decreasing
order of bond angle?
A. BF3, SiH4, CO2, SF6
B. BF3, SF6, CO2, SiH4
C. CO2, BF3, SiH4, SF6
D. SF6, CO2, SiH4, BF3
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21. (N00) Which molecule has the largest bond angle?
A. BF3
B. CF4
C. NF3
D. OF2
22. (M02) The shape of the triiodide ion, I3-, is best described as
A. bent
B. linear
C. T-shaped
D. triangular
23. (M02) The delocalisation of electrons is most likely to be significant in
A. CO2
B. SO2
C. HCOOH
D. TiO2
24. (M02) The carbon-carbon bond angle in CH3CHCH2 is closest to
A. 180
C. 120
C. 109
D. 90
25. (M02) When a Lewis structure for HCOOCH3 is drawn, how many bond pairs and how many lone
pairs of electrons are present
Bond pairs
A.
B.
C.
D.
Lone pairs
8
7
7
5
4
5
4
5
26. (N02) Which intermolecular forces exist in dry ice, CO2 (s)?
A. Covalent bonds
B. Dipole-dipole attractions C. Van der Waals forces D. Hydrogen bonding
27. (N02) The elements X and Y have the following electronic configurations:
X 1s2 2s2 2p6 3s2 3p6 4s2
Y 1s2 2s2 2p6 3s2 3p5
What is the formula of the compound formed between X and Y?
A. XY2
B. X5Y2
C. X2Y5
D. XY5
28. (N02) Which statements about the following molecule are correct?
(CH3)2CHCH=CHC≡CCH=CH2
I. Three carbon atoms are sp3 hybridized.
II. Three carbon atoms are sp2 hybridized.
III. Two carbon atoms are sp hybridized.
A. I and II only
B. I, II and III
C.
II and III only
D. I and III only
29. (M01) Which molecule has the longest nitrogen-nitrogen bond length?
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A. N2
PAPER 2
B. N2F2
1. (M08/TZ2)
2. (M07/TZ1)
3. (M07/TZ1)
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C. N2H4
D. N2H2
4. (M06)
5. (N05) This question is about Period 3 elements and their compounds.
(a) Explain, in terms of their structure and bonding, why the element sulfur is a non-conductor of
electricity and aluminium is a good conductor of electricity.
[4]
(b) Explain, in terms of its structure and bonding, why silicon dioxide, SiO2, has a high melting point.
[2]
(c) Silicon tetrachloride, SiCl4, reacts with water to form an acidic solution.
(i) Explain why silicon tetrachloride has a low melting point.
[2]
(ii) Write an equation for the reaction of silicon tetrachloride with water.
[1]
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6. (N06)
7.
(a) Explain the meaning of the term hybridization.
[1]
(b) State the type of hybridization shown by the carbon atom in the H–C≡N molecule, and
the number of σ and π bonds present in the C≡N bond.
[2]
(c) Describe how σ and π bonds form.
[4]
(N03)
(a) The boiling points of the hydrides of group 6 elements increase in the order
H2S  H2Se  H2Te  H2O
Explain the trend in the boiling points in terms of bonding.
[3]
(b) (i) Draw the Lewis structures for carbon monoxide, carbon dioxide and the carbonate ion, CO 32-.
[3]
(ii) Identify the species with the longest carbon-oxygen bond and explain your answer.
[3]
(iii) Draw the Lewis structure of ClF3 and predict its shape.
[2]
(c) Hydrazoic acid, N3H, can be represented by two possible Lewis structures in which the atoms
can be arranged as NNNH.
(i) Draw the two possible Lewis structures of N3H.
[2]
(ii) Predict the N-N-N and H-N-N bond angles in each case and give your reasoning.
[6]
(iii) Predict the hybridization of the N atom bonded to the hydrogen atom in each case.
[2]
8. (N01)
(a) (i) Draw Lewis structures to represent BF3, NF3 and BF4- .
[3]
(ii) Use the principles of the Valence Shell Repulsion Electron Pair (VSEPR) theory to predict the
shapes of the above three species. Compare and account for the bond angles in NF3 and
BF4- in terms of VSEPR theory.
[5]
(iii) Explain the meaning of the term hybridization. State the type of hybridization shown by the
central atoms in BF3 and NF3.
[3]
(iv) Explain the term polar bond. Predict and explain the polarity of the bonds within BF3 and
NF3. State whether BF3 and NF3 are polar molecules. Explain your answer.
[5]
(b) (i) Explain what is meant by a sigma () and a pi ( ) bond. Describe a double bond and a triple
bond in terms of  and  bonds.
[4]
(ii) Define the term delocalisation.
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[1]
9. (M01)
(a) For each of the molecules C2H2, C2Cl4 and SF4, draw their Lewis structure, and use the
VSEPR theory to predict their shape and bond angles.
[10]
(b) State the type of hybridization in C2H2 and C2Cl4.
[2]
(c) Draw two resonance structures for each of the ethanoate ion (CH3CO2-).
[2]
(d) Comment on the carbon to oxygen bond length in the ethanoic acid molecule and the
ethanoate ion.
[4]
10. (N02)
(a) Draw Lewis structures of the following species.
BF4-
H2CO
NO2+
NO2-
[4]
(b) Predict the shape and bond angle of each of the species in (a), explaining your choice
using VSEPR.
[8]
(c) Explain the term hybridization and state the type of hybridization in each of the species
in (a).
[3]
(d) Ethanoic acid contains two different types of carbon to oxygen bond. Explain how the
carbon and oxygen atoms combine to form each of those bonds and compare their
bond lengths and strengths.
[7]
(e) The structure of the ethanoate ion can be written as shown below
The stability of the ethanoate ion suggests a different type of carbon to oxygen bond.
(i) Describe the actual carbon to oxygen bond in the ethanoate ion.
[2]
(ii) Predict a value for the bond length of the carbon to oxygen bon din the ethanoate
ion.
[1]
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Mark scheme TOPIC 14
Paper 1
1
2
3
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6
D
C
C
D
A
B
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Paper 2
1. (M06)
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B
B
B
A
C
C
13
14
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D
D
C
A
A
A
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A
C
A
B
B
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C
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2. (N05)
3. (N06)
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6. (M01)
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7.
24 | P a g e
Topic 14
25 | P a g e
Topic 14
26 | P a g e
Topic 14