4.2.7 & 4.2.8 Shapes, and bond angles for molecules with two, three and four negative charge centers
The shape of a molecule has an important part to play in determining its chemical (e.g. reactivity and pH) and physical properties (e.g. mpt and
bpt). A negative charge center or region refers to the number of pairs electrons around the central bonded atom. This includes both the lone
(non bonding pairs) pairs and bonded pairs of electrons in single, double or triple bonds. Each double and triple bond counts as one negative
charge center. Watch out for non octet examples.
Number of negative
charge centers
Shape with respect to the
number of negative charged
centers
Bond Angle with respect
to negative charge centers
2
Linear
180°
3
Trigonal planar
120°
4
Tetrahedral
109.5°
Valence shell electron pair repulsion theory (VSEPR)
GN Lewis proposed that chemical bonds resulted when two atoms shared a pair of electrons. The Lewis concept allowed for "electron dot
bookkeeping" in the form of a Lewis dot digram to show how atoms could share electrons to achieve their quota as a noble gas or an octet of
electrons. Lewis originally did not set out to describe the shape of the molecule but it soon became apparent to him that the electron pairs
around the central atom, being like charged, would repel each other. This lead to his VSEPR theory which is still used today to describe the
shape of a molecule.
In VSEPR theory lone pairs of electrons are closer to the central atom and therefore closer to one another than the bonding pairs of electrons
because they are being attracted by the positive protons in the nucleus of one atom instead of two. The lone pairs and bonding pairs of
electrons are arranged around the central atom so as to minimize the repulsion between the lone and bonding pairs of electrons. The relative
magnitude of the electron pair repulsions is:
Lone pair / lone pair > bonded pair / lone pair > bonded pair / bonded pair repulsion
The overall shape according to the VSEPR theory depends on the number of bonding pairs and lone pairs of electrons around the central atom
of a molecule. The overall bond angle is the angle between the bonded atoms attached to the central atom.
1
To determine the overall shape and bond angle of a molecule by the VSEPR theory the number of bonding and lone pairs of electrons around the
central atom need to be identified.
Number of
negative
charge
centers
Number of
bonding
pairs of
electrons
Number of
lone pairs of
electrons
Overall Shape
(according to
VSEPR)
Overall Bond
Angle
(according to
VSEPR)
2
2
0
Linear
180°
3
3
0
Trigonal planar
120°
4
4
0
Tetrahedral
109.5°
4
3
1
Trigonal pyramid
107º
4
2
2
Bent / V shape
104.5º
Reason for the shape
(according to VSEPR)
To minimize the repulsion between
them, the two bonding pairs repel
each other equally
To minimize the repulsion between
them, the three bonding pairs repel
each other equally
To minimize the repulsion between
them, the four bonding pairs repel
each other equally
To minimize the repulsion
bonded pair-lone pair > bonded pairbonded pair repulsion
To minimize the repulsion
Lone pair-lone pair > bonded pairlone pair > bonded pair- bonded pair
repulsion
Examples
CO2, HCN, BeCl2
BF3, C2H4
CH4, SiCl4, CCl4
NH3, NF3
H2O, H2S
NOTE: The number of negative charge centers may not be reflected in the molecular shape. If the electrons are not being used for bonding then
they cannot be seen and the molecule is described as if they were not there. In other words the shape and angle of the molecule is determined
with respect to the attached atoms.
2
Complete the
column.
Formula and
name
HCN
Hydrogen
cyanide
BeCl2
Beryllium
chloride
table below leaving out the last
Structural
formula in
correct shape
(To do this
you may need
to the Lewis
dot diagram
first)
H-C≡N
Cl-Be-Cl
Overall
Number
Bond
negative
Angle
charge
(according
centers
to VSEPR)
2
2
180
180
Overall
Shape
(according
to VSEPR
theory)
Explanation for the overall shape using VSEPR theory
linear
2 bonding pairs of electrons & 0 lone pairs around the
central atom. In order to minimize the repulsion the
bonding pairs of electrons repel each other equally.
linear
2 bonding pairs of electrons & 0 lone pairs around the
central atom. In order to minimize the repulsion the
bonding pairs of electrons repel each other equally.
3
Polar or non
polar
covalent
Formula and
name
CO2
H2O
CHCl3
NH3
Structural
formula
O=C=O
Number
negative
charge
centers
2
4
4
4
Bond
Angle
180
104.5
109.5
107
Overall
Shape
Explanation for shape using VSEPR theory
linear
2 bonding pairs of electrons & 0 lone pairs of electrons
around the central atom. Bonding pairs of electrons
repel each other equally in order to minimize the
repulsion between them. OR 2 negative charge
centers repel each other as much as possible to
minimize the repulsion between them.
bent
2 bonding pairs of electrons and 2 lone pairs of
electrons. In order to minimize the repulsion between
then the 2 lone pairs of electrons exert a greater
repulsion than the bonding pair and lone-bonding pair
repulsion.
4 bonding pairs of electrons and 0 lone pairs of
electrons around the central atom. To minimize the
tetrahedral
repulsion between them, the four bonding pairs repel
each other equally
3 bonding pairs of electrons and 1 lone pair of
electrons around the central atom. To minimize the
repulsion between them the bonded pair / lone
pair > bonded pair / bonded pair repulsion
trigonal
pyramid
4
Polar or non
polar
Formula and
name
Structural
formula
Number
negative
charge
centers
Bond
Angle
Overall
Shape
Explanation for shape using VSEPR theory
H2S
CH4
BF3
SiCl4
5
Polar or non
polar
Formula and
name
Structural
formula
Number
negative
charge
centers
Bond
Angle
Overall
Shape
Explanation for shape using VSEPR theory
CClF3
Cl2O
C2H2
ethyne
H2CO
methanal
6
Polar or non
polar
Formula and
name
Structural
formula
Number
negative
charge
centers
Bond
Angle
Overall
Shape
Explanation for shape using VSEPR theory
SO2
Sulfur dioxide
C2H4
ethene
POCl3
O3
7
Polar or non
polar
CH3NH2
Aminomethane
(C and N)
(C and N)
( with respect to C and N)
Extra Problems
1.
Predict the bond angle and shape around the carbon atom and the oxygen atom attached to the hydrogen atom in methanoic acid
(HCOOH).
2.
Compare the bond length and strength between the carbon and oxygen atoms in carbon dioxide and carbon monoxide.
3.
State and compare the difference in the bond angle in CH4 and NF3.
8
4.
Compare the N-N-H bond angle between the two possible Lewis structures of N3H.
5.
State the bond angle around the carbon and nitrogen atoms in aminomethane, CH 3NH2
6.
Draw the structural formula for ethanoic acid and determine its relative molecular mass.
9
ANSWERS
Formula
HCN
Hydrogen
cyanide
BeCl2
Beryllium
chloride
Structural
formula
H-C≡N
Cl-Be-Cl
Number
negative
charge
centers
2
2
Shape with
respect to
negative
charge
centers
linear
linear
Bond
Angle
180º
180º
Overall
Shape
Explanation for shape using VSEPR
theory
Polar or
non
polar
linear
2 bonding pairs of electrons & 0 lone
pairs around the central atoms. In order
to minimize repulsion the bonding pairs
of electrons repel each other equally.
polar
linear
2 bonding pairs of electrons & 0 lone
pairs around the central atoms. Bonding
pairs of electrons repel each other
equally.
Non
polar
CO2
O=C=O
2
linear
180º
linear
C2H2
ethyne
H-C≡C-H
2
linear
180º
linear
3
Trigonal
planar
120º
Trigonal
planar
BF3
10
2 bonding pairs of electrons & 0 lone
pairs of electrons around the central
atom. Bonding pairs of electrons repel
each other equally. OR 2 negative
charge centers arranged as far apart as
possible.
2 bonding pairs of electrons & 0 lone
pairs of electrons around the central
atom. Bonding pairs of electrons repel
each other equally. OR 2 negative
charge centers arranged as far apart as
possible.
3 bonding pairs of electrons & 0 lone
pairs of electrons around the central
atom. Bonding pairs of electrons repel
each other equally. OR 3 negative
charge centers arranged as far apart as
possible.
Non
polar
Non
polar
H2CO
methanal
3
SO2
Sulfur dioxide
3
C2H4
ethene
3 for
each C
atom
CH4
methane
4
SiCl4
4
ClF3
H2O
4
4
3 bonding pairs of electrons & 0 lone
pairs of electrons around the central
Trigonal
Trigonal
atom. Bonding pairs of electrons repel
120º
planar
planar
each other equally. OR 3 negative
charge centers arranged as far apart as
possible.
2 bonding pairs & 1 lone pair of electrons
Trigonal
around the central atom. Lone pair bent
~117º
planar
bonding pair repulsion is greater than
bonding pair – bonding pair repulsion.
3 bonding pairs & 0 lone pairs of
electrons around each C atom. 3
Trigonal
120º
bonding pairs of electrons repel each
planar for between
Trigonal
other equally in order to minimize the
each C
each H
planar
repulsion.
OR
atom
atom
3 negative charge centers on each
carbon atom arranged as far apart as
possible to minimize repulsion.
4 bonding pairs & 0 lone pairs of
electrons. In order to minimize the
Tetrahedral 109.5º Tetrahedral
repulsions the 4 bonding pairs of
electrons repel each other equally.
Tetrahedral
Tetrahedral
Tetrahedral
Polar
Polar
Non
Polar
Non
Polar
109.5º
4 bonding pairs & 0 lone pairs of
Tetrahedral electrons. 4 bonding pairs of electrons
repel each other equally.
Non
Polar
109.5º
4 bonding pairs & 0 lone pairs of
Tetrahedral electrons. 4 bonding pairs of electrons
repel each other equally.
Polar
104.5º
2 bonding pairs of electrons and 2 lone
pairs of electrons. The 2 lone pairs of
electrons exert a greater repulsion than
the bonding pair and lone-bonding pair
repulsion.
Polar
Bent
11
NH3
ammonia
4
POCl3
3 or 4
Tetrahedral
Trigonal
pyramid
107º
Tetrahedral 100-112 Tetrahedral
(C and N)
CH3NH2
Aminomethane
4
3 bonding pairs and 1 lone pair of
electrons. Lone pair – bonding pair
repulsion is greater than bonding pair
repulsion.
Tetrahedral
C
(109.5º)
N
(107º)
Polar
Carbon - 4 bonding pairs & 0 lone pairs
of electrons. 4 bonding pairs of electrons
repel each other equally.
C
Tetrahedral
N
N - 3 bonding pairs and 1 lone pair of
Trigonal
electrons. Lone pair – bonding pair
pyramid
repulsion is greater than bonding pair
repulsion.
12
Polar
Polar
IB Problems
1.
Predict the bond angle around the carbon atom and the oxygen atom attached to the hydrogen atom in methanoic acid
(HCOOH).
Carbon – 3 bonding pairs and 0 lone pairs of electrons. Bond angle 120º (trigonal planar shape)
Oxygen – 2 bonding pairs of electrons and 2 lone pairs. Bond angle 104.5º (bent shape)
2.
Compare the bond length and strength between the carbon and oxygen atoms in carbon dioxide and carbon monoxide.
The more bonding electrons between the nuclei of the two bonded atoms, The greater the attraction the protons in the nucleus
of each atoms have for the bonding electrons the shorter the bond and the more energy will be required to break it. Therefore
the C=O bond in CO2 is longer and has a lower bond energy than the C≡O bond in CO.
3.
State and compare the difference in the bond angle in CH4 and NF3.
CH4 – the bond angle between the H atoms is 109.5º due to 4 bonding pairs & 0 lone pairs of electrons. 4 bonding pairs of
electrons repel each other equally in order to minimize the repulsion between them..
NF3 - 107º due to 3 bonding pairs and 1 lone pair of electrons. Lone pair – bonding pair repulsion is greater than bonding pair
repulsion, which results in the bonding pairs of electrons being pushed closer together than in CH 4, decreasing the bond angle
between the attached atoms.
4.
Compare the N-N-H bond angle between the two possible Lewis structures of N3H.
13
5.
State the bond angle around the carbon and nitrogen atoms in aminomethane, CH 3NH2
C - 109.5º
N - 107º
6.
Draw the structural formula for ethanoic acid and determine its relative molecular mass.
14