Download CHAPTER 3: SHAPES A D SIZES

 Dr Roger Nix (Queen Mary, University of London) - 3.1
CHAPTER 3: SHAPES AD SIZES
A knowledge of shape and size is essential to an understanding of molecular properties. It is therefore
necessary to know how to manipulate bond length and bond angle information. The basic
mathematical technique that will be required is trigonometry.
Trigonometric Functions
Trigonometric Functions - Defined using a Right-Angled Triangle
The fundamental notions of trigonometry are enshrined in the following length - angle relationships
derived from a right-angled triangle ( a right-angled triangle has one of its three angles equal to 90° )
r
The side opposite the right-angle (side length r ) is
known as the hypotenuse
y
θ
x
The basic trigonometric functions are defined as follows:
Sine (usually abbreviated as sin ) of an angle
i.e.
Sine of θ
=
Cosine of θ
=
Tangent of θ
Note :
=
=
length of side adjacent to angle
length of hypotenuse
=
length of side opposite angle
length of side adjacent to angle
r
cos θ =
x
r
Tangent (usually abbreviated as tan ) of an angle
i.e.
length of side opposite angle
length of hypotenuse
sin θ = y
Cosine (usually abbreviated as cos ) of an angle
i.e.
=
tan θ =
y
x
the triangle must be right-angled to use these definitions.
 Dr Roger Nix (Queen Mary, University of London) - 3.2
Relationships between Trigonometric Functions
From a consideration of the same
right-angled triangle ....
r
y
θ
x
sin θ = y
cos θ =
r
⇒
tan θ =
x
r
tan θ =
y
x
sin θ
cos θ .
Values of the Trigonometric Functions for Common Angles
In order to be able to estimate certain common values of trigonometric functions the following
triangles are useful:
45°
√2
30° 30°
2
1
45°
60°
60°
1
1
(i) the isosceles right-angled triangle
2
√3
2
1
(ii) based on the equilateral triangle,
The lengths of √2 and √3 can be calculated using the Pythagoras Theorem (see later in this section).
It follows that:
sin 30° =
1
2
sin 45° =
1
2
sin 60° =
3
2
cos 30° =
3
2
cos 45° =
1
2
cos 60° =
1
2
tan 30° =
1
tan 45° = 1
tan 60° =
sin 0° = 0
cos 0° = 1
tan 0° = 0
sin 90° = 1
cos 90° = 0
tan 90° = ∞
3
Note also the following:
3
 Dr Roger Nix (Queen Mary, University of London) - 3.3
Trigonometric Functions - Extension to Angles greater than 90°
So far the values of trigonometric functions have only been defined for the remaining angles in a rightangled triangle (i.e. for angles less than 90°). It is however possible to liberate the concept of an angle
from this constraint and consider a more general situation.
y - axis
A (x,y)
y
θ
O
x - axis
x
B
Consider a point A at some coordinates (x,y) with respect to an origin O . The angle θ is defined as
shown and increases from θ = 0°, when OA lies along the + x axis, to any larger (positive) value
simply by rotating the line in an anticlockwise direction.
Since OB̂A is a right angle, then, by analogy with our previous definitions, the trigonometric
functions for θ can be defined as follows :
and
sin θ =
BA
y
=
OA
r
tan θ =
y sin θ
=
x cos θ
and
cos θ =
OB x
=
OA r
The angle θ may now take any value (including negative values, which correspond to a clockwise
rotation from the x-axis, and values over 360°, which correspond to more than a complete rotation).
By inspection of the diagram it should be clear that :
1. if θ changes from positive to negative, then x does not change but y does change sign, hence
sin (–θ ) = – sin θ
whereas
cos (–θ ) = cos θ
2. the values of both the sine and cosine functions will start to repeat after θ exceeds 360° ; the
functions are said to be periodic (cyclic) with a period of 360°.
3. as θ ranges from 0° to 360° the signs of the trigonometric functions change in accord with the
changes in the sign of x and y , i.e.
 Dr Roger Nix (Queen Mary, University of London) - 3.4
θ
x
+
–
–
+
0 - 90°
90 - 180°
180 - 270°
270 - 360°
sin θ
+
+
–
–
y
+
+
–
–
cos θ
+
–
–
+
tan θ
+
–
+
–
It should also be clear (since the values of x and y can never exceed the length r ) that the values of the
sine and cosine functions can never be greater than +1, nor less than –1. By contrast the tangent
function is discontinuous and varies from –∞ to +∞ every 180°.
These latter points are exemplified by a graphical representation of the functions :
f (θ ) = sin θ
1.0
0.0
-90.0
0.0
90.0
180.0
270.0
360.0
450.0
θ
90.0
180.0
270.0
360.0
450.0
θ
90.0
180.0
270.0
360.0
450.0
θ
-1.0
f (θ ) = cos θ
1.0
0.0
-90.0
0.0
-1.0
f (θ ) = tan θ
3.0
2.0
1.0
0.0
-90.0
0.0
-1.0
-2.0
-3.0
 Dr Roger Nix (Queen Mary, University of London) - 3.5
Trigonometric Functions - Additional Properties/Relationships
From consideration of the graphs on the preceding page, certain relationships may be readily
established for the trigonometric functions.
For example,
sin (–θ ) = – sin θ
cos (–θ ) = cos θ
sin (180 – θ ) = sin θ
cos (180 – θ ) = – cos θ
sin (90 – θ ) = cos θ
cos (90 – θ ) = sin θ
By using the Pythagoras Theorem (see later in this section) it can also be shown that:
sin2 θ + cos2 θ = 1
It is also possible to establish many more complex relationships, but it is not necessary to memorize
these since they can always be found in a textbook if they are needed.
Examples include,
sin (2θ ) = 2 sin θ cos θ
sin (x + y) = sin x cos y + cos x sin y
Inverse Trigonometric Functions
If :
y = sin θ , then what is θ ?
To find the value of θ you need to apply the inverse of the sine function to both sides of the equation
- this is written as sin–1 or arcsin .
i.e.
θ = sin–1 y = arcsin y
Similarly,
cos–1 ( arccos ) is the inverse of cos
tan–1 ( arctan ) is the inverse of tan
Important : in this special case the superscripted –1 is used to represent the inverse of the function
and not "to the power of –1" as it would otherwise be interpreted.
 Dr Roger Nix (Queen Mary, University of London) - 3.6
Trigonometric Functions - Application to on-Right Angled Triangles
Although the trignonometric functions have been defined using a right-angled triangle, there are useful
relationships between these functions that apply equally to all (i.e. not just right-angled) triangles.
In this instance capital letters have been used to label the apices (corners) and corresponding lower
case letters for the opposite sides.
The Sine Rule
a
b
c
=
=
sin A sin B
sin C
B
c
Proof : in the triangle ABC construct a line from B to a
point X such that BXA = BXC = 90° (i.e. a line
perpendicular from AC to B.) Let the angle at a vertex of the
triangle be denoted by the letter for that point.
a
x
A
C
X
Then
⇒
⇒
⇒
x
c
x = c sin A
sin A =
x
a
x = a sin C
and sin C =
and
a sin C = c sin A
a
c
=
sin A
sin C
etc.
The Cosine Rule
a2 = b2 + c2 – 2 b c cos A
B
Proof :
c
and
a2 =
=
=
=
=
(c2 –
c2 +
c2 +
c2 +
c2 +
c2 = x2 + m2
a
x
A
a2 = x2 + n2
m
X
⇒
n
C
b
m
c
2
= c + b2 – 2bc cos A
cos A =
but
⇒
m2 ) + n 2
n 2 – m2
(n + m)(n – m)
b (b – 2m)
2
b – 2bm
a2
You do not need to remember the proofs, but you do need to know how to use the rules. The cosine
rule is actually the more useful of the two rules when it comes to calculating distances and angles in
molecules.
 Dr Roger Nix (Queen Mary, University of London) - 3.7
Degrees and Radians - Alternative Measures of the Size of an Angle
So far angles have been defined on the arbitrary scale that a right angle is 90° (90 degrees) and a
complete rotation about a point 360° (360 degrees).
A more rational scale is based on the radian; in this case the magnitude of an angle is defined as the
length of the arc formed on a circle of unit radius.
Thus one radian is the angle subtended by an arc of unit length, on a circle of unit radius.
This also establishes the following general relationship between the radius, angle and arc length.
 θ 
Arclength = r × 

 radians 
θ
r
For a complete circle, the arc length is equal to the circumference, which is equal to 2 π r
Hence, the angle corresponding to 360° is given by :
 θ 
2πr = r × 

 radians 
⇒
θ = 2π radians ( = 2π c )
Consequently,
 360 o
1 radian = 
 2π

 = 57.30°

When using the trigonometric functions for simple geometrical calculations the argument of the
trigonometric functions may normally be specified in either degrees or radians, so for example
sin 30° = sin (π/6)
cos 45° = sin (π/4)
However, there are many other occasions (e.g. calculus) where it is necessary to always use radians.
Important : when using a calculator and functions it is very important to ensure that the calculator is
in the required mode, i.e. to know whether it is working in degrees or radians.
For all calculations involving the geometry of molecules you should work in degrees but there are
occasions when it is essential to work with radians.
 Dr Roger Nix (Queen Mary, University of London) - 3.8
Trigonometric Functions - Use for Representing Oscillatory Phenomena
As noted previously the trigonometric functions are periodic; the sine and cosine functions are also
continuous functions.
The motion of a range of oscillating objects may therefore be described using either the sine or cosine
functions.
Formulae for oscillations as a function of time
e.g.
 2π t 
f (t ) = A sin(ω t ) = A sin( 2πν t ) = A sin

 τ 
Amplitude, A
f(t)
Time, t
Period, τ
where:
ω
ν
-
"angular frequency"
frequency of oscillation
Formulae for oscillations as a function of distance
e.g.
 2π x 
f ( x ) = A sin

 λ 
Amplitude, A
f(x)
Distance, x
Wavelength, λ
 Dr Roger Nix (Queen Mary, University of London) - 3.9
Pythagoras Theorem
Pythagoras showed that for any right-angled triangle
r
x2 + y2 = r2
y
θ
x
Example : consider a square, with side-length a
Length of diagonal, d , is given by:
d
a2 + a2 = d2
a
a
⇒
d2 = 2 a2
⇒
d = 2a 2 = 2 .a
Relationships between Trigonometric Functions
.
From Pythagoras :
x2 + y2 = r2
⇒
x2 y2
+
=1
r2 r2
⇒
sin2θ + cos2θ = 1
where sin2θ means sin θ × sin θ
i.e. (sin θ ) 2
Extension of Pythagoras Theorem to 3-Dimensions (3D)
The Pythagoras theorem can be readily extended to three-dimensions.
In 3-dimensions :
r2 = x2 + y2 + z2
 Dr Roger Nix (Queen Mary, University of London) - 3.10
Proof :
C
Consider the 3-dimensional object shown - in which
the angles OAB and ABC are both right angles.
z
The required length, r , is OC.
B
Step 1 : consider the triangle OAB on the base of
the object - this is a right-angled triangle so we can
apply the simple form of Pythagoras to give :
O
x
y
A
OB 2 = OA 2 + AB 2 = x 2 + y 2
Step 2 : now consider the shaded-triangle OBC. This is also a right-angled triangle since the angle
OBC is also a right angle. We can again apply the simple form of Pythagoras to give :
OC 2 = OB 2 + BC 2 = ( x 2 + y 2 ) + z 2
⇒
r2 = x2 + y2 + z2
Example : what is the length of the diagonal of a cube with side length a ?
C
d
OC 2 = d 2 = a 2 + a 2 + a 2
a
a
O
a
⇒
d 2 = 3a 2
⇒
d = 3 .a
 Dr Roger Nix (Queen Mary, University of London) - 3.11
Formulae for Calculating Perimeters, Areas & Volumes
of Common 2D and 3D Shapes
Two Dimensional (2-D) Shapes
Triangle :
Area = ½ × base (b) × height (h)
h
b
Parallelogram :
Area
= a×h
b
h
θ
= a b sin θ
a
( Area of rectangle = ab )
Circle :
Circumference
Area
= 2πr = πd
= π r2
=
πd
r
d(=2r)
2
4
Three Dimensional (3-D) Shapes
Cube :
Volume = a3
a
a
a
Sphere :
Surface Area = 4 π r2
Volume
=
r
4
/3 π r
3
d(=2r)
 Dr Roger Nix (Queen Mary, University of London) - 3.12
Distances and Angles in Molecules
Applications of Trigonometric Functions & Trigonometric Relationships
Example
Water is a bent molecule. The HOH angle is 104° and the O–H bond distance is 97 pm. What is the
H--H non-bonded distance ?
Approach 1 :
Consider the triangle OMH where M is the midpoint of
the H--H line. Since the two O–H distances are the
same, then the angle OMH is 90°.
H
Thus
x
sin β =
r
⇒
x = r sin β
r = 97 pm
⇒
⇒
O
r
r
β
x
M
H
x
d (H--H) = 2 x = 2 r sin β
β = (104/2)° = 52°
d (H--H) = 153 pm
Approach 2 :
Use the cosine rule :
r
d 2 = r 2 + r 2 – 2 r r cos θ
⇒
d 2 = 2 r 2 – 2 r 2 cos θ
⇒
d 2 = 2 r 2 ( 1 – cos θ )
r = 97 pm
⇒
H
O
θ
d
r
H
θ = 104°
d (H--H) = 153 pm
Applications of the Pythagoras Theorem
The Pythagoras theorem is very useful for calculating internuclear distances in molecules in those
cases where the angle between two bonds is exactly 90° , e.g. in square planar complexes and
octahedral molecules.
As previously noted, the Pythagoras theorem can be readily extended to three-dimensions and this is
very useful for calculating internuclear distances in solids exhibiting cubic symmetry.
 Dr Roger Nix (Queen Mary, University of London) - 3.13
Regularly Shaped Molecules
Regular 2-Dimensional Polygons
Some molecules are flat and have geometries based on regular polygons. A regular polygon is a
many-sided figure in which the sides have equal lengths.
triangle
e.g. cyclopropane
C 3 H6
square
e.g. cyclobutane *
C 4 H8
hexagon
e.g. benzene
C 6 H6
pentagon
e.g. cyclopentadienyl anion
C 5 H5 –
heptagon
e.g. tropylium cation
C 7 H7 +
octagon
etc.
( * = the cyclobutane molecule is actually not quite flat )
It is easy to show that the sum of the interior angles for an n-sided polygon is 180(n – 2)°
- it follows that for a regular polygon in which the n interior angles are all the same, that the interior
angle (in degrees) is given by:
θ=
θ
∼
∼
180(n − 2)
 360 
= 180 − 

n
 n 
( n-sides )
Summary :
n
3
4
5
6
7
Sum of Interior Angles (°)
(any type of polygon)
180
360
540
720
900
Interior Angle for
Regular Polygon (°)
60
90
108
120
128.6
 Dr Roger Nix (Queen Mary, University of London) - 3.14
Regular 3-Dimensional Shapes
Most molecules are not flat and it is most important to appreciate their three-dimensional shape.
Fortunately some very simple shapes are also very common in chemistry. The cube, the tetrahedron
and the octahedron are examples of Regular Archimedian Solids (there are five altogether). Each solid
is characterised by having a set of identical faces which are regular polygons.
Shape
No. of faces
No. of vertices
tetrahedron
4
4
cube
6
8
octahedron
8
6
Note that the name of the solid shape derives from the number of faces and not from the number of
vertices (corners). Thus for an inorganic complex in which six ligands are coordinated to a central
atom or ion, the positions of the outer atoms define an octahedron (see left hand diagram) and the
complex is said to be octahedral.
Note also that an octahedron can be constructed inside a cube by locating the six vertices in the middle
of each face of the cube (see right hand diagram). This explains why we talk about octahedral holes
or octahedral interstices inside face centred cubic (and other related) lattices. If you imagine how
cubes may be stacked together, you should also be able to see how you can easily get a regular array of
(corner-sharing) octahedra linked by their vertices.
Tetrahedral Coordination
Most MX4 type molecules (e.g. methane, CH4 ) have a tetrahedral bonding geometry. You should
know that the tetrahedral bonding angle is around 109° - but how do we prove this ?
The tetrahedral geometry is normally drawn as shown below on the left. It arises from placing the
central atom at the centre of a tetrahedron, and the four outer atoms at the vertices of the tetrahedron,
as shown in the right hand diagram.
X
X
M X
X
 Dr Roger Nix (Queen Mary, University of London) - 3.15
What is less widely recognised is that the tetrahedron (and a molecule with tetrahedral geometry) can
be considered within the context of a cube and this can be used to simplify calculations of internuclear
distances and angles. The central atom is placed at the centre of the cube, two atoms are place on
either end of a diagonal of one face of the cube, and the remaining two atoms on either end of the
cross-diagonal of the opposite face, as shown below:
C
M
θ
B
X
O
Let the length of the cube side be a .
OB = length of the diagonal of one face of the cube =
2 .a
(by Pythagoras, see 3.9)
2 .a
a
=
2
2
OX = ½ OB =
MX = ½ a (since M is at the centre of the cube)
a
OX
=
MX
2
a
2
a
2
2
=
= 2
a
2
⇒
tan θ =
⇒
θ = arctan 2 = 54.736°
⇒
Tetrahedral angle, 2θ = 109.47°
=
2
×
( )
(Note - there are several other ways of using trigonometry to arrive at the same answer)