452 Chapter 16Vectors
A system of electric dipoles with momentsμ
1, μ
2, =, μ
N, has total dipole moment
(16.17)
The total dipole moment of a molecule is sometimes interpreted as the (vector) sum
of ‘bond moments’; that is, a dipole moment is associated with each bond. In some
cases of high symmetry, these bond moments may cancel to give zero total dipole
moment. For example, the methane molecule in its stable state has its four hydrogens
at the vertices of a regular tetrahedron, with the carbon at the centre. If one of the
vertices is placed at the ‘111–position’, withr
11 = 1 (a,a,a), the positions of the other
vertices arer
21 = 1 (a, −a, −a),r
31 = 1 (−a,a, −a), andr
41 = 1 (−a, −a,a). The length of each
of these bond vectors is the CH bondlength, and the dipole moment of each
bond lies along the direction of the bond, and is therefore a multiple of the bond
vector,μ
i1 = 1 kr
i(i 1 = 1 1, 2, 3, 4). The total dipole moment is then
μ
1 = 1μ
11 + 1
μ
21 + 1
μ
31 + 1
μ
4= 1 k(r
11 + 1 r
21 + 1 r
31 + 1 r
4)
= 1 k(a 1 + 1 a 1 − 1 a 1 − 1 a,a 1 − 1 a 1 + 1 a 1 − 1 a,a 1 − 1 a 1 − 1 a 1 + 1 a)
= 1 k(0, 0, 0)
= 10
0 Exercise 12
Base vectors
The cartesian unit vectors i,jand klie along the x, y, and zdirections and have
components,
i 1 = 1 (1, 0, 0), j 1 = 1 (0, 1, 0), k 1 = 1 (0, 0, 1) (16.18)
The rules (16.10) and (16.11) then confirm that every vector in the three-dimensional
space can be expressed as a linear combination of these three base vectors:
a 1 = 1 a
xi 1 + 1 a
yj 1 + 1 a
zk
= 1 a
x(1, 0, 0) 1 + 1 a
y(0, 1, 0) 1 + 1 a
z(0, 0, 1)
(16.19)
= 1 (a
x, 0, 0) 1 + 1 (0, a
y, 0) 1 + 1 (0, 0, a
z)
= 1 (a
x, a
y, a
z)
EXAMPLE 16.7Givena 1 = 12 i 1 + 13 j 1 + 1 kandb 1 = 1 i 1 − 12 j, find (i)d 1 = 12 a 1 + 13 b, (ii) a vector
perpendicular to b, (iii) a vector perpendicular to d.
3 a,
μμμ=+++ =μμμμμμμ
=∑
121NNii