Derive the irreducible representations for the s, p, and d orbitals for SF4. Show your work. Confirm that your answers match those given in the character table.
Answers:
First, draw an axis system onto the molecule for reference:
The z axis is set by the C2 axis and the x and y axes are arbitrary.
Now find the reducible representations for each set of orbitals:
C2v |
E |
C2 |
|
|
|
|
a1 |
1 |
1 |
1 |
1 |
z |
x2, y2, z2 |
a2 |
1 |
1 |
–1 |
–1 |
Rz |
xy |
b1 |
1 |
–1 |
1 |
–1 |
x, Ry |
xz |
b2 |
1 |
–1 |
–1 |
1 |
y, Rx |
yz |
s |
1 |
1 |
1 |
1 |
|
|
p |
3 |
–1 |
1 |
1 |
|
|
d |
5 |
1 |
1 |
1 |
|
|
Using the reduction formula for each reducible representation:
s orbital:
n(a1) = [(1)(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 1
No further calculations are required because the reducible rep is 1-dimensional and that irreducible rep has been found.
The s orbital transforms as an a1. This is the same as x2 + y2 + z2 found in the far right column of the character table.
p orbitals:
n(a1) = [(3)(1)(1) + (–1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 1
n(a2) = [(3)(1)(1) + (–1)(1)(1) + (1)(1)(–1) + (1)(1)(–1)]/4 = 0
n(b1) = [(3)(1)(1) + (–1)(1)(–1) + (1)(1)(1) + (1)(1)(–1)]/4 = 1
n(b2) = [(3)(1)(1) + (–1)(1)(–1) + (1)(1)(–1) + (1)(1)(1)]/4 = 1
The p orbitals transform as a1 + b1 + b2. This is the same as found the second to last column of the character table, where z = pz is a1, x = px is b1, and y = py is b2.
d orbitals:
n(a1) = [(5)(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 2
n(a2) = [(5)(1)(1) + (1)(1)(1) + (1)(1)(–1) + (1)(1)(–1)]/4 = 1
n(b1) = [(5)(1)(1) + (1)(1)(–1) + (1)(1)(1) + (1)(1)(–1)]/4 = 1
n(b2) = [(5)(1)(1) + (1)(1)(–1) + (1)(1)(–1) + (1)(1)(1)]/4 = 1
The d orbitals transform as 2a1 + a2 + b1 + b2. This is the same as found the last column of the character table, where z2 = dz2 and x2 – y2 = dx2–y2 are a1, xy = dxy is a2, xz = dxz is b1 and yz = dyz is b2.
v(xz)