CHM 501

25. 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.

Answer

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

σv(xz)

σv(xz)

 

 

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.