## 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.

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.