25. Derive the irreducible representations for the s, p, and d orbitals for SF_{4}. 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 C_{2} axis and the x and y axes are arbitrary.
Now find the reducible representations for each set of orbitals:
C_{2v} 



a_{1} 

a_{2} 

b_{1} 

b_{2} 








s 



p 



d 


Using the reduction formula for each reducible representation:
s orbital:
n(a_{1}) = [(1)(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 1
No further calculations are required because the reducible rep is 1dimensional and that irreducible rep has been found.
The s orbital transforms as an a_{1}. This is the same as x^{2} + y^{2} + z^{2} found in the far right column of the character table.
p orbitals:
n(a_{1}) = [(3)(1)(1) + (–1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 1
n(a_{2}) = [(3)(1)(1) + (–1)(1)(1) + (1)(1)(–1) + (1)(1)(–1)]/4 = 0
n(b_{1}) = [(3)(1)(1) + (–1)(1)(–1) + (1)(1)(1) + (1)(1)(–1)]/4 = 1
n(b_{2}) = [(3)(1)(1) + (–1)(1)(–1) + (1)(1)(–1) + (1)(1)(1)]/4 = 1
The p orbitals transform as a_{1} + b_{1} + b_{2}. This is the same as found the second to last column of the character table, where z = p_{z} is a1, x = p_{x} is b_{1}, and y = p_{y} is b_{2}.
d orbitals:
n(a_{1}) = [(5)(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1)]/4 = 2
n(a_{2}) = [(5)(1)(1) + (1)(1)(1) + (1)(1)(–1) + (1)(1)(–1)]/4 = 1
n(b_{1}) = [(5)(1)(1) + (1)(1)(–1) + (1)(1)(1) + (1)(1)(–1)]/4 = 1
n(b_{2}) = [(5)(1)(1) + (1)(1)(–1) + (1)(1)(–1) + (1)(1)(1)]/4 = 1
The d orbitals transform as 2a_{1} + a_{2} + b_{1} + b_{2}. This is the same as found the last column of the character table, where z^{2} = d_{z}2 and x^{2} – y^{2} = d_{x}2_{–y}2 are a_{1}, xy = d_{xy} is a_{2}, xz = d_{xz} is b_{1} and yz = d_{yz} is b_{2}.