The molecule belongs to the C_2v point group. Use the p orbitals perpendicular to the page as the basis set. The z axis is along the C-Cl bond and let the x axis be in the plane of the paper. Let p₁ be on the C connected to Cl and number clockwise around the ring.

C2v	E	C2	σv(xz)	σv'(yz)
a1	1	1	1	1
a2	1	1	–1	–1
b1	1	–1	1	–1
b2	1	–1	–1	1
p orbitals	3	–1	–3	1
R(p1)	p1	–p1	–p1	p1
R(p2)	p2	–p3	–p2	p3

n(a₁) = [(1)(1)(3) + (1)(1)(–1) + (1)(1)(–3) + (1)(1)(1)]/4 = 0

n(a₂) = [(1)(1)(3) + (1)(1)(–1) + (1)(–1)(–3) + (1)(–1)(1)]/4 = 1

n(b₁) = [(1)(1)(3) + (1)(–1)(–1) + (1)(1)(–3) + (1)(–1)(1)]/4 = 0

n(b₂) = [(1)(1)(3) + (1)(–1)(–1) + (1)(–1)(–3) + (1)(1)(1)]/4 = 2

a₂ + 2b₂

From the results of the symmetry operations on each orbital, p₁ is never changed with any other orbital, so its representation is found by simply inspecting the characters (1, –1, –1, 1) and is one of the b₂ orbitals.

Using the projection operators gives:

P(a₂ p₂) = (1)p₂ + (1)(–p₃) + (–1)(–p₂) + (–1)(p₃) = 2[p₂ – p₃]

Ψ(a₂) = 2^½[p₂ – p₃]

P(b₂ p₂) = (1)p₂ + (–1)(–p₃) + (–1)(–p₂) + (1)(p₃) = 2[p₂ + p₃]

Ψ(b₂) = 2^½[p₂ + p₃]