SME
Inorganic Chemistry · Problem type
Reducing a representation (C₂ᵥ)
Use this for “The 3N motions of CH₂Cl₂ give Γ = 15, −1, 3, 3. Reduce Γ and find how many vibrations are IR active.”
Diagram
n(A₁)
=
(
1
h
)
Σ
g
χΓ
χ(A₁)
n(A₁) = (9 × 1 + (−1) × 1 + 1 × 1 + 3 × 1) ÷ 4
n(A₁) = 3
Γ = 3A₁ + A₂ + 2B₁ + 3B₂
Tap a species to multiply its row by Γ.
This is an example. Type your own values to start a new problem.
χ(E)
Character under E
example
χ(C₂)
Character under C₂
example
χ(σᵥ)
Character under σᵥ(xz)
example
χ(σᵥ′)
Character under σᵥ′(yz)
example
n(A₁)
Count of A₁
calculated
n(A₂)
Count of A₂
calculated
n(B₁)
Count of B₁
calculated
n(B₂)
Count of B₂
calculated
nIR
IR-active vibrations
worked out (not typed)
Formulas
n(A₁) = (χ(E) + χ(C₂) + χ(σᵥ) + χ(σᵥ′)) ÷ 4
3
=
9 + (−1) + 1 + 3
4
n(A₂) = (χ(E) + χ(C₂) − χ(σᵥ) − χ(σᵥ′)) ÷ 4
1
=
9 + (−1) − 1 − 3
4
n(B₁) = (χ(E) − χ(C₂) + χ(σᵥ) − χ(σᵥ′)) ÷ 4
2
=
9 − (−1) + 1 − 3
4
n(B₂) = (χ(E) − χ(C₂) − χ(σᵥ) + χ(σᵥ′)) ÷ 4
3
=
9 − (−1) − 1 + 3
4
nIR = (n(A₁) − 1) + (n(B₁) − 2) + (n(B₂) − 2)
3 = (3 − 1) + (2 − 2) + (3 − 2)
Assumptions
•
C₂ᵥ has order 4: each count is Γ times that species’ row of the character table, summed, then ÷ 4.
•
A planar molecule lies in the yz plane, so σᵥ′(yz) is the plane holding every atom.
•
For all 3N motions, take away translations (A₁ + B₁ + B₂) and rotations (A₂ + B₁ + B₂); the vibrations left in A₁, B₁ and B₂ are IR active.
•
The picture: the character table with Γ under it; each count is that species’ row times Γ, summed, ÷ 4.
Step-by-step
WE KNOW
χ(E) = 9, χ(C₂) = −1, χ(σᵥ) = 1, χ(σᵥ′) = 3
FIND
count of A₁ (n(A₁)), count of A₂ (n(A₂)), count of B₁ (n(B₁)), count of B₂ (n(B₂)), IR-active vibrations (nIR)
Step 1 · Find count of A₁ (n(A₁))
Use n(A₁) = (χ(E) + χ(C₂) + χ(σᵥ) + χ(σᵥ′)) ÷ 4
A₁’s characters are all 1: add the four characters of Γ, then divide by the order, 4.
n(A₁)
=
9 + (−1) + 1 + 3
4
n(A₁)
=
12
4
n(A₁) = 3
Step 2 · Find count of A₂ (n(A₂))
Use n(A₂) = (χ(E) + χ(C₂) − χ(σᵥ) − χ(σᵥ′)) ÷ 4
A₂ has −1 under both planes, so those two characters are taken away.
n(A₂)
=
9 + (−1) − 1 − 3
4
n(A₂)
=
4
4
n(A₂) = 1
Step 3 · Find count of B₁ (n(B₁))
Use n(B₁) = (χ(E) − χ(C₂) + χ(σᵥ) − χ(σᵥ′)) ÷ 4
B₁ has −1 under C₂ and σᵥ′(yz), so those two are taken away.
n(B₁)
=
9 − (−1) + 1 − 3
4
n(B₁)
=
8
4
n(B₁) = 2
Step 4 · Find count of B₂ (n(B₂))
Use n(B₂) = (χ(E) − χ(C₂) − χ(σᵥ) + χ(σᵥ′)) ÷ 4
B₂ has −1 under C₂ and σᵥ(xz), so those two are taken away.
n(B₂)
=
9 − (−1) − 1 + 3
4
n(B₂)
=
12
4
n(B₂) = 3
Step 5 · Find IR-active vibrations (nIR)
Use nIR = (n(A₁) − 1) + (n(B₁) − 2) + (n(B₂) − 2)
Translations use one A₁, one B₁ and one B₂, rotations one more B₁ and B₂; what is left vibrates, and A₁, B₁, B₂ are IR active.
nIR = (3 − 1) + (2 − 2) + (3 − 2)
nIR = 3 − 1 + 1
nIR = 3 → Vibrations: 2A₁ + B₂.
Check
Put the numbers back in. Both sides should match.
3
=
9 + (−1) + 1 + 3
4
✓
1
=
9 + (−1) − 1 − 3
4
✓
2
=
9 − (−1) + 1 − 3
4
✓
3
=
9 − (−1) − 1 + 3
4
✓
3 = (3 − 1) + (2 − 2) + (3 − 2) ✓
Topic check
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