# Reduction Formula

When the system gets far too complicated, i.e. more than 3 peripheral atoms, then it is hard to reduce the reducible representation to irreducible representations. To overcome this, a formula can be used to calculate the irreducible representations. The formula is as followed: (# of irreducible representations of a given type) = (1/order)*sum of [ (# operations in class) * (character of reducible representation) * (character of irreducible representation)]

where order = sum of operation of all classes

# Example: Ammonia

Below is the sigma framework of ammonia peripheral atoms.

Ammonia belongs to C3v point group.

C3v point group character table is given below.

The reducible representation are as followed:

Because there are three orbitals going in, there are three SALCs coming out.

Now the reduction formula can be applied to find the irreducible representations of the SALCs. nA1 = 1/6 [(1)(3)(1) + (2)(0)(1) + (3)(1)(1)] = 1 This means that there is one A1 SALC in the irreducible represenation The order of this point group is 6 because the coefficient of E, C3, SigmaV are 1, 2, and 3, respectively, and the sum of these coefficients is 6. 1. The first 1 in first term represents the coefficient of E in the character table.

2. The 3 in first term represents the character in the reducible representation

3. The second 1 in first term represents the character in the character table

4. The 2 in second term represents the coefficient of C3 in the character table.

5. The 0 in the second term represents the character in the reducible representation

6. The 1 in the second term represents the character in the irreducible representation

7. The 3 in the third term represents the coefficient of SigmaV in the character table

8. The first 1 in the third term represents the character in the reducible representation

9. The second 1 in the third term represents the character in the irreducible representation

The number of irreducible representation of A2 and E can be found with the same fashion. nA2 = 1/6 [(1)(3)(1) + (2)(0)(1) + (3)(1)(-1) ] = 0 nE = 1/6 [(1)(3)(2) + (2)(0)(-1) + (3)(1)(0) ] = 1

Although there is A1 and only one E, this E SALC actually composed of two SALCS. As the result, in total, we have three SALCs.

# Projection Operator

Now that we know that ammonia molecule is made up by one A1 and one E SALC sets. How do we know how they look like? Projection operation is a tool to help us visualize the SALCS. It maps the SALCs into its correct orientation according to its symmetry.

The formula for projection operator is as followed:

The SALCs in ammonia, for example, can be found by using this projection operator formula. There are a few rules to followed in order to use this formula correctly

1. Name the atoms alphabetically.

2. Decide whether the primary axis of rotation is counterclockwise or clockwise.

3. Expand all the symmetry operation

＝Example: Ammonia=

The axis of rotation in this case is along the z-axis counterclockwise. Hydrogen A is align with the x-axis. It is important to remember to put one peripheral atom along x or y axis in a three-fold symmetry for convenience.

Now, we first determine how A1 should look like. (You should be able to immediately visualize A1. However, here is a simple demonstration)

(Note that the symmetry operation is expanded to its unique operation.)

First row is focusing on hydrogen A atom. 1. When E operation is done on hydrogen A, hydrogen A is returned. Then A is multiply by the character from the irreducible representation, which is 1.

2. When C3 operation counterclockwise is done on hydrogen A, hydrogen B is returned. Then B is multiply by the character from the irreducible representation, which is 1.

3. When C32 operation counterclockwise is done on hydrogen A, hydrogen C is returned. Then C is multiply by the character from the irreducible representation, which is 1.

4. When vertical reflection plane operation along hydrogen A is done on hydrogen A, hydrogen A is returned. Then A is multiply by the character from the irreducible representation, which is 1.

5. When vertical reflection plane operation along hydrogen B is done on hydrogen A, hydrogen C is returned. Then C is multiply by the character from the irreducible representation, which is 1.

6. When vertical reflection plane operation along hydrogen C is done on hydrogen A, hydrogen B is returned. Then B is multiply by the character from the irreducible representation, which is 1.

Using similar operation focusing on hydrogen B and C, the outcome is the same. (Note: this is because A1 is singly degenerate)

The sum of these operations is 2A + 2B + 2C

Which can be further reduced to A + B + C As the result, the SALC looks exactly like:

Now, we will determine how the E SALCs would look like, which is not as trivial as the A1 SALC.

Below is the projection operation of the E SALCs

(Note that the coefficients are different because the coefficients from the character table for E is 2 -1 0)

We need to figure out two unique E SALCs from these three equations. This is why we need to put a peripheral atom along one of the axis. Since hydrogen A is along the x-axis, the first equation is a unique equation. However, the second unique equation should not contain hydrogen A. In order to get an equation without hydrogen A, the other two equations must subtract each other. As the result, one of the unique equations is 2A – B – C, while the other one being 3B -3C, which further reduces to B – C. Now we are ready to draw the E SALCs.

Below is the complete molecular orbital diagram of ammonia.

In ammonia, the 2S orbital of nitrogen not only forms 1a1 molecular orbital with the A1 SALC but also mixes with 2Pz to form the 2a1 molecular orbital. This orbital is hence known as sp hybridization. Because 2a1 molecular orbital is formed by mixing with the 2S orbital, the orbital is stabilized and hence has lower orbital energy.

The electrons in the 2a1 molecular orbital are the lone pair on ammonia. This sp hybridized orbital has an inversion center. This inversion center can readily flip the nitrogen as the figure shown below. As the result, there is never a chiral center in amine groups, even though four different constituents are attached to nitrogen.