In the previous post we arrived at two simultaneous equations (*We are using two different colors in order to distinguish between the two equations*)-

c_{i}α + c_{j}β = c_{i}E – Equation I

c_{i}β + c_{j}α = c_{j}E – Equation ii

Now we take like terms on one side of the equation –

c_{j}β = c_{i}E – c_{i}α

c_{i}β = c_{j}E – αc_{j}

Taking out the common term-

c_{j}β = c_{i }(E–α) c_{i}β = c_{j }(E–α)

Reversing the sign and getting all terms on one side, we get our **secular equations** –

c_{i }(α–E) + c_{j}β = 0 c_{i}β + c_{j }(α–E)=0.

The above two are now our working equations to form a matrix. We have two equations and so we will get a 2×2 matrix. The following is also known as **secular determinant**.

In the above matrix, the first square bracket holds the determinant (*Refer to post # 111 to know more about determinants)* and the second bracket holds the co-efficients –

*[ Note* –

*Doing a reverse operation i.e getting equations from the matrix , we just have to –*

*I) Multiply the first row first column element i.e ***(α-E)*** with the first row coefficient-***c _{i}**

*.*

*II)Multiply the first row second column element i.e*

**β**

*with the second row coefficient –*

**c**

_{j}*.*

*III)Add the two terms and equal them to zero to get –*

**c**

_{i }(α – E ) + c_{j}β = 0 .*Similarly get the second equation.]*

**Mathematically, if a non – zero determinant times some coefficients equals zero , then the determinant is zero.** So,

To simplify the above matrix, let us divide all terms by β –

Let us assume that (α-E / β) = x. Then the above matrix can be written as –

We already know how to calculate determinant of a 2×2 matrix (refer post #111). Here, ad-bc will be ,

x^{2} – 1^{2} = 0.

x^{2} – 1 = 0.

This is quadratic equation and will have two solutions, when we solve the equation –

x^{2} – 1 = 0.

x^{2}= 0.

x= ±1.

When x =+1 , (α-E / β) =1 . **E = α- β**

When x =-1 , (α-E / β) =-1 . **E = α+ β**

So, we have two energies **– **

E = α + β or E = α – β

Now that we can calculate energy with the above equations, we can construct an energy level diagram as follows –

β is a negative number and so α+β is lower energy than α-β.

The two π electrons will occupy the lower energy (α+β) bonding orbital.

Constructing a matrix for conjugated systems is relatively simple. The trick is as follows-

**The diagonal elements will always be α.**

2. **The near diagonal elements will be β . If its a ring/cyclic structure , then there will be two β’s at the top right corner and bottom left corner of the matrix as shown below- **

3. All the other elements in the matrix are zero (for non-neighbouring atoms).

## Thus, matrices in HMO theory , are tridiagonal matrices. There are two

β’s at top right and bottom left if the structure of the molecule is cyclic.

In the next post we will draw energy level diagrams for some more molecules and try to make better sense of the HMO theory w.r.t understanding aromaticity.

Till then ,

Be a perpetual student of life and keep learning….

Good day !

References and further reading –

1.https://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2007/lecture-notes/lecture31.pdf