Applications of Molecular Orbit Theory

Electronic configuration and Molecular Behavior:

• The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule.

Important information of configuration:

1. Stability of Molecules:

1. The molecule is stable if  NB is greater than NA
2. The molecule is unstable if NB is less than NA
   Where, NB → Bonding e- s NA → Anti-Bonding e- s

2. Bond order:

• Bond order is defined as one half the difference between the number of electrons present in the bonding and the anti-bonding orbitals.Bond order = ½ [NB - NA]

• The rules discussed in the above regarding the stability of the molecule can be restated in terms of Bond order as follows.
• A positive Bond order means a stable molecule while negative (or) zero bond order means an unstable molecule.
• Nature of Bond: Integral bond order values of 1, 2 (or) 3 correspond to single, double (or) triple bonds respectively.

3. Bond Length:

• The bond order between two atoms in a molecule may be taken as an approximate measure of the Bond length.
• Bond length decreases bond order increases.

4. Magnetic Nature:

• If all the molecular orbitals in a molecule are doubly occupied the substance is diagrammatic.
• If one (or) more molecular orbitals are singly occupied it is paramagnetic.
  EX: O2 Molecule

Bonding in some Homonuclear diatomic molecules

1. Hydrogen Molecule (H2):

• There are 2 electrons in hydrogen molecule which are present in σ1S molecular orbital so electronic configuration of hydrogen molecule is H2 : (σ1S)2
  The bond order of H2 molecule can be calculated as given below
  Bond order = [NB - NA] / 2
• Since there is no unpaired electron in hydrogen molecule, therefore it is diamagnetic.
  

2. Helium molecule (He2):

• In (He2) molecule there are 4 electrons. These electrons will be accommodated in σ1S and  σ*1S molecular orbital leading to electronic configuration.
  He2 : (σ1S)2 (σ*1S)2
• Bond order He2  is ½ (2-2) = 0
• He2  Molecule is therefore unstable and does not exist.
  

3. Lithium Molecule (Li2):

• The electronic configuration of lithium is 1S2,2S1. There are six electrons in Li2 the electron configuration of Li2 molecule therefore is Li2 : (σ1S)2 (σ*1S)2 (σ2S)2
• The above configuration is also written as KK (σ2S)2 where KK represents the closed K shell structure (σ1S)2 (σ*1S)2  
• There are four electrons present in bonding molecular orbitals and two electrons in anti-bonding molecular orbitals.
  Bond order ½ [4 - 1] = 1
• It has no unpaired electrons so it is diamagnetic.
  

4. Oxygen molecule (O2):

• The electronic configuration of oxygen atom is 1S2 2S2 2P4 each oxygen atom has 8 electrons hence, in O2 molecule there are 16 electrons. The electronic configuration of O2 molecule therefore
  O2 : (σ1S) (σ*1S)2 (σ2S)2 (σ*2S)2 (σ2PZ)2 (π2P2x= π2P2y) (π*2P1x = π*2P1y)
• From the electronic configuration of O2 molecule, t is clear that ten electrons are present in anti-bonding molecular orbitals and six electrons are present in anti-bonding molecular orbitals.
  Its bond order = ½ [NB - NA] = ½ [10 - 6] = 4
• It may be noted that it contains two unpaired electrons π*2Px and π*2Py molecular orbitals.
• O2 Molecule should be paramagnetic.
  

5. Nitrogen molecule (N2):

• The electronic configuration of Nitrogen is 1S2 2S2 2P3, there are 14 electrons in N2. The electronic configuration of
  N2 : σ1S2 < σ*1S2 < σ2S2 < σ*2S2 < (π2P2x=π2P2y) < σ2P2z
• Bond order of N2 = ½ [10 -4] = 3
• There is no unpaired electron in the electronic configuration of N2
• It is diamagnetic in nature

Law of conservation of energy

Energy is either created or destroyed but converts from one form to another.

At point A:

A body of mass ‘m’ is thrown up with a velocity ‘u’ from the ground level. Let us take the point as A. therefore it has only kinetic energy and potential energy is equal to ‘0’ at A.

Energy at A

K.E = ½ mu2

P.E = 0.

Total Energy = ½ mu2

At point B:

Now let the velocity of a body after reaching a height ‘x’ from the ground be ‘u’. Let us take that point as B.

So it is

K.E at x is ½ m (u’)2

P.E is mgx

Total Energy = ½ m (u’)2  + mgx.

At point c:

Now let us assume the body has reached the maximum height of ‘Hmax’. Let this point be ‘C’.

Here P.E is mgHmax

K.E is 0

At C i.e., at maximum Height

K.E = 0

P.E = mgHmax

According to law of conservation of energy at A, B & C must be equal. As, the gravitational force is a conservative force.

TA = TB = TC

½ mu2 = ½ mu2 + mgx = mgHmax

INVERSE MATRICES

Evaluation adjoint and inverse Matrix using determinants and elementary transformations

MINORS AND COFACTORS

MINOR: Let A = [aij] be a square matrix of order n. Then the minor Mij of aij in A is the determinant of the square sub-matrix of order (n-1) obtained by leaving ith row and jth column of A

1. If , we have

M11 = Minor of A11 = 2, M12 = Minor of A12 = —3,

M21 = Minor of A21 = —7, M22 = Minor of A22 = 4

COFACTORS: Let A = [aij] be a square matrix of order n. Then the cofactor Cij of aij in A is equal to (—1) i+j. times the determinant of the sub-matrix of order (n-1) obtained by leaving ith row and jth column of A.

If follows from this definition that

Cij =Cofactor of aij in A

= (-1) i+j Mij, where Mij is minor of aij in A

Thus, Cij = Mij if i+j is even

And, Cij = - Mij if i+j is odd.

2. If , then we have

C11 = (-1)1+1 M11= M11 = = 2,

C12 = (-) 1+2M12 = - M12 = = 7

C13 = (-1)1+3 M13 = M13 = = 8,

C23 = (-1)2+3 M23 = - M23 = = 8 etc.

ADJOINT OF A SQUARE MATRIX

DEFINITION

Let A = [aij] be a square matrix of order n and let Cij be cofactor of aij in A. Then the transpose of the matrix of cofactors of elements of A is called the adjoint of A and is denoted by adj A.

Thus, adj A = [Cij] T.

(adj A) ij = Cij = Cofactor of aji in A.

Find the adjoint of matrix.

SOLUTION

We have,

Cofactor of A11 = s,

Cofactor of A12 = —r,

Cofactor of A21 = -q

And, Cofactor of A22 = p.

It is evident from this example that the adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal elements and changing signs of off-diagonal elements.

If , then by the above rule, we have,

Find the adjoint of matrix

SOLUTION

Let Cij be cofactor of aji in A. Then cofactors of elements of A are given by

PROPERTIES OF ADJOINT The following are some properties of adjoint of a square matrix which are states as theorems.

Let A be a square matrix of order n. Then

A (adj A) = |A|In = (adj A) A

Let A be a non-singular square matrix of order n. Then

|adj A| = |A|n-1

If A and B are non-singular square matrices of the same order, then

adj AB = (adj B) (adj A)

If A is an invertible square matrix, then

adj AT = (adj A) T.

If A is a non-singular square matrix, then

adj ( adj A) = |A|n-2 A

If A is a symmetric matrix, then adj A is also a symmetric matrix.

INVERSE OF A MATRIX

DEFINITION A square matrix of order n is invertible if there exists a square matrix B of the same order such that

AB = I„ = BA

In such a case, we say that the inverse of A is B and we write,

A-1= B

PROPERTIES OF INVERSE

The inverse of a matrix has the following properties

Every invertible matrix possesses a unique inverse.

A square matrix is invertible iff it is non-singular.

The inverse of A is given by

1. AB=AC => B=C
2. BA=CA => B=C

Clearly, AB = BC but B ≠ C.

(Reversal law) if A and B are invertible matrices of the same order, then AB is invertible and

(AB)-1 = B-1 A-1

If A is an invertible square matrix, then

AT is also invertible and (AT)-1 = (A-1) T.

If the non-singular matrix A is symmetric, then A-1 is also symmetric.

If A is a non-singular matrix, then