Energy level diagram for Molecular orbitals

Energy level diagram for Molecular orbitals

The first ten molecular orbitals may be arranged in order of energy as follow:

σ(1s) <σ^∗(1s) < σ(2s) <σ^∗(2s) < π(2p_x) = π(2p_y) < σ(2p_z) < π^∗(2p_x) =π^∗(2p_y) <π^∗( 2p_z)

Relationship between electronic configuration and Molecular behaviour

1) Stability of molecules in terms of bonding and antibonding electrons

Number of electrons present in the bonding orbitals is represented by N_b and the number of electrons present in antibonding orbitals by Na.

1) If N_b > Na ,the molecule is stable because greater number of bonding orbitals are occupied than antibonding orbital, resulting in a net force of attraction.

2) If N_b < Na , the molecule is unstable because the antibonding influence is greater than the bonding influence, resulting in net force of repulsion.

3) If N_b = Na ,the molecule is again unstable because influence of electrons in the antibonding molecular orbital is greater than the bond influence of electron in the bonding molecular orbitals.

 

2) Stability of molecules in terms of bond order

Bond order is defined as half of the difference between the number of electrons present in the bonding and antibonding orbitals.

Bond Order = ½ ( N_b – Na)

The molecule is stable if N_b > Na ie. bond order is  positive. The molecule is unstable if  N_b < Na i.e. the bond order is negative or zero.

3) Relative stability of molecule in terms of bond order

For diatomic molecules ,the stability is directly proportional to the bond order.

A molecule with the bond order of 3 is more stable than a molecule with bond order of 2 and so on.

4) Nature of bond in terms of bond order :

Bond order 1 ,2 and 3 mean single ,double and triple bond.

5) Bond length in terms of bond order:

Bond length is found to be inversely proportional to the bond order. Greater the bond order, shorter is the bond length.

6) Diamagnetic and paramagnetic nature of the molecules

If all the electrons in the molecule are paired, it is diamagnetic in nature.

If the molecules has some unpaired electrons ,it is paramagnetic in nature.

Greater the number of unpaired electrons present in the molecular or ion, greater is its paramagnetic nature.

Electronic configuration of Homonuclear Diatomic Molecules

 

1) H₂⁺

The electronic configuration of  H₂⁺ is ( σ(1s) ) ¹

1) N_b =1 , Na = 0

Bond Order = 1

2) Positive bond order means it is stable.

3) One unpaired electron is present. Hence it is paramagnetic.

2) H₂ 

The electronic configuration of  H₂ is ( σ(1s) )²

N_b = 2 , Na =0

Bond order = 1

Positive value of bond order indicates that H₂ molecule is stable.

Bond order value of 1 means that two hydrogen atoms are connected by a single bond.

Greater value of bond order for H₂ molecule than H₂⁺ ion shows that two H₂ molecule is more stable than H₂⁺.

Bond length of H₂ is smaller than that of H₂⁺ ion.

As no unpaired electron is present, the H₂ molecule should be diamagnetic.

3) H₂^–

The electronic configuration of H₂^– is ( σ(1s) )²  (σ^∗(1s)) ¹

1) Bond order= ½

2) Smaller positive value of bond order indicates that it is somewhat stable.

3) Since it has one unpaired ,it is paramagnetic.

4) He₂

 

Electronic configuration of helium is (σ(1s))²  (σ^∗(1s))²

N_b= 2,  Na =2

Bond Order =0

No bond is formed between two helium therefore He₂ does not exist.

5) He₂⁺

The electronic configuration of He₂⁺ is (σ(1s))²  (σ^∗(1s))¹

N_b= 2, Na =1

Bond Order= ½

It is paramagnetic in nature.

6) Li₂

The electronic configuration of Li₂ is (σ(1s))²  (σ^∗(1s))²  (σ(2s))²

As inner closed closed shells do not participate in bonding therefore the configuration is written as:

KK  (σ(2s))²

As the K closed shells do not take part in bonding, they are called non-bonding orbitals.

1)N_b=2, Na= 4

Bond Order= 1

2)Li₂ is a stable molecule.

3)It has no unpaired electron ,it should be diamagnetic.

7) Be₂

The electronic configuration of Be₂ is KK  (σ(2s))²  (σ^∗(2s))²

N_b= 2, Na =2

Bond order= 0

Be₂ does not exist

8) B₂

The electronic configuration of B₂ is KK (σ(2s))²  (σ^∗(2s))²  π(2p_x)¹  π(2p_y)¹

N_b= 4 Na =2

Bond order = 1

The molecule is paramagnetic

9) C₂

 

The electronic configuration of C₂ is KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²

N_b= 6, Na =2

Bond order= 2 

The molecule is diamagnetic

The double bond in C₂ consist of both Pi bonds because the four electrons are present in the two pi molecular orbitals.

10) N₂

The atomic number of nitrogen is 7

The electronic configuration of N2 is KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²

N_b= 8, Na= 2

Bond Order= 3

Bond order value of 3 means that N₂ contains a triple bond.

High value of bond order implies that it should have highest bond dissociation energy.

Presence of no unpaired electron indicates it to be diamagnetic.

11) N₂⁺ ion

The electronic configuration is KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z)¹

N_b= 7, Na =2

Bond order = 2.5

It has one unpaired electron , therefore it is paramagnetic in nature.

12) N₂⁻

The electronic configuration is KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))¹

N_b= 8 , Na =3

Bond Order= 2.5

It has one unpaired electron , therefore it is paramagnetic in nature.

13) N₂^2-

The electronic configuration is KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))¹  (π^∗(2p_y))¹

N_b= 8, Na =4

Bond order = 2

It has two unpaired electron , therefore it is paramagnetic in nature.

 

As bond dissociation energies are directly proportional to the bond order, therefore, the dissociation energies of these molecular species are in the order:

N₂ > N₂⁺ = N₂⁻ > N₂^2-

As bond length is inversely proportional to bond order, therefore, bond length will be in the order:

N₂^2- > N₂^– = N₂⁺ > N₂

14) O₂

The electronic configuration of O₂ is  KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))¹  (π^∗(2p_y))¹

N_b=8 , Na=4

Bond order= 2

The bond order of 2 justifies the presence of double bond in O₂ molecule.

The presence of 2 unpaired electrons makes the molecule paramagnetic.

15) O₂⁺

 

The electronic configuration of O₂⁺ is  KK (σ(2s))²  (σ^∗(2s))²  (σ(2p_z))² (π(2p_x))²  (π(2p_y))²   (π^∗(2p_y))¹

N_b= 8, Na= 3

Bond order=2.5

The presence of unpaired electron makes the molecule paramagnetic.

16) O₂^– ion

The electronic configuration of O₂^– is  KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))²  (π^∗(2p_y))¹

N_b= 8, Na= 5

Bond order=1.5

The presence of unpaired electron makes the molecule paramagnetic.

17) O₂^2- ion

The electronic configuration of O₂^– is  KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))²  (π^∗(2p_y))²

N_b= 8, Na= 6

Bond order=1

The presence of no unpaired electron makes the molecule diamagnetic.

As bond dissociation energies are directly proportional to the bond order, therefore, the dissociation energies of these molecular species are in the order:

O₂⁺ > O₂> O₂^– > O₂ ^2-

As bond length is inversely proportional to bond order, therefore, bond length will be in the order:

O₂ ^2- > O₂^– > O₂ > O₂ ⁺

18) Fluorine molecule (F₂)

The electronic configuration of O₂ is  KK (σ(2s))²  (σ^∗(2s))^{2  (}π(2p_x))²  (π(2p_y))²  (σ(2p_z))²  (π^∗(2p_x))² (π^∗(2p_y))²

N_b= 8, Na= 6

Bond order=1

The presence of no unpaired electron makes the molecule diamagnetic.