chemistry-class-11-unit-04-chapter-06-chemical-bonding-molecular-structure-l-6-7-gs3vpkalvpe

Because of the number of interactions and the electron electron repulsion present in the molecule , the energy level cannot be calculated
Then , we must go for an approximation method
Linear combination of atomic orbital - LCAO is one of the approximation method
$H_2^{+}$ ( $H_A$ and $H_B$ ) - there is only one electron
- The atomic orbitals of $H_A$ can be described by wave function $\phi_A$
- The atomic orbitals of $H_B$ can be described by wave function $\phi_B$
- If the electron is present near the $H_A$ then the molecule can be described by $\phi_A$
- If the electron is present near the $H_B$ then the molecule can be described by $\phi_B$
- If the electron is present in between the $H_A$ and $H_B$ then the molecule can be described by $\phi_A \pm \phi_B$

Bonding
- $\psi_B$ = $\phi_A+\phi_B$
- $\phi_A$ $and$ $\phi_B$ reinforce each other
- Lower in energy
Antibonding
- $\psi_a$ = $\phi_A-\phi_B$
- $\phi_A$ $and$ $\phi_B$ cancel each other
- Higher in energy

$\Delta$ E - bond energy of H-H bond
The bonding molecular orbital for $H_2$ is $\sigma$ while the antibonding is $\sigma^*$
The same way - $\sigma \hspace{1mm} , \sigma^* \hspace{1mm}, \pi \hspace{1mm} , \pi^* \hspace{1mm} , \delta$
$\phi_A+\phi_B$ Bonding
Cylindrically symmetrical

The combination of $P_x$ and $P_x$ orbital will form $\pi$ bond - $P_x+P_x$ = $\pi$ bond , or
The combination of $P_y$ and $P_y$ orbital will form $\pi$ bond - $P_y+P_y$ = $\pi$ bond

Number of electrons in bonding orbital is $N_b$
Number of electrons in anti - bonding orbital is $N_b$
Bond order(b.o) = $\frac{N_b-N_a}{2}$
Bond order should be +ve for stable molecule
if , bond order = 0 (unstable molecule)
- bond order = 1 single bond
- bond order = 2 double bond
- bond order = 3 triple bond

$Li_2$
- $1s^2 2s^1$
- $(\sigma_{1s})^2$ $(\sigma^*_{1s})^2$
- B.O = $\frac{2-0}{2}$ = 1
- Li-Li (stable molecule )