Question

Using MOT, If bond order of $O_2$ = a Bond order of $N_2^+$ = b then find the value of (a + b)

Answer 1

4.5

Answer 2

To find the value of (a + b), we first need to calculate the bond order of $O_2$ (denoted as 'a') and the bond order of $N_2^+$ (denoted as 'b') using Molecular Orbital Theory (MOT).

1. Bond order of $O_2$ (a):
   The total number of electrons in $O_2$ is $8 + 8 = 16$.
   The molecular orbital configuration for molecules with more than 14 electrons (like $O_2$) is:
   $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1$
   Number of bonding electrons ($N_b$) = $2 + 2 + 2 + 2 + 2 = 10$
   Number of anti-bonding electrons ($N_a$) = $2 + 2 + 1 + 1 = 6$
   Bond order of $O_2$ (a) = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 6) = \frac{1}{2}(4) = 2$.
   So, a = 2.

2. Bond order of $N_2^+$ (b):
   The total number of electrons in $N_2$ is $7 + 7 = 14$.
   $N_2^+$ has lost one electron, so the total number of electrons is $14 - 1 = 13$.
   The molecular orbital configuration for molecules with 14 or fewer electrons (like $N_2^+$) has a different order of energy levels for the 2p orbitals compared to molecules with more than 14 electrons. The order is $\pi(2p_x) = \pi(2p_y) < \sigma(2p_z)$.
   The molecular orbital configuration for $N_2^+$ (13 electrons) is:
   $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^1$
   Number of bonding electrons ($N_b$) = $2 + 2 + 2 + 2 + 1 = 9$
   Number of anti-bonding electrons ($N_a$) = $2 + 2 = 4$
   Bond order of $N_2^+$ (b) = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(9 - 4) = \frac{1}{2}(5) = 2.5$.
   So, b = 2.5.

3. Calculate (a + b):
   a + b = 2 + 2.5 = 4.5.