Q.Which of the following options represents the correct bond order :

The key idea is to determine the bond order for each species using Molecular Orbital (MO) theory. Bond order is calculated as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$).

1. Calculate total electrons:

   • $O_2$: $8 \times 2 = 16$ electrons
   • $O_2^+$: $16 - 1 = 15$ electrons
   • $O_2^-$: $16 + 1 = 17$ electrons

2. Write MO configurations and calculate bond order:

   The general MO configuration for diatomic molecules like oxygen (with $>14$ electrons) is:

   $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1 ...$

   • For $O_2$ (16 electrons):

     $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1$

     $N_b = 2+2+2+2+2 = 10$

     $N_a = 2+2+1+1 = 6$

     Bond Order ($O_2$) $= \frac{1}{2}(10 - 6) = 2.0$

   • For $O_2^+$ (15 electrons): (Remove one electron from $\pi^*$ orbital of $O_2$)

     $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1$

     $N_b = 10$

     $N_a = 2+2+1 = 5$ …

In chemistry, the bond order is a fundamental concept that describes the number of chemical bonds between a pair of atoms. It is a direct measure of bond strength and stability: a higher bond order implies a stronger, shorter, and more stable bond. For diatomic molecules and ions, we determine bond order using Molecular Orbital (MO) theory, which describes how atomic orbitals combine to form molecular orbitals.

Let's apply this principle to $O_2^-$, $O_2$, and $O_2^+$.

1. Determine the total number of electrons for each species.

   An oxygen atom ($O$) has 8 electrons.

   • For $O_2$: $2 \times 8 = 16$ electrons.
   • For $O_2^+$: $16 - 1 = 15$ electrons (one electron removed).
   • For $O_2^-$: $16 + 1 = 17$ electrons (one electron added).

2. Recall the Molecular Orbital (MO) filling order.

   For diatomic molecules with more than 14 electrons (like $O_2$), the energy order of molecular orbitals is:

   $\sigma 1s, \sigma^* 1s, \sigma 2s, \sigma^* 2s, \sigma 2p_z, (\pi 2p_x = \pi 2p_y), (\pi^* 2p_x = \pi^* 2p_y), \sigma^* 2p_z$.

   We can simplify the calculation by focusing only on valence electrons, as the core $1s$ orbitals form filled bonding ($\sigma 1s$) and antibonding ($\sigma^* 1s$) orbitals, which cancel out in the bond order calculation ($N_b=2, N_a=2 \implies BO=0$).

   For oxygen, the valence electrons are in the $2s$ and $2p$ orbitals. The valence MO filling order is:

   $\sigma 2s, \sigma^* 2s, \sigma 2p_z, (\pi 2p_x = \pi 2p_y), (\pi^* 2p_x = \pi^* 2p_y), \sigma^* 2p_z$.

3. Write the MO configuration and calculate the bond order for each species.

   • For $O_2$ (16 electrons):

     There are $2 \times 6 = 12$ valence electrons.

     The MO configuration is:

     $(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$

     Number of bonding electrons ($N_b$) = $2 (\sigma 2s) + 2 (\sigma 2p_z) + 2 (\pi 2p_x) + 2 (\pi 2p_y) = 8$.

     Number of antibonding electrons ($N_a$) = $2 (\sigma^* 2s) + 1 (\pi^* 2p_x) + 1 (\pi^* 2p_y) = 4$.

     Bond Order ($BO$) = $\frac{8 - 4}{2} = \frac{4}{2} = 2$.

   • For $O_2^+$ (15 electrons):

     This ion is formed by removing one electron from $O_2$. The electron is removed from the highest occupied molecular orbital (HOMO), which is one of the $\pi^* 2p$ antibonding orbitals.

     The MO configuration is:

     $(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^0$

     Number of bonding electrons ($N_b$) = $8$. …