Question

Which of the following options represents the correct bond order ?

• A. ${ O-_2 > O_2 < O_2^+}$
• B. ${ O-_2 < O_2 > O_2^+}$
• C. ${ O-_2 > O_2 > O_2^+}$
• D. ${ O-_2 < O_2 < O_2^+}$

Answer 1

The Correct Option is D

To determine the bond order of the given species, we need to understand the concept of molecular orbital theory. Bond order indicates the stability of a molecule and is calculated using the formula:

\text{Bond Order} = \frac{1}{2} ( \text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals} )

The electronic configurations for the following oxygen species are:

• O_2\;(16\:\text{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) )
• O_2^+\;(15\:\text{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 )
• O_2^-\;(17\:\text{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}^2 = \pi^*_{2p_y}^1 )

Now, calculating the bond order for each:

1. O_2: \frac{1}{2} \times (10 - 6) = 2
2. O_2^+: \frac{1}{2} \times (10 - 5) = 2.5
3. O_2^-\: \frac{1}{2} \times (10 - 7) = 1.5

From these calculations, we can deduce:

• O_2^+ has the highest bond order of 2.5.
• O_2 has a bond order of 2.
• O_2^-\ has the lowest bond order of 1.5.

Therefore, the correct order of bond order is:

O_2^- < O_2 < O_2^+

This justifies why the correct answer is: O_2^- < O_2 < O_2^+