Question

What is the bond order of \(O_2^{2-}\)?

• A. 1
• B. 2
• C. 2.5
• D. 1.5

Hint:

Work out how many extra electrons the 2- charge adds, and where they land in the molecular orbital diagram.

Answer 1

The Correct Option is A

Step 1: Count the total number of electrons in \(O_2^{2-}\). A neutral oxygen atom has 8 electrons, so neutral \(O_2\) has \(8 \times 2 = 16\) electrons. The peroxide ion \(O_2^{2-}\) carries 2 extra electrons, giving a total of \[16 + 2 = 18 \text{ electrons}\]
Step 2: Fill these 18 electrons into the molecular orbitals of oxygen in order of increasing energy: \[\sigma1s^2\ \sigma^{*}1s^2\ \sigma2s^2\ \sigma^{*}2s^2\ \sigma2p_z^2\ \pi2p_x^2\ \pi2p_y^2\ \pi^{*}2p_x^2\ \pi^{*}2p_y^2\]
Adding these up: \(2+2+2+2+2+2+2+2+2 = 18\) electrons, which checks out.
Step 3: Count bonding and antibonding electrons. Bonding electrons: \(\sigma1s(2) + \sigma2s(2) + \sigma2p_z(2) + \pi2p_x(2) + \pi2p_y(2) = 10\). Antibonding electrons: \(\sigma^{*}1s(2) + \sigma^{*}2s(2) + \pi^{*}2p_x(2) + \pi^{*}2p_y(2) = 8\).
Step 4: Bond order is given by \[\text{Bond order} = \dfrac{N_b - N_a}{2} = \dfrac{10 - 8}{2} = \dfrac{2}{2} = 1\]
\[\boxed{\text{Bond order} = 1}\]