Question

According to Molecular Orbital Theory, the bond order of O$_2$ is:

• A. 1
• B. 1.5
• C. 2
• D. 3

Hint:

A quick trick for finding bond orders of second-period diatomic species (total electrons $10$ to $18$):
$14$ electrons (like $\text{N}_2$) $\implies$ Bond Order = $3.0$
Each addition or subtraction of $1$ electron decreases the bond order by $0.5$.
For $\text{O}_2$ ($16$ electrons): $3.0 - (2 \times 0.5) = 2.0$.
For $\text{O}_2^+$ ($15$ electrons): $3.0 - 0.5 = 2.5$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question is from "Chemical Bonding and Molecular Structure," focusing on Molecular Orbital Theory (MOT).
We are required to determine the bond order of the diatomic oxygen molecule ($\text{O}_2$).

Step 2: Key Formula or Approach:
According to Molecular Orbital Theory, the bond order of a homonuclear diatomic molecule is:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \] where $N_b$ is the number of electrons in bonding molecular orbitals, and $N_a$ is the number of electrons in antibonding molecular orbitals.

Step 3: Detailed Explanation:

• Each oxygen atom contains $8$ electrons, so the diatomic oxygen molecule ($\text{O}_2$) contains a total of $16$ electrons.

• According to the electronic filling rules of MOT, the molecular orbital configuration for $\text{O}_2$ is written as:
\[ \sigma_{1s}^2\ \sigma_{1s}^{*2}\ \sigma_{2s}^2\ \sigma_{2s}^{*2}\ \sigma_{2p_z}^2\ (\pi_{2p_x}^2 = \pi_{2p_y}^2)\ (\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}) \]

• Let us count the bonding electrons ($N_b$), which are in orbitals without an asterisk ($*$):
- $\sigma_{1s}$: 2 electrons
- $\sigma_{2s}$: 2 electrons
- $\sigma_{2p_z}$: 2 electrons
- $\pi_{2p_x}$ and $\pi_{2p_y}$: 4 electrons
\[ N_b = 2 + 2 + 2 + 4 = 10 \]

• Let us count the antibonding electrons ($N_a$), which are in orbitals with an asterisk ($*$):
- $\sigma_{1s}^*$: 2 electrons
- $\sigma_{2s}^*$: 2 electrons
- $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$: 2 electrons
\[ N_a = 2 + 2 + 2 = 6 \]

• Now we substitute these values into the bond order formula:
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \]

• A bond order of 2 represents a double bond consisting of one $\sigma$ bond and one $\pi$ bond.

• Note that because of the presence of two unpaired electrons in the degenerate $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ antibonding orbitals, $\text{O}_2$ is paramagnetic.



Step 4: Final Answer:
The bond order of $\text{O}_2$ is $2$, which is option (C).