In molecular orbital theory, what is the bond order of the oxygen molecule O2 in its ground state?

Introduction / Context:
Bond order is an important concept in chemical bonding that gives a simple measure of the number of chemical bonds between a pair of atoms. For diatomic molecules such as O2, bond order is related to bond strength, bond length, and stability. Understanding the bond order of oxygen is especially important in molecular orbital theory and helps explain why O2 has a double bond and is paramagnetic. This question focuses on the numerical bond order for O2 in its ground state.

Given Data / Assumptions:
- The molecule considered is O2, a homonuclear diatomic molecule. - We are interested in the ground state, not excited or ionised species. - Bond order is defined using the molecular orbital electron configuration. - We use the standard bond order formula based on bonding and antibonding electrons.

Concept / Approach:
In molecular orbital theory, electrons in a molecule occupy bonding and antibonding molecular orbitals. Bond order can be calculated using the formula: bond order = (number of electrons in bonding orbitals minus number of electrons in antibonding orbitals) divided by 2. For O2, we fill the molecular orbitals in order of increasing energy, considering the total number of valence electrons from both oxygen atoms. The resulting configuration leads to a bond order of 2, which corresponds to a double bond between the two oxygen atoms.

Step-by-Step Solution:
Step 1: Each oxygen atom has 6 valence electrons, so O2 has 12 valence electrons in total. Step 2: In the molecular orbital scheme for O2, these valence electrons fill the sigma 2s, sigma 2s star, sigma 2p z, pi 2p x and 2p y, and pi 2p star x and 2p star y orbitals in order. Step 3: Count bonding electrons. The bonding molecular orbitals (sigma 2s, sigma 2p z, pi 2p x, pi 2p y) contain a total of 8 electrons. Step 4: Count antibonding electrons. The antibonding molecular orbitals (sigma 2s star and pi 2p star x and 2p star y) together contain 4 electrons. Step 5: Apply the bond order formula: bond order = (8 minus 4) divided by 2 = 4 divided by 2 = 2. Step 6: Conclude that the O2 molecule has a bond order of 2, consistent with a double bond.

Verification / Alternative check:
You can verify this result by comparing with simple Lewis structure ideas. The Lewis structure of O2 is often written with a double bond between the two oxygen atoms, each also having two lone pairs. A double bond corresponds to a bond order of 2. Furthermore, experimental data such as bond length and bond energy indicate stronger bonding in O2 compared to a single bond and weaker than a hypothetical triple bond, which is consistent with bond order 2. Paramagnetism of O2 arises from unpaired electrons in antibonding orbitals but does not change the bond order calculation.

Why Other Options Are Wrong:
Option A (1.5) would correspond to a weaker bond, more typical of some resonance averaged structures such as in ozone, not O2 in its ground state. Option C (2.5) would imply more bonding than a double bond, closer to a bond between two atoms with partial triple bond character, which does not match experimental O2 data. Option D (1) suggests a single bond and would predict a much longer and weaker bond than observed for O2, so it is not correct.

Common Pitfalls:
A common mistake is to confuse O2 with other oxygen species such as O2 2 minus or O2 plus, which have different bond orders. Another pitfall is to rely only on Lewis structures without understanding that paramagnetism requires molecular orbital theory for complete explanation. Students may also incorrectly count electrons in bonding and antibonding orbitals. Careful counting of valence electrons and systematic filling of molecular orbitals helps avoid such errors.

Final Answer:
The correct answer is: 2.