In molecular orbital theory, what is the bond order of the molecular ion N2+, obtained by removing one electron from the nitrogen molecule N2?

Introduction / Context:
Bond order is an important concept in molecular orbital (MO) theory that helps predict the relative stability, bond strength, and bond length of diatomic molecules and ions. It is defined in terms of the number of electrons occupying bonding and antibonding molecular orbitals. Nitrogen molecule N2 is a classic example with a strong triple bond. The molecular ion N2+ is formed by removing one electron from N2, and its bond order changes accordingly. This question asks you to calculate the bond order of N2+ using the basic formula from MO theory.

Given Data / Assumptions:

• N2 is a homonuclear diatomic molecule of a second period element.
• N2 has a well known bond order of 3 in molecular orbital theory.
• N2+ is formed by removing one electron from N2.
• Bond order in MO theory is given by (number of bonding electrons minus number of antibonding electrons) divided by 2.
• We assume the standard ordering of molecular orbitals for molecules from boron to nitrogen.

Concept / Approach:
For diatomic molecules, the bond order can be computed using the formula: bond order = (N_b − N_a) / 2, where N_b is the number of electrons in bonding molecular orbitals and N_a is the number in antibonding molecular orbitals. For N2, the resulting bond order is 3, corresponding to a very strong triple bond. When one electron is removed to form N2+, that electron is removed from the highest occupied molecular orbital. For N2, the highest occupied molecular orbitals are bonding orbitals. Removing one bonding electron reduces the net bonding by half a bond order unit, so the bond order in N2+ becomes 2.5.

Step-by-Step Solution:
Step 1: Recall the MO electron configuration for N2, which has 14 electrons. Using the ordering for molecules from boron to nitrogen, the valence MO configuration leads to a bond order of 3. Step 2: Apply the bond order formula for N2. The number of bonding electrons N_b is 10 and the number of antibonding electrons N_a is 4, so bond order = (10 − 4) / 2 = 3. Step 3: To obtain N2+, remove one electron from the highest occupied bonding molecular orbital, which decreases N_b by 1 while N_a remains the same. Step 4: For N2+, N_b becomes 9 and N_a stays 4, giving bond order = (9 − 4) / 2 = 5 / 2 = 2.5. Step 5: Recognise that this value is between 2 and 3, implying that the N2+ bond is slightly weaker and longer than the N2 bond but still quite strong.

Verification / Alternative check:
Another way to think about this is to remember that each pair of bonding electrons minus antibonding electrons contributes one bond. Removing a bonding electron reduces the total bonding by half a bond. Since N2 has bond order 3, taking one electron from a bonding orbital reduces the bond order to 2.5. Spectroscopic and experimental data show that N2+ indeed has a bond length slightly longer than N2 and a lower bond energy, which is consistent with a reduction in bond order from 3 to 2.5. This cross check supports the calculation.

Why Other Options Are Wrong:
A bond order of 3 would be correct for N2, not for N2+, because the removal of an electron must change the bond order. A bond order of 2 would require the removal of two bonding electrons or the addition of antibonding electrons to reduce the net bonding further, which is not the case here. A bond order of 1.5 would correspond to a much weaker bond and would require even more change in the electron configuration. A bond order of 0.5 would indicate a very weak or unstable bond; N2+ is more stable than that. Therefore, only 2.5 correctly represents the slight reduction from the strong triple bond of N2.

Common Pitfalls:
A common mistake is to assume that the bond order of N2+ remains 3 because students remember only the value for N2 and forget to recalculate after ionisation. Another error is to guess that taking away one electron reduces the bond order from 3 directly to 2, without using the bond order formula. To avoid these pitfalls, always apply the formula bond order = (N_b − N_a) / 2 and pay attention to whether the removed electron comes from a bonding or antibonding orbital. Removing a bonding electron reduces bond order by 0.5, while removing an antibonding electron would increase bond order by 0.5.

Final Answer:
The bond order of the molecular ion N2+ is 2.5.