Can someone tell me why N2^2+ is a diamagnetic molecule based on MO theory and diagrams? I get a bond order of 2, which is paramagnetic.

N2 has pi2py pi2pz and sigma2p filled. The two electrons in sigma 2p are at a higher energy level than the pi bonds. No antibonding electrons are present in the 2p orbitals. The bond order is 3 since that is 6 electrons/2 = 3.

The two sigma 2p electrons occupy the highest energy. When the 2+ ion is made, those two electrons disappear leaving the 2+ ion. In terms of energy level, those two electrons are the outside two electrons and those are the ones to be ionized because it requires less energy to remove them and the others. Hund's rule states (for MO theory also) that electrons occupy the same energy level singly before pairing. It would take MORE energy to remove ONE of the sigma 2p electrons PLUS one from the lower pi2px or pi2pz bonding set. Removing both sigma 2p electrons leaves all the remaining electrons paired and that will be a diamagnetic molecule. We initially had 6 electrons, we now have 6-2 or 4 electrons and that is a bond order of 4/2 = 2.

To determine whether a molecule is diamagnetic or paramagnetic based on molecular orbital (MO) theory, we need to consider its molecular orbital diagram and calculate its bond order. In the case of N2^2+, the molecular orbital diagram can help us understand its magnetic properties.

Here's a step-by-step process to evaluate the diamagnetic or paramagnetic nature of N2^2+ using MO theory:

1. Start by drawing the molecular orbital diagram for N2^2+. The molecule has a total of 14 valence electrons, as there are two nitrogen atoms with five valence electrons each. We can arrange these electrons in the molecular orbital diagram by following the Aufbau principle and Hund's rule.

2. The molecular orbital diagram for N2^2+ consists of a combination of σ and π orbitals. The σ orbitals form from the overlap of the atomic orbitals along the internuclear axis, while the π orbitals form from the overlap of atomic orbitals perpendicular to the internuclear axis.

3. We need to fill the molecular orbitals in the diagram with the valence electrons of the molecule. Starting from the lowest energy level, fill the molecular orbitals one by one, pairing the electrons in accordance with Hund's rule. Remember that each orbital can accommodate a maximum of two electrons with opposite spins.

4. After filling the molecular orbital diagram, count the number of electrons in bonding (lower energy) orbitals and anti-bonding (higher energy) orbitals separately.

5. The bond order (BO) is then calculated as half the difference between the number of electrons in the bonding and anti-bonding orbitals. The formula for bond order is: BO = (Number of bonding electrons - Number of anti-bonding electrons)/2.

Now, let's apply this process to N2^2+:

1. The molecular orbital diagram of N2^2+:
σ2s(sigma) ↑↓↓
σ2s*(sigma) ↓
σ2px(sigma) ↑↓↓
σ2py(sigma) ↑↓↓
π2py(pi) ↓↓
π2pz(pi) ↑↓
σ2pz(sigma) ↑↓↓

2. Place the 14 valence electrons of N2^2+ in the molecular orbitals, following Hund's rule and the Aufbau principle:
σ2s(sigma) ↑↓↓
σ2s*(sigma) ↓
σ2px(sigma) ↑↓↓
σ2py(sigma) ↑↓↓
π2py(pi) ↑↓
π2pz(pi) ↑↓
σ2pz(sigma) ↑↓↓

3. Count the number of electrons in bonding (lower energy) and anti-bonding (higher energy) orbitals:
- Bonding electrons = 10 (3 from σ2s, 4 from σ2px, and 3 from σ2py)
- Anti-bonding electrons = 4 (2 from σ2s* and 2 from σ2pz)

4. Calculate the bond order using the formula:
BO = (Number of bonding electrons - Number of anti-bonding electrons) / 2
BO = (10 - 4) / 2
BO = 6/2
BO = 3

The calculated bond order for N2^2+ is 3.

Based on the bond order, we can conclude that N2^2+ is actually paramagnetic, not diamagnetic. A bond order of 3 indicates the presence of unpaired electrons, which results in paramagnetism.