Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

The nitrogen atom has 7 electrons. Their electron configuration in the ground state can be expressed as follows: \(1s^2 2s^2 2p^3\). In a nitrogen molecule (N₂), there are a total of 14 electrons. The electron configuration of N₂ in the ground state can be written as: \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^4\)

To determine the first excited state of N₂, we need to move the highest-energy electron to the next available energy level. The highest-energy electron resides in the 1\(\pi_{u}\) molecular orbital. We'll move one electron from the 1\(\pi_{u}\) MO to the next available energy level, which is the 1\(\pi_{g}\) MO. This gives us the following electron configuration: \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\)

A molecule is diamagnetic if all its electrons are paired, and it's paramagnetic if there's at least one unpaired electron. In the first excited state of N₂, the electron configuration shows that there are unpaired electrons in the 1\(\pi_{u}\) and 1\(\pi_{g}\) MOs (\(1\pi_u^3 1\pi_g^1\)). Therefore, the first excited state of N₂ is paramagnetic.

To determine the bond strength, we look at the bond order. The bond order can be calculated using the formula: \(Bond\:Order=\frac{1}{2}(number\:of\:electrons\:in\:bonding\:orbitals - number\:of\:electrons\:in\:antibonding\:orbitals)\) For the ground state electron configuration (\(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^4\)): Bond Order = \(\frac{1}{2}(10 - 4) = 3\) For the first excited state electron configuration (\(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\)): Bond Order = \(\frac{1}{2}(10 - 5) = 2.5\) Since the bond order for the ground state is higher than the first excited state (3 vs. 2.5), the bond strength in the ground state is stronger compared to that in the first excited state.