To determine the correct order of energies of molecular orbitals for the $N_2$ molecule, we need to understand the molecular orbital theory for diatomic molecules.$\\$For $N_2$ (a homonuclear diatomic molecule with 14 electrons), the molecular orbital configuration is influenced by the energy levels of the $2p$ orbitals.$\\$In molecules like $N_2$, the energy order of molecular orbitals is typically:$\\ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < \left( \pi^* 2p_x = \pi^* 2p_y \right) < \sigma^* 2p_z \\$This order is due to the fact that in $N_2$ and other diatomic molecules with atomic numbers less than 8, the $\pi$ orbitals are lower in energy than the $\sigma 2p_z$ orbital.$\\$Let's analyze the given options:$\\$Option 1: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < \left( \pi^* 2p_x = \pi^* 2p_y \right) < \sigma^* 2p_z \\$Option 2: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \left( \pi 2p_x = \pi 2p_y \right) < \left( \pi^* 2p_x = \pi^* 2p_y \right) < \sigma^* 2p_z \\$Option 3: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \sigma^* 2p_z < \left( \pi 2p_x = \pi 2p_y \right) < \left( \pi^* 2p_x = \pi^* 2p_y \right) \\$Option 4: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \left( \pi^* 2p_x = \pi^* 2p_y \right) < \sigma 2p_z < \sigma^* 2p_z \\$The correct order for $N_2$ is given in Option 1, where the $\pi 2p_x$ and $\pi 2p_y$ orbitals are lower in energy than the $\sigma 2p_z$ orbital.$\\$Therefore, the correct answer is Option 1.