Question

Question was provided via image upload.

Explanation:

To determine the correct electron - dot structure for diatomic nitrogen ($N_2$) with a triple bond, we start by recalling the electron configuration of nitrogen. A nitrogen atom has 5 valence electrons. In the $N_2$ molecule, the two nitrogen atoms share electrons to achieve a stable octet configuration.

For a triple bond to exist between the two N atoms, there should be three pairs of shared electrons (a triple bond) and one pair of non - shared (lone) electrons on each N atom. Let's analyze the options:

• Option A: The structure shows three pairs of shared electrons (which corresponds to a triple bond) between the two N atoms, and each N atom has one lone pair of electrons (2 electrons) and one unpaired electron? Wait, no, let's count the valence electrons. Each N in option A: around the first N, we have 2 (from the triple bond) + 3 (lone electrons)? Wait, no, let's do a proper count. The correct Lewis structure for $N_2$ is :N≡N:, which in electron - dot (Lewis) structure terms has three pairs of shared electrons (the triple bond) and one lone pair (2 electrons) on each N atom. Looking at the options, the last option (let's assume it's D, but from the given image, the last one is :N::N: with the correct triple bond and lone pairs) - Wait, the correct structure for $N_2$ has a triple bond (three pairs of shared electrons) and one lone pair on each N. So the structure with three pairs of shared electrons between the N atoms and one lone pair on each N is the correct one. Among the given options, the last one (the fourth one, let's say the one with :N::N: with the triple bond and lone pairs) is correct. Wait, from the image, the last option is :N::N: (with the triple bond as three pairs of dots between N and N, and one lone pair on each N). So the correct option is the last one (assuming the options are A, B, C, D and the last is D). But from the user's image, the last structure is :N::N: (with three pairs of shared electrons and one lone pair on each N), which is the correct Lewis structure for $N_2$ with a triple bond.

Answer:

The correct option is the last one (the fourth structure in the given image, with the triple bond and appropriate lone pairs on each N atom, i.e., the structure :N::N:). If we assume the options are labeled and the last is, for example, D (but from the user's image, the last structure is the correct one for $N_2$'s Lewis structure with a triple bond). So the answer is the last electron - dot structure (the one at the bottom of the given image) which represents diatomic nitrogen with a triple bond.