Question

1. which of the electron dot structures correctly illustrates diatomic nitrogen correctly?

Explanation:

To determine the correct Lewis dot structure for diatomic nitrogen ($\ce{N2}$), we start by recalling the electron configuration of nitrogen. A nitrogen atom has 5 valence electrons. In $\ce{N2}$, the two nitrogen atoms form a triple bond to satisfy the octet rule (each nitrogen needs 8 valence electrons).

• A triple bond consists of 3 pairs of shared electrons (6 electrons total), and each nitrogen atom also has one lone pair (2 electrons) to complete its octet (5 valence electrons + 3 shared from the triple bond = 8).

Looking at the options:

• The first option shows a triple bond (three pairs of electrons between the N atoms) and each N has one lone pair (2 electrons) and also some non - bonding electrons around? Wait, no, let's count the valence electrons. Each N should have 5 valence electrons. In the first structure, between the two N atoms, there are three pairs (triple bond), and each N has two non - bonding electrons (one lone pair) and also some other electrons? Wait, no, let's do the electron count. The total number of valence electrons in $\ce{N2}$ is $2\times5 = 10$.

In a Lewis structure, the number of valence electrons is the sum of lone pair electrons and shared electrons. For a triple bond (6 shared electrons) and two lone pairs (2 electrons on each N, so 4 lone pair electrons), total is $6 + 4=10$, which matches.

Looking at the first option: The two N atoms have a triple bond (three pairs of electrons between them) and each N has one lone pair (2 electrons) and also some other electrons? Wait, no, the first structure (the top one) has the triple bond (three pairs between N and N) and each N has two non - bonding electrons (one lone pair) and also some extra? Wait, no, let's visualize. The first option: N with a lone pair (2 electrons), triple bond (3 pairs) to the other N, which also has a lone pair (2 electrons). Wait, no, the dots: let's count the valence electrons for each N. In the first structure, each N has 5 valence electrons? Wait, the first structure: around the first N, there are 2 (top) + 2 (middle) + 1 (bottom) + 2 (left) = 7? No, maybe I'm mis - counting. Wait, the correct Lewis structure for $\ce{N2}$ is :N≡N:, where each N has one lone pair (2 electrons) and 3 shared electrons (from the triple bond), so total valence electrons per N: $2 + 3=5$, which is correct.

Now looking at the options: The first option (top) has the triple bond (three pairs between N and N) and each N has a lone pair (2 electrons) and some other electrons? Wait, no, the first option's dot structure: let's count the electrons around each N. For the first N: left (1), bottom (1), middle (2), top (2) → total 6? No, maybe the first option is the correct one because it shows a triple bond (three pairs of electrons between the two N atoms) and each N has a lone pair (2 electrons) and the remaining electrons? Wait, no, the correct Lewis structure for $\ce{N2}$ has a triple bond and one lone pair on each N. So the first option (the top one) with the triple bond (three pairs between N and N) and each N having a lone pair (2 electrons) and some other electrons? Wait, maybe the first option is the correct one as it represents the triple bond and the correct number of lone pairs. The other options do not have a triple bond. The second option has a single bond? No, the second option has two pairs between N and N (a double bond), and the third option has a single bond? No, the first option has three pairs (triple bond) between N and N, which is correct for $\ce{N2}$ as nitrogen forms a triple bond to achieve the octet.

Answer:

The first option (the top - most option with the structure $\boldsymbol{\cdot \cdot N\colon \colon \colon N\cdot \cdot}$ (represented by the dots as in the first choice) is the correct one. So the answer is the first option (the one with the triple bond between the two N atoms and appropriate lone pairs). If we assume the first option is labeled as, say, Option A, then the answer is A. (Since the original problem's options are not labeled with A, B, C, but from the image, the top option is the correct one.)