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Homework Help: Lewis structure for NSF

  1. Nov 16, 2012 #1
    Hi there. I have to determine the Lewis structure for NSF.

    So first I thought of something like this:

    N=S-F, with the nitrogen with four free electrons, the sulfur with two, and the fluorine with six. The formal charges gives -1 for Nitrogen, +1 For sulfur, and 0 for Fluorine. But the octet rule is accomplished. The sulfur, which has the lowest electronegativity it's central, as is proposed by the general rule.

    Then I thought of this:
    With this configuration S has four free electrons, N two, and F 6 again. Formal charges now are zero, but now the sulfur is a terminal atom, so I don't have the lowest electronegativity at the center.

    Then I looked for this at the internet, and found that the actual formula is:

    With this configuration formal charges again gives zero. But the octet rule is not accomplished. Now I know this must be the real formula. But what law should I follow when I have options like this? should I always use the expanded octet in this cases?
  2. jcsd
  3. Nov 17, 2012 #2


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    Gold Member

    Octet rule is a rule for Grade 6 students, as it is applicable for upto Neon. Sulphur can, and it does, expand its octet.

    To determine the molecular structure, first we need to decide the central atom, which you know how to do so. Then, the work is mainly instinct-based. Bond the remaining atoms, complete their bonding, like N prefers 3 bonds, O prefers 2 bonds, S prefers 2, 4, or 6 bonds, etc.
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