To my knowledge, there is no stable SF2^- ion. For one thing, it would be a free radical, meaning that it has an odd number of electrons. Free radicals tend to be quite reactive because of the single unpaired electron.
Fluorine has 7 valence electrons, not six. You seem to have trouble counting electrons. Assuming the anion exists, it would have 21 electrons, six for sulfur and seven each for the fluorine atoms, plus one more because of its "charge". It would have no double bonds. There does exist a neutral sulfur difluoride molecule. It has a bent geometry with single bonds to each fluorine and two lone pairs on the sulfur.
Your probably nonexistent SF2^- would have to have one more single electron on sulfur, giving it an expanded octet, and making the anion a free radical. Not likely.
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The dream is a little unclear about how to draw Lewis structures. To accurately draw the Lewis structure you MUST count the valence electrons.
Hemiketal is having trouble counting electrons, too. The question says, "SF2 1-" which I take to mean one sulfur, two fluorines, and a charge of -1. Count the electrons: 6 + 2(7) + 1 = 21.
Electronegativity was first developed by Linus PAULING. Pauli, as in Wolfgang Pauli, was a pioneer in quantum theory.