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$2Li + S rarr Li_(2)S$

And if we wanted to (which we do) we could split this up into half equations:

$2Li rarr 2Li^(+) + 2e^(-)$ $"Oxidation"$

$S+2e^(-) rarr S^(2-)$ $"Reduction"$

And in terms of electron configuration, sulfur gains 2 valence electrons to give 8 valence electrons.

And lithium loses 1 valence electron to give a $[He]^+$ core.

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