I was predicting a very simple chemical equation $$\ce{Ba + S ->}$$ And I had solved it as $$\ce{Ba + S -> BaS}$$ The answer should make sense as barium has a charge of +2, and sulfide has a charge of −2; however, the book had said that the answer was $$\ce{Ba + S -> Ba2S2}$$ I do indeed know how and why they got the answer, but I just want to know why my answer isn't correct, or if it is and I'm just acting silly.


As orthocresol pointed out, $\ce{Ba_2S_2}$ does not exist as it's own substance. Perhaps you've misread the question and it instead asks for something else? Otherwise it is wrong, or at least not simplified. Your answer is in effect the correct one, although I'm not sure if sulfur and barium would react like that exactly.

It might in addition generate multiple polysulfides, such as $\ce{BaS_2}$, $\ce{BaS_3}$, following the rule of $\ce{BaS_{$x$}}$. However, I believe your answer is correct in that the initial reaction between barium and sulfur would produce $\ce{BaS}$, while further reactions might add more sulfurs onto it.

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    $\begingroup$ BaS2 does exist, BaS3 and BaS4 too (also @ortho). You should not answer like that, I'm tempted to downvote you. $\endgroup$
    – Mithoron
    Jul 6 '16 at 23:28
  • $\begingroup$ @Mithoron Alright. I can see what you mean, however, I do not know what exactly it would yield. My answer does not add any information on what the actual product would be, so I will likely remove it soon if I cannot come up with a better one. $\endgroup$
    – ChemBird
    Jul 6 '16 at 23:37
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    $\begingroup$ Obligatory poke towards using \ce{...} $\endgroup$
    – Jan
    Jul 7 '16 at 4:14
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    $\begingroup$ "Exist" is not quite the right word. Why don't we write $\ce{BaS}$ as $\ce{Ba2S2}$, really? Sure, it does not contain isolated particles $\ce{Ba2S2}$, but it does not contain $\ce{BaS}$ particles either. It is not that $\ce{Ba2S2}$ is "not true"; it is that writing it like that has no point. $\endgroup$ Jul 7 '16 at 9:14
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    $\begingroup$ @Dan I would say no, not normally. As Ivan has pointed out it is true, but it is like saying the answer to 20 divided by 4 is $10/2$. This is correct, but not simplified. $\endgroup$
    – ChemBird
    Jul 7 '16 at 22:30

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