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This question already has an answer here:

Among NH₃,PH₃, AsH₃, SbH₃ the increasing order of volatility given in my textbook is as follows.

NH₃< SbH₃< AsH₃< PH₃

I know that among all these hydrides NH₃ alone is capable of forming Hydrogen bonds thus it is least volatile, but I’m not sure how to compare others.

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marked as duplicate by Mithoron, pentavalentcarbon, Tyberius, airhuff, M.A.R. Feb 19 '18 at 12:38

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Recall that volatility is inversely proportional to the boiling point. The boiling point itself is directly proportional to the intermolecular forces. In the above case, we have two major intermolecular forces: H-bonds and dipole-dipole attractions.

Now, it is factual info that ammonia's intermolecular forces surpass even stibine's intermolecular attractions, so its volatility is the least.

Thus, we are only left with phosphine, arsine, and stibine. Here, recall that intermolecular forces are van Der Waal's forces, and thus their magnitude is directly to their size. (read the "why" of it here). Hence, stibine is the second-least volatile (as it has maximum van der waal's force due to its bigger size), followed by asrine, and finally phosphine.

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