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a. $\ce{PH3}$

b. $\ce{HBr}$

c. $\ce{SO2}$

d. $\ce{N2}$

I believe that it is $\ce{N2}$, but I'm not too sure, since technically all of them would act the same at STP.

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    $\begingroup$ They would act the same if they all were ideal, which they aren't. $\endgroup$
    – Ivan Neretin
    Apr 23, 2018 at 20:56

2 Answers 2

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One of the defining characteristics of ideal gases is that ideal gas particles do not interact with each other, and there are no intermolecular forces between them. If a certain type of gas molecule has a molecular bipole, then each molecule will have ends with opposite partial charges. In a collection of gas particles, if the particles have a molecular dipole, these particles will attract and repel each other and thus will experience intermolecular forces. Because the two nitrogens in $\ce{N2}$ are equally electronegative, the bond and each $\ce{N2}$ gas particle will have minimal partial charges and thus minimal intermolecular forces. Thus, $\ce{N2}$ behaves very ideally.

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Ideal Gas Law disregards intermolecular interaction. So polar molecules will most likely differ from ideal gases to a higher degree than nonpolar molecules

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