Answer
$I_2$
Work Step by Step
Since both molecules are non-polar, we look to london dispersion forces. $I_2$ is much larger than $N_2$, so it has greater london dispersion forces and thus a higher boiling point.
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 9 - Liquids, Solids, and Materials - Questions for Review and Thought - Topical Questions - Page 421b: 33
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