When forming an ion, iodine will do which of the following?
1 answer:
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Answer:
a)
Explanation:
In this question we have the following answer choices:
a)
b)
c)
d)
e)
We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:
a) (Van der waals interactions)
b) (Hydrogen bonding)
c) (Hydrogen bonding)
d) (Dipole-dipole interaction)
e) (Hydrogen bonding)
For molecules b, c and e we have <u>hydrogen bond to a heteroatom</u> (O, N, S, or P). In this case oxygen, therefore we will have <u>hydrogen bonding </u>interactions (a very strong interaction). So, we can discard these ones.
In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a <u>dipole</u> (due to the <u>electronegativity difference</u>). If we have a dipole, we will have a <u>dipole-dipole interaction</u> (a strong interaction, less than hydrogen bonding but still is a strong interaction).
In molecule a, we have only <u>Van der Waals interactions</u> because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a n<u>on-polar molecule</u>. These interactions are the weakest interactions of all the molecules given. So, <u>if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure. </u>