# What is the dominant intermolecular force present in the Group 15 hydrides: "ammonia"; "phosphine"; "arsine"?

Dec 1, 2016

For ammonia, it's easy. The dominant intermolecular force is hydrogen bonding...........

#### Explanation:

For carbon dioxide, which has no dipole moment, dispersion forces operate to some extent.

Hydrogen bonding as an intermolecular force is quite significant. If we compare the boiling points of water to hydrogen sulfide to hydrogen selenide, we find that the normal boiling points vary respectively as, $100$ ""^@C, $- 60$ ""^@C, and $- 41.2$ ""^@C.

Clearly, the hydrogen oxygen bond is the most polar, the most charge separated, of the series, and thus gives rise to the most significant hydrogen bonding. While hydrogen bonding operates to some extent in hydrogen sulfide, and hydrogen selenide, the reduced bond dipole, reduces intermolecular interaction, and DEPRESSES boiling point. On the other hand, the larger, more polarizable hydrogen selenide has a greater opportunity for dispersion forces, and thus it is LESS volatile than ${H}_{2} S$.

And now this is your job. Find the boiling points of ammonia, $N {H}_{3}$, and compare this with phosphine, $P {H}_{3}$, and with $\text{arsine}$, $A s {H}_{3}$. Can you rationalize the boiling points on the basis of intermolecular interaction?

Dec 1, 2016

The intermolecular forces of attraction between carbon dioxide & ammonia is of Induced-Dipole nature.

#### Explanation:

According to VSEPR Theory, $N {H}_{3}$ is having a pyramidal shape. So, the ammonia molecule is polar in nature.

Now, the $C {O}_{2}$ molecule is having a linear shape. So, the carbondioxide molecule is non-polar in nature.

So, the intermolecular forces of attraction between the polar $N H 3$ & nonpolar $C {O}_{2}$ molecules is an Induced-Dipole one.

Hope it Helps :)