# Despite the large difference in electronegativity between "Si" and "F", "SiF"_4 has a boiling point lower than that of "NH"_3 (-86^@ "C", vs. -33^@ "C"). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

## Note: the $\text{Si"-"F}$ bond length is $\text{155.4 pm}$, and the $\text{N"-"H}$ bond length is $\text{101.2 pm}$.

Jul 9, 2017

1.) Because of the hydrogen bonds (precisely interactions)

2.) Because of the tetrahedral arrangement of Si-F

#### Explanation:

Though the large difference in electronegativity between
$\text{Si" and "F}$ the ${\text{SiF}}_{4}$ (and even the bond length is more) has a boiling point lower than $N {H}_{3}$ precisely because of hydrogen bonding.

This is because that boiling is completely dependent on intermolecular forces and not on intramolecular forces.

${\text{SiF}}_{4}$ is nonpolar though $\text{Si"-"F}$ is polar because the tetrahedral arrangement of four Si-F cancel out the dipoles rendering the ${\text{SiF}}_{4}$ of zero dipole. This means that ${\text{SiF}}_{4}$ is nonpolar and London Dispersion Forces are the force than act on these molecules

And in ${\text{NH}}_{3}$ there are hydrogen bonds which are a special case of dipole-dipole forces.

Because that London Dispersion Forces are temporary because electron density is always changing all across the atom dipole-dipole forces are much stronger because they are permanent and always in alignment.

Despite the large difference in electronegativity between $\text{Si}$ and $\text{F}$, ${\text{SiF}}_{4}$ the electronegativity doesn't matter because of the tetrahedral of four Si-F make the the compound nonpolar.

And more stronger the intermolecular force more the boiling point