Question

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 `"Si"-"F"` bond length is `"155.4 pm"`, and the `"N"-"H"` bond length is `"101.2 pm"`.

Answer 1

There are two answers

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
`"Si" and "F"` the `"SiF"_4` (and even the bond length is more) has a boiling point lower than `NH_3` precisely because of hydrogen bonding.

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

`"SiF"_4` is nonpolar though `"Si"-"F"` is polar because the tetrahedral arrangement of four Si-F cancel out the dipoles rendering the `"SiF"_4` of zero dipole. This means that `"SiF"_4` is nonpolar and London Dispersion Forces are the force than act on these molecules

And in `"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 `"Si"` and `"F"`, `"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