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".

1 Answer
Professor Sam
Jul 9, 2017

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

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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

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