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
Jul 9, 2017

Answer:

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