Despite the large difference in electronegativity between $Si$ $"Si"$ and $F$ $"F"$ , ${SiF}_{4}$ $"SiF"_4$ has a boiling point lower than that of ${NH}_{3}$ $"NH"_3$ ( $- 86^{\circ} C$ $-86^@ "C"$ , vs. $- 33^{\circ} C$ $-33^@ "C"$ ). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

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Despite the large difference in electronegativity between $Si$ $"Si"$ and $F$ $"F"$ , ${SiF}_{4}$ $"SiF"_4$ has a boiling point lower than that of ${NH}_{3}$ $"NH"_3$ ( $- 86^{\circ} C$ $-86^@ "C"$ , vs. $- 33^{\circ} C$ $-33^@ "C"$ ). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

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

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Answer 1 · 2017-07-09T17:33:51

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
$Si and F$ $"Si" and "F"$ the ${SiF}_{4}$ $"SiF"_4$ (and even the bond length is more) has a boiling point lower than $N H_{3}$ $NH_3$ precisely because of hydrogen bonding.

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

${SiF}_{4}$ $"SiF"_4$ is nonpolar though $Si - F$ $"Si"-"F"$ is polar because the tetrahedral arrangement of four Si-F cancel out the dipoles rendering the ${SiF}_{4}$ $"SiF"_4$ of zero dipole. This means that ${SiF}_{4}$ $"SiF"_4$ is nonpolar and London Dispersion Forces are the force than act on these molecules

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And in ${NH}_{3}$ $"NH"_3$ 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 $Si$ $"Si"$ and $F$ $"F"$ , ${SiF}_{4}$ $"SiF"_4$ 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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Despite the large difference in electronegativity between $Si$ $"Si"$ and $F$ $"F"$ , ${SiF}_{4}$ $"SiF"_4$ has a boiling point lower than that of ${NH}_{3}$ $"NH"_3$ ( $- 86^{\circ} C$ $-86^@ "C"$ , vs. $- 33^{\circ} C$ $-33^@ "C"$ ). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

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

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Explanation:

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Despite the large difference in electronegativity between "Si"Si"Si" and "F"F"F", "SiF"_4SiF4"SiF"_4 has a boiling point lower than that of "NH"_3NH3"NH"_3 (-86^@ "C"−86∘C-86^@ "C", vs. -33^@ "C"−33∘C-33^@ "C"). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

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

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Explanation:

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Despite the large difference in electronegativity between $Si$ $"Si"$ and $F$ $"F"$ , ${SiF}_{4}$ $"SiF"_4$ has a boiling point lower than that of ${NH}_{3}$ $"NH"_3$ ( $- 86^{\circ} C$ $-86^@ "C"$ , vs. $- 33^{\circ} C$ $-33^@ "C"$ ). Why is that? Could it be due to the symmetry? Hydrogen-bonding?

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