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The melting point of XeOF4 (diagram E) is greater than the melting point of XeO3F2 (diagram D). Identify the type(s) of intermolecular force(s) that the two substances have in common. Explain the difference in melting points based on the types and relative strengths of intermolecular forces.

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XeOF4 had a square pyramidal structure as it has 1 lone pair and 5 bond pairs. The 4 Fluorine atoms dipole about the equatorial axis is all cancelled out. However, the single oxygen atom on the axial positions have a dipole. Hence, XeOF4 has a net dipole moment. It is a polar molecule.
However, XeO3F2 has a trigonal bipyramidal having 5 bond pairs. The molecule is completely symmetrical and hence, all of its dipole moments cancel each other out. It is a non polar molecule.
Polar molecules such as XeOF4 has instantaneous dipole- induced dipole interactions (id-id) and also permanent dipole- permanent dipole interactions (pd-pd).
Where as non polar molecules such as XeO3F2 only have id-id intermolecular forces of attraction.
Additionally, pdpd interactions are much stronger than idid interactions. Since XeOF4 have stronger intermolecular forces of attraction, it takes a larger amount of energy to overcome the intermolecular forces and separate the molecules. Hence, XeOF4 had a greater melting point.
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