Question

Question #18d2e

Answer 1

`CF_4<CHF_3<H_2O<NaBr<Cu<SiC`

Explanation:

Im order to answer this question, we will need to list the type of intermolecular interactions that every compound can carry on:

1. `H_2O` is a polar molecular and therefore, it carries, in addition to the London Dispersion Force ( LDF ), a dipole-dipole interaction (d-d) which is the strongest and is called H-bond .
2. `CF_4` the geometry around the carbon atom is tetrahedral. Since all four substituents are identical, the dipole moments of the `C-F` bonds will cancel each other and the molecule is non polar. Therefore, the only possible type of interaction is LDF .
3. `NaBr` is made from a metal `Na` and a non metal `Br` and therefore, it is an ionic compound. Therefore, the intermolecular interaction is ion-ion interaction .
4. `SiC` will form a solid network and the type of intermolecular interaction is directional covalent bond .
5. `CHF_3` the geometry around the carbon atom is tetrahedral. Since it is connected to three identical substituents only, this molecule will be polar. The type of intermolecular interaction possible are LDF and d-d interactions .
6. `Cu` is a metal and the type of intermolecular interaction is a non directional covalent bond .

Thus the increasing order of the melting point would be:
`CF_4<CHF_3<H_2O<NaBr<Cu<SiC`