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Chapter 11: Problem 78

For each of the following pairs of substances, predict which will have the higher melting point, and indicate why: (a) HF, \(\mathrm{HCl} ;\) (b) C (graphite), \(\mathrm{CH}_{4}\); (c) \(\mathrm{KCl}, \mathrm{Cl}_{2}\); (d) \(\mathrm{LiF}, \mathrm{MgF}_{2}\).

Short Answer

Expert verified
The higher melting points for each pair are as follows: (a) HF, due to stronger hydrogen bonding (b) C (graphite), due to a strong covalent bond network structure (c) KCl, due to a strong ionic bond (d) MgF2, due to stronger electrostatic forces between ions.

Step by step solution

01

Identify the type of bond in each substance.

We start by identifying the principles types of bonds in each substance: (a) HF - Polar covalent bond (due to hydrogen bonding), HCl - Polar covalent bond (due to dipole-dipole interactions) (b) C (graphite) - Covalent bond, CH4 - Nonpolar covalent bond (due to dispersion forces) (c) KCl - Ionic bond, Cl2 - Nonpolar covalent bond (due to van der Waals forces) (d) LiF - Ionic bond, MgF2 - Ionic bond
02

Compare and predict the higher melting point for each pair.

We will compare the bonds and predict the higher melting point considering the factors mentioned in the analysis. (a) HF vs HCl: HF has stronger hydrogen bonding than the dipole-dipole interaction in HCl. Thus, **HF** will have a higher melting point. (b) C (graphite) vs CH4: Graphite has a strong covalent bond network structure, whereas CH4 has weaker dispersion forces. Therefore, **C (graphite)** will have a higher melting point. (c) KCl vs Cl2: KCl has a strong ionic bond, while Cl2 has weak van der Waals forces. Hence, **KCl** will have a higher melting point. (d) LiF vs MgF2: Both are ionic compounds; however, MgF2 has a +2 charge on the Mg ion and -1 charge on each F ion, resulting in stronger electrostatic forces compared to LiF with a +1 Li ion charge and -1 F ion charge. Consequently, **MgF2** will have a higher melting point.

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