Question

Question: The boiling point of pentane (${\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{12}}$) is slightly less than the melting point of rubidium. Rank the four substances in problem 28 from lowest to highest in melting point.

Short answer

The order of melting point is:

$C_5{H}_{12}<N{a}_{2}HP{O}_4<Rb<B$

Step-by-step solution

The bonding nature of different given compounds.

The difference in electronegativities makes the compound Sodium Phosphate ionic. It has a given melting point of $250°C$.

${\text{Na}}^{+}-{\text{PO}}_4^{2-}-{\text{H}}^{+}$

The melting point of the boron is somewhat near $2000°C$. The bonding between boron atoms is covalent as boron valency is 3; it neither accepts nor donates electrons. Example:$Boro{n}_{12}$

Pentane is a molecular crystal, just like many organic compounds attached through Van der Waals forces. It is the weakest kind of bonding among all with the lowest melting point.

The interactions between the stable ions of rubidium that keep the metal together, and strong is metallic bonding. There is a sea of mobile electrons found around the fixed ions. Metal is electrically neutral as the density of electrons is similar to that of positively charged ions. The melting point of rubidium is around $40°C$.

The rank from lowest to highest in melting point.

The strongest bonding is the bonding between boron, followed by the ionic bonding of Sodium Phosphate, the metallic bonding of rubidium, and finally, the Van der Waals forces of pentane.

This makes the order:

$C_5{H}_{12}<Rb<N{a}_{2}HP{O}_4<B$