Question

Given that the ions in LiF and MgO are of similar size, which compound has stronger ionic bonding? Use Coulomb’s law in your explanation.

Short answer

MgO the compound has stronger ionic bonding.

Step-by-step solution

Define Coulomb’s law

Coulomb’s law is defined as the interaction force between two ions (cation and anion) in an ionic bond is directly proportional to the product of charges of ions and inversely proportional to the distance between two ions.

$\mathbf{F}\mathbf{=}\mathbf{\alpha }\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here,

Fis the interaction between two ions.

${\mathrm{q}}_1$is the charge of an ion (cation).

${\mathbf{q}}_{\mathbf{2}}$is the charge of an ion (anion).

r is the distance between two ions.

Apply the Coulomb’s law

In , MgO the charge of ${\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}$cation is +2 while the charge of ${\mathit{O}}^{\mathbf{2}}$anion is -2. In the case of LiF , the charge of ${\mathbf{Li}}^{\mathbf{+}}$cation is +1 while the charge of F anion is -1. So, the product of charges of two ions in MgO is greater than that of two ions in . Hence,MgOthe compound has stronger ionic bonding.