Question

Why is the $\mathrm{CsCl}$ structure more stable than the $\mathrm{NaCl}$ structure?

Answer 1

$\mathbf{CsCl}$ structure

• In the compound cesium chloride $\left(\mathrm{CsCl}\right)$ structure, each cesium ion $\left({\mathrm{Cs}}^{+}\right)$ is surrounded by $8$ chloride ions $\left({\mathrm{Cl}}^{-}\right)$ and each chloride ion is surrounded by $8$ cesium ions.
• Therefore, the coordination number of $\mathrm{CsCl}$ is $8:8$.
• Structure of $\mathrm{CsCl}$ is as follows:

$\mathbf{NaCl}$ Structure

• In a Sodium chloride $\left(\mathrm{NaCl}\right)$ structure, there are six chloride ions $\left({\mathrm{Cl}}^{-}\right)$ around each sodium ion $\left({\mathrm{Na}}^{+}\right)$ and six sodium ions $\left({\mathrm{Na}}^{+}\right)$ surrounding each chloride ion $\left({\mathrm{Cl}}^{-}\right)$.
• Therefore, the coordination number of $\mathrm{NaCl}$ will be $6:6$.
• Structure of $\mathrm{NaCl}$ is as follows:

Compare the stability of $\mathbf{NaCl}$and $\mathbf{CsCl}$ structure

• $\mathrm{CsCl}$ is more stable than $\mathrm{NaCl}$.
• This is due to the fact that the larger the coordination number, the stronger the forces of attraction between the cations and anions in the close-packed arrangement.
• $\mathrm{CsCl}$ is more stable because its coordination number is $8:8$, compared to $\mathrm{NaCl}$ which has coordination number $6:6$.