Q81 E - Chemistry (OpenStax)

Question

Which compound in each of the following pairs has the larger lattice energy? Note: \({\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}\) and \({{\rm{K}}^{\rm{ + }}}\) have similar radii; \({{\rm{S}}^{{\rm{2 - }}}}\) and \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) have similar radii. Explain your choices. 

(a) \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) or \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\) 

(b) \({{\rm{K}}_{\rm{2}}}{\rm{S}}\) or \({\rm{BaS}}\) 

(c) \({\rm{KCl}}\) or \({\rm{BaS}}\) 

(d) \({\rm{BaS}}\) or \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}\)

Step-by-Step Solution

Answer

1. \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\) has greater lattice energy.
2. \({\rm{BaS}}\) has greater lattice energy.
3. \({\rm{BaS}}\) has greater lattice energy.
4. \({\rm{BaS}}\) has greater lattice energy

Step 1: Concept Introduction

The energy required to dissociate the ions in a crystal lattice into individual gaseous ions is known as lattice energy.

Step 2: Larger Lattice Energy between \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) and \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\)

(a) 

Since \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) has a smaller radius than \({{\rm{K}}^{\rm{ + }}}\).This leads to \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\) having larger lattice energy than \({{\rm{K}}_{\rm{2}}}{\rm{O}}\).

Therefore, \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\) has lattice energy than \({{\rm{K}}_{\rm{2}}}{\rm{O}}\).

 

Step 3: Larger Lattice Energy between \({{\rm{K}}_{\rm{2}}}{\rm{S}}\) and \({\rm{BaS}}\)

(b)

Since there are higher charges on \({\rm{Ba}}\) than \({\rm{K}}\), whereas radii of the ions are almost similar. 

Therefore, \({\rm{BaS}}\) has larger lattice energy than \({{\rm{K}}_{\rm{2}}}{\rm{S}}\).

Step 4: Larger Lattice Energy between \({\rm{KCl}}\) and \({\rm{BaS}}\)

(c)

Since there are higher charges on \({\rm{Ba}}\) and \(S\), whereas radii of the ions are almost similar. 

Therefore, \({\rm{BaS}}\) has larger lattice energy than \({\rm{KCl}}\).

Step 5: Larger Lattice Energy between \({\rm{BaS}}\) and \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}\)

(d)

Since there are higher charges on \({\rm{S}}\) than \({\rm{Cl}}\), whereas radii of the ions are almost similar. 

Therefore, \({\rm{BaS}}\) has larger lattice energy than \({\rm{BaC}}{{\rm{l}}_2}\).