Question

using the concept of electrostatic force, predict which of the following salts will have the highest lattice energy and justify your prediction:

1. **na₃p or na₂o

1. *bes or mgs

Explanation:

Part 1: $\boldsymbol{Na_3P}$ or $\boldsymbol{Na_2O}$

Step1: Recall Lattice Energy Factors

Lattice energy ($U$) depends on ion charge and ionic radius, related by Coulomb's law: $U \propto \frac{|q_+ q_-|}{r_+ + r_-}$, where $q_+$/$q_-$ are ion charges, $r_+$/$r_-$ are radii.

Step2: Analyze Ion Charges

• For $\ce{Na3P}$: $\ce{Na^+}$ ($q_+ = +1$), $\ce{P^{3-}}$ ($q_- = -3$). Product of charges: $|(+1)(-3)| = 3$.
• For $\ce{Na2O}$: $\ce{Na^+}$ ($q_+ = +1$), $\ce{O^{2-}}$ ($q_- = -2$). Product of charges: $|(+1)(-2)| = 2$.

Step3: Analyze Ionic Radii

$\ce{P^{3-}}$ and $\ce{O^{2-}}$ are isoelectronic (same electron configuration), but $\ce{P^{3-}}$ has more protons? No—$\ce{O}$ (Z=8) vs $\ce{P}$ (Z=15)? Wait, no: $\ce{O^{2-}}$: 8 protons, 10 electrons. $\ce{P^{3-}}$: 15 protons, 18 electrons. Wait, ionic radius: for anions, more electrons (higher charge magnitude) and same period? No, $\ce{O}$ is in period 2, $\ce{P}$ in period 3. So $\ce{O^{2-}}$ has smaller radius than $\ce{P^{3-}}$. But charge product for $\ce{Na3P}$ is 3, vs 2 for $\ce{Na2O}$. The charge effect is stronger here.

Step4: Compare Lattice Energies

Since $U \propto |q_+ q_-|$, and 3 > 2, $\ce{Na3P}$ has higher lattice energy (despite $\ce{P^{3-}}$ being larger, the charge product dominates here).

Step1: Recall Lattice Energy Factors

Again, $U \propto \frac{|q_+ q_-|}{r_+ + r_-}$. Both salts have $\ce{S^{2-}}$ ($q_- = -2$). Cations: $\ce{Be^{2+}}$ ($q_+ = +2$) and $\ce{Mg^{2+}}$ ($q_+ = +2$) (same charge).

Step2: Analyze Ionic Radii

$\ce{Be^{2+}}$ (period 2, group 2) and $\ce{Mg^{2+}}$ (period 3, group 2). Ionic radius: $\ce{Be^{2+}} < \ce{Mg^{2+}}$ (smaller period = smaller radius).

Step3: Compare Lattice Energies

Since $|q_+ q_-|$ is same (2×2=4 for both), the ionic radius term dominates. Smaller cation radius ($\ce{Be^{2+}}$) means smaller $r_+ + r_-$, so higher lattice energy (as $U \propto \frac{1}{r_+ + r_-}$).

Answer:

$\ce{Na3P}$ has the highest lattice energy. Justification: Lattice energy depends on ion charge magnitude (product) and ionic radius. $\ce{Na3P}$ has $\ce{Na^+}$ and $\ce{P^{3-}}$ (charge product = 3), while $\ce{Na2O}$ has $\ce{Na^+}$ and $\ce{O^{2-}}$ (charge product = 2). The larger charge product in $\ce{Na3P}$ (despite $\ce{P^{3-}}$’s slightly larger radius) leads to stronger electrostatic attraction, hence higher lattice energy.

Part 2: $\boldsymbol{BeS}$ or $\boldsymbol{MgS}$