The lattice energy of a salt is related to the energy required to separate the ions. For which of th — Bonding Chemistry Question

STEPS:

1. Identify the governing chemical principle: The energy required to separate ions in a gaseous state (lattice energy) is governed by Coulomb's Law:
$$E \propto \frac{q_1 q_2}{d}$$
where $q_1$ and $q_2$ represent the charges of the individual ions, and $d$ is the internuclear distance between their centers (which is equal to the sum of their ionic radii). The energy required to separate a pair of ions is directly proportional to the magnitude of their charges and inversely proportional to the distance separating them.
2. Compare ionic charges (the dominant factor): First, examine the charges of the ions in each of the given options:
* Pair A ($\text{Na}^+$ and $\text{Cl}^-$): Charge magnitudes are $+1$ and $-1$ (product of charges $|q_1 q_2| = 1$).
* Pair B ($\text{Cs}^+$ and $\text{Br}^-$): Charge magnitudes are $+1$ and $-1$ (product of charges $|q_1 q_2| = 1$).
* Pair C ($\text{Mg}^{2+}$ and $\text{O}^{2-}$): Charge magnitudes are $+2$ and $-2$ (product of charges $|q_1 q_2| = 4$).
* Pair D ($\text{Ca}^{2+}$ and $\text{O}^{2-}$): Charge magnitudes are $+2$ and $-2$ (product of charges $|q_1 q_2| = 4$).
Because the product of charges for $\text{Mg}^{2+}/\text{O}^{2-}$ and $\text{Ca}^{2+}/\text{O}^{2-}$ is four times greater than that of the $+1/-1$ pairs, their electrostatic attractions are vastly stronger, and they require significantly more energy to separate. This immediately eliminates Options A and B.
3. Compare ionic sizes (distances) for the remaining pairs: Now, compare Pair C and Pair D, which have identical charges ($+2$ and $-2$). According to Coulomb's Law, when charges are equal, the pair with the smaller internuclear distance $d$ (smaller ionic radii) will experience a stronger attractive force and require more energy to separate:
* Both pairs share the same anion, $\text{O}^{2-}$.
* Compare the sizes of the cations: Magnesium ($\text{Mg}$) is in Period 3 of the periodic table, while Calcium ($\text{Ca}$) is in Period 4.
* Because $\text{Mg}^{2+}$ has fewer occupied electron shells than $\text{Ca}^{2+}$, $\text{Mg}^{2+}$ has a smaller ionic radius than $\text{Ca}^{2+}$.
4. Determine the correct option: Since $\text{Mg}^{2+}$ is smaller than $\text{Ca}^{2+}$, the distance $d$ between the centers of $\text{Mg}^{2+}$ and $\text{O}^{2-}$ is smaller than the distance between $\text{Ca}^{2+}$ and $\text{O}^{2-}$. This smaller distance leads to a stronger electrostatic attraction, meaning $\text{Mg}^{2+}(g)$ and $\text{O}^{2-}(g)$ require the largest amount of energy to separate, which corresponds to Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect: The ions $\text{Na}^+$ and $\text{Cl}^-$ carry only $+1$ and $-1$ charges. Because ionic charge is the dominant factor in Coulombic attraction, their lower charges result in a much smaller lattice energy compared to the $+2/-2$ pairs.
• Option B is incorrect: Like Option A, $\text{Cs}^+$ and $\text{Br}^-$ only have $+1$ and $-1$ charges. Additionally, both $\text{Cs}^+$ and $\text{Br}^-$ have very large atomic radii due to being lower down on the periodic table, which increases the internuclear distance and further weakens the electrostatic attraction between them.
• Option D is incorrect: Although $\text{Ca}^{2+}$ and $\text{O}^{2-}$ have the same highly favorable $+2$ and $-2$ charges as magnesium oxide, the $\text{Ca}^{2+}$ ion has an extra shell of core electrons and is therefore larger than the $\text{Mg}^{2+}$ ion. This larger size increases the internuclear distance $d$, resulting in a weaker attraction and less energy required to separate them than $\text{Mg}^{2+}$ and $\text{O}^{2-}$.