[VISUAL] 19. Based on Coulomb’s law and the information in the table above, which of the following a — Bonding Chemistry Question

STEPS:

1. Identify the underlying chemistry concepts being tested: This question tests your understanding of Coulomb’s law and how it governs ion-dipole attractions between dissolved anions and polar water molecules in an aqueous solution.
2. Recall Coulomb’s law:
The electrostatic force of attraction ($F$) or potential energy of interaction ($E$) between two charged species is directly proportional to the magnitude of their charges ($q_1$ and $q_2$) and inversely proportional to the distance between their centers ($d$):
$$F \propto \frac{q_1 q_2}{d^2}$$
3. Analyze the effect of charge ($q$) on the interaction strength:
* Water is a polar molecule with a partial positive charge ($\delta^+$) on its hydrogen atoms. Anions attract these $\delta^+$ hydrogen atoms.
* The charges ($q_1$) of the given anions are:
* Chloride ($\text{Cl}^-$): $-1$
* Iodide ($\text{I}^-$): $-1$
* Sulfide ($\text{S}^{2-}$): $-2$
* Telluride ($\text{Te}^{2-}$): $-2$
* Because charge magnitude has a direct, proportional relationship with force, the divalent anions ($\text{S}^{2-}$ and $\text{Te}^{2-}$) with a $-2$ charge will have significantly stronger electrostatic attractions with polar water molecules than the monovalent anions ($\text{Cl}^-$ and $\text{I}^-$) with a $-1$ charge.
4. Analyze the effect of distance ($d$) among the divalent anions:
* For ions with the same charge, we must compare their ionic radii.
* A smaller ionic radius means the outer shell of the anion is closer to its nucleus, which allows the surrounding polar water molecules to get closer to the center of negative charge. This minimizes the distance $d$ in the denominator of Coulomb’s law, maximizing the attractive force.
* Comparing the ionic radii of the two $-2$ anions:
* Sulfide ($\text{S}^{2-}$): $184\text{ pm}$
* Telluride ($\text{Te}^{2-}$): $221\text{ pm}$
* Because $\text{S}^{2-}$ has a significantly smaller ionic radius than $\text{Te}^{2-}$, it has a much higher charge density. The water molecules can get closer to its charge center, yielding a stronger electrostatic interaction.
5. Conclude:
Sulfide ($\text{S}^{2-}$) has the optimal combination of a high charge magnitude ($-2$) and a relatively small ionic radius ($184\text{ pm}$), making it the anion most likely to have the strongest interactions with nearby water molecules. This corresponds to Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect ($\text{Cl}^-$): Although $\text{Cl}^-$ has the smallest ionic radius of all the options ($181\text{ pm}$), its charge magnitude is only $-1$. Because charge magnitude dominates over small differences in radius in Coulomb's law, its interaction with water is weaker than that of the $-2$ charged ions.
• Option B is incorrect ($\text{I}^-$): Iodide has both a low charge magnitude ($-1$) and a very large ionic radius ($216\text{ pm}$). Both factors work together to minimize the electrostatic force of attraction, making its interactions with water the weakest among the choices.
• Option D is incorrect ($\text{Te}^{2-}$): While telluride has the favorable high charge magnitude of $-2$, its ionic radius ($221\text{ pm}$) is much larger than that of sulfide ($184\text{ pm}$). This greater distance weakens the electrostatic attraction compared to the smaller, more charge-dense sulfide ion.