Question

based on coulombs law and the information in the table above, which of the following cations is most likely to have the weakest interaction with an adjacent water molecule in an aqueous solution?
a li+
b na+
c ca2+
d in3+

Explanation:

Step1: Recall Coulomb's Law

The force of interaction $F$ between two charged particles is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is a constant, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them. In the context of cations and water - molecules, the charge of the cation ($q_1$) and the partial charge on the water - molecule ($q_2$) are involved, and $r$ is the distance from the center of the cation to the point of interaction on the water - molecule.

Step2: Analyze the factors affecting interaction

The smaller the charge of the cation or the larger the ionic radius (which increases $r$), the weaker the interaction with water. Among the given cations, $Li^+$ has a charge of + 1 and an ionic radius of 60 pm, $Na^+$ has a charge of + 1 and an ionic radius of 95 pm, $Ca^{2 + }$ has a charge of + 2 and an ionic radius of 99 pm, and $In^{3+}$ has a charge of + 3 and an ionic radius of 81 pm.

Step3: Compare the cations

Since $Na^+$ has a + 1 charge (the same as $Li^+$) and a larger ionic radius than $Li^+$, and compared to $Ca^{2+}$ and $In^{3+}$ which have higher charges, $Na^+$ will have the weakest interaction with an adjacent water molecule according to Coulomb's Law.

Answer:

B. $Na^+$