Question

in which of the following ionic compounds is the force of attraction between cation and anion the weakest? choose 1 answer: a kbr b rbi c mgo d cao

Explanation:

Step1: Recall Coulomb's law for ionic compounds

The force of attraction ($F$) between cation and anion in an ionic compound 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 ions and $r$ is the distance between the ions. Smaller charges and larger ionic - radii lead to weaker attraction.

Step2: Analyze the charges of ions in each compound

In $KBr$, $K^+$ has a charge of + 1 and $Br^-$ has a charge of - 1. In $RbI$, $Rb^+$ has a charge of + 1 and $I^-$ has a charge of - 1. In $MgO$, $Mg^{2 + }$ has a charge of + 2 and $O^{2 - }$ has a charge of - 2. In $CaO$, $Ca^{2+}$ has a charge of + 2 and $O^{2 - }$ has a charge of - 2. Compounds with ions of + 1 and - 1 charges have weaker attraction than those with + 2 and - 2 charges. So we can rule out $MgO$ and $CaO$.

Step3: Compare ionic radii of remaining compounds

The ionic radius of $Rb^+$ is larger than that of $K^+$, and the ionic radius of $I^-$ is larger than that of $Br^-$. According to Coulomb's law, larger ionic radii ($r$) result in a weaker force of attraction. So, the force of attraction in $RbI$ is weaker than that in $KBr$.

Answer:

B. $RbI$