Question

which of these pairs of particles will experience the greatest coulombic attraction?
be and o
be and f
mg and s
mg and cl

Explanation:

Step1: Recall Coulomb's Law

The Coulombic attraction force is given by:
$$F = k\frac{q_1 q_2}{r^2}$$
where $k$ is a constant, $q_1,q_2$ are the charges of the particles, and $r$ is the distance between their nuclei.

Step2: Determine ionic charges

• Be forms $\text{Be}^{2+}$, O forms $\text{O}^{2-}$; $q_1 q_2 = (2)(2) = 4$
• Be forms $\text{Be}^{2+}$, F forms $\text{F}^-$; $q_1 q_2 = (2)(1) = 2$
• Mg forms $\text{Mg}^{2+}$, S forms $\text{S}^{2-}$; $q_1 q_2 = (2)(2) = 4$
• Mg forms $\text{Mg}^{2+}$, Cl forms $\text{Cl}^-$; $q_1 q_2 = (2)(1) = 2$

Step3: Compare ionic radii

For pairs with equal charge products:

• $\text{Be}^{2+}$ has a smaller radius than $\text{Mg}^{2+}$, and $\text{O}^{2-}$ has a smaller radius than $\text{S}^{2-}$. So the distance $r$ for Be and O is smaller than for Mg and S.

Step4: Compare attraction strength

Smaller $r$ and equal $q_1 q_2$ means larger $F$. Be and O have the largest charge product and smallest nuclear distance.

Answer:

Be and O