Question

the table below shows the distance between the atoms in an atom - pair. all four pairs of atoms have the same amount of electric charge.

pair of atoms	distance between the atoms
c & d	6 units
p & q	2 units
x & y	8 units

which pair of atoms are repelled from each other with the most force?
a. pair p & q
b. pair x & y
c. pair c & d
d. pair a & b

Explanation:

Step1: Recall Coulomb's law

The electrostatic force $F$ between two charged particles is inversely - proportional to the square of the distance $r$ between them, i.e., $F=\frac{kq_1q_2}{r^{2}}$ (where $k$ is a constant and $q_1$ and $q_2$ are the charges). Since the charges are the same for all pairs, the force is mainly determined by the distance. The smaller the distance, the larger the force.

Step2: Compare the distances

We have the following distances: $r_{AB} = 4$ units, $r_{CD}=6$ units, $r_{PQ}=2$ units, $r_{XY}=8$ units.

Step3: Determine the pair with the most force

Since the force is inversely proportional to the square of the distance and $r_{PQ}$ is the smallest among all the distances, the pair of atoms $P$ and $Q$ will experience the most repulsive force.

Answer:

A. pair P & Q