a
Step \(1:\) Net force on \(Cl ^{-}\)ion at the centre

From Figure, Position of \(Cl ^{-}\)is symmetric at the centre, and distances of all \(Cs ^{+}\) ions from center is same, so magnitude of all attraction forces on \(Cl ^{-}\)will also be same.

Therefore, all \(Cs ^{+}\)ions will attract \(Cl ^{-}\)at center with equal forces in opposite directions. So, all their forces will cancel out. Hence Net force on \(Cl ^{-}\)ion will be \(zero\).

Therefore correct answer is \(D\)

Step \(2\): Detailed Explanation of forces on \(Cl ^{-}\)

The forces on \(Cl ^{-}\)will be get cancelled due to equal and opposite forces.

Forces due to \(1\) and \(7\) will cancel

Forces due to \(2\) and \(8\) will cancel

Forces due to \(3\) and \(5\) will cancel

Forces due to \(4\) and \(6\) will cancel

So, Net force on \(Cl ^{-}\)will be zero