Question

1. the ionic radii of the ions $s^{2-}$, $cl^{-}$, and $k^{+}$ are 184, 181, 138 pm, respectively. explain why these ions have different sizes even though they contain the same number of electrons.

Explanation:

These ions (\(S^{2-}\), \(Cl^-\), \(K^+\)) are isoelectronic (same number of electrons, electron configuration \([Ne]3s^23p^6\)). The difference in ionic radii comes from the nuclear charge. \(S^{2-}\) has 16 protons, \(Cl^-\) has 17 protons, and \(K^+\) has 19 protons. As nuclear charge (number of protons) increases, the attractive force on the same number of electrons increases, pulling the electron cloud closer to the nucleus. So \(K^+\) (highest nuclear charge) has the smallest radius, \(Cl^-\) is smaller than \(S^{2-}\) (lower nuclear charge than \(Cl^-\)), and \(S^{2-}\) has the largest radius due to the weakest nuclear attraction among the three.

Answer:

The ions \(S^{2-}\), \(Cl^-\), and \(K^+\) are isoelectronic (same electron count). Their ionic radii differ because of varying nuclear charges: \(S^{2-}\) (16 protons), \(Cl^-\) (17 protons), \(K^+\) (19 protons). Higher nuclear charge (more protons) exerts a stronger pull on the same electron cloud, shrinking the ion’s radius. Thus, \(K^+\) (highest nuclear charge) has the smallest radius, \(Cl^-\) is smaller than \(S^{2-}\), and \(S^{2-}\) (lowest nuclear charge) has the largest radius.