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 (\( \text{S}^{2-} \), \( \text{Cl}^- \), \( \text{K}^+ \)) are isoelectronic (same number of electrons, \( 18 \) electrons, electron configuration \( 1s^2 2s^2 2p^6 3s^2 3p^6 \)). The difference in ionic radii comes from the nuclear charge (\( Z \)): \( \text{S}^{2-} \) has \( Z = 16 \), \( \text{Cl}^- \) has \( Z = 17 \), \( \text{K}^+ \) has \( Z = 19 \). Higher nuclear charge (more protons) pulls the same electron cloud more strongly, shrinking the ionic radius. So \( \text{K}^+ \) (highest \( Z \)) has the smallest radius, \( \text{S}^{2-} \) (lowest \( Z \)) the largest.

Answer:

These ions are isoelectronic (same electron count, 18 electrons) but have different nuclear charges (proton numbers: \( \text{S}^{2-} = 16 \), \( \text{Cl}^- = 17 \), \( \text{K}^+ = 19 \)). Higher nuclear charge (more protons) exerts a stronger pull on the identical electron cloud, reducing the ionic radius. Thus, \( \text{K}^+ \) (largest \( Z \)) has the smallest radius, \( \text{S}^{2-} \) (smallest \( Z \)) the largest, explaining the size differences.