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:

Step1: Identify Isoelectronic Species

The ions \( S^{2-} \), \( Cl^{-} \), and \( K^{+} \) are isoelectronic, meaning they have the same number of electrons (18 electrons, as \( S^{2-} \): \( 16 + 2 = 18 \), \( Cl^{-} \): \( 17 + 1 = 18 \), \( K^{+} \): \( 19 - 1 = 18 \)).

Step2: Analyze Nuclear Charge

The nuclear charge (number of protons) for \( S^{2-} \) is \( Z = 16 \), for \( Cl^{-} \) is \( Z = 17 \), and for \( K^{+} \) is \( Z = 19 \).

Step3: Relate Nuclear Charge to Ionic Radius

The attractive force between the nucleus and the valence electrons depends on the nuclear charge. A higher nuclear charge means a stronger pull on the same number of electrons.

• \( K^{+} \) has the highest nuclear charge (19 protons), so it pulls the 18 electrons most strongly, resulting in the smallest ionic radius (138 pm).
• \( Cl^{-} \) has a nuclear charge of 17, so the electron pull is less than in \( K^{+} \), leading to a larger radius (181 pm) than \( K^{+} \).
• \( S^{2-} \) has the lowest nuclear charge (16), so the electron cloud is pulled least, resulting in the largest radius (184 pm).

Answer:

These ions (\( S^{2-} \), \( Cl^{-} \), \( K^{+} \)) are isoelectronic (same number of electrons) but have different nuclear charges (16, 17, 19 protons respectively). A higher nuclear charge exerts a stronger pull on the electron cloud, shrinking the ionic radius. Thus, \( K^{+} \) (highest \( Z \)) has the smallest radius, \( Cl^{-} \) (intermediate \( Z \)) has a larger radius, and \( S^{2-} \) (lowest \( Z \)) has the largest radius.