Question

na$_2$s vs. k$_2$s
which choice is a true statement about these two particles?
the k⁺ is larger than the na⁺ because electrons occupy energy levels further from the nucleus.
the attractive force between na⁺ and s²⁻ is less than that between k⁺ and s²⁻ due to size effects.
the attractive force between na⁺ and s²⁻ is less than that between k⁺ and s²⁻ due to charge effects.
the k⁺ is larger than the na⁺ because it has more protons and electrons.

Explanation:

• For ionic radius: $K^+$ has an additional electron shell (n=4) compared to $Na^+$ (n=3), so its electrons are in energy levels farther from the nucleus, making it larger. The claim about more protons/electrons is incorrect because the extra proton in $K^+$ increases nuclear pull, but the additional electron shell dominates size.
• For ionic attractive force (Coulomb's law: $F = \frac{kq_1q_2}{r^2}$): Both pairs have the same ion charges ($+1$ and $-2$), so charge effects are identical. Since $Na^+$ is smaller than $K^+$, the distance $r$ between $Na^+$ and $S^{2-}$ is smaller, leading to a stronger attractive force than between $K^+$ and $S^{2-}$. This invalidates the two force-related options.

Answer:

The K⁺ is larger than the Na⁺ because electrons occupy energy levels further from the nucleus.