Periodicity II: the ionic radius (Answers)

Question 1: Can you do the comparison for the following particles and give reasons for your answers?

1. Cu, \( \ce{Cu^+} \), \( \ce{Cu^{2+}} \)
2. Mg, \( \ce{Mg^+} \), \( \ce{Mg^{2+}} \)
3. N, \( \ce{N^-} \), \( \ce{N^{2-}} \), \( \ce{N^{3-}} \)
4. Cl, \( \ce{Cl^-} \), \( \ce{Cl^{2-}} \)

Answer:

1. \( \ce{Cu^{2+}} \lt \ce{Cu^{+}} \lt \ce{Cu} \) because 1) the ions have one less energy level as compared to the atom; 2) \( \ce{Cu^{2+}} \) has a greater positive charge than \( \ce{Cu^{+}} \).
2. \( \ce{Mg^{2+}} \lt \ce{Mg^{+}} \lt \ce{Mg} \) because 1) \( \ce{Mg^{2+}} \) has one less energy level; 2) \( \ce{Mg^{+}} \) has a greater positive charge than Mg atom.
3. \( \ce{N} \lt \ce{N^{-}} \lt \ce{N^{2-}} \lt \ce{N^{3-}} \) because 1) they have the same number of energy levels; 2) their radii increase with their negative charges due to the rising repulsions among electrons.
4. \( \ce{Cl} \lt \ce{Cl^{-}} \lt \ce{Cl^{2-}} \) because 1) \( \ce{Cl^{2-}} \) has one more energy level while the other two particles have the same number of energy levels; 2) negatively charged anion has additional repulsion among electrons such that it is larger in radius than the neutral atom.

Question 2: Can you sort the following elements in a sequence of increasing ionic radii of the common ions:

P, S, Cl, K, Ca ?

Answer:

Particle	P	S	Cl	K	Ca
Ions	\( \ce{P^{3-}} \)	\( \ce{S^{2-}} \)	\( \ce{Cl^{-}} \)	\( \ce{K^{+}} \)	\( \ce{Ca^{2+}} \)
Electron configuration	\( \ce{[Ar]} \) or \( \ce{[Ne] 3s^2 3p^6} \)	\( \ce{[Ar]} \)	\( \ce{[Ar]} \)	\( \ce{[Ar]} \)	\( \ce{[Ar]} \)
Energy levels	3	3	3	3	3
Charge	-3	-2	-1	+1	+2

Since all of them have the same number of energy levels, their charges determine their sizes. More negatively charged particles have larger radii, and more positively charged ones would have smaller radii. Hence we could arrange them in a sequence of increasing ionic radii as following:

\[ \ce{Ca^{2+}} \lt \ce{K^{+}} \lt \ce{Cl^{-}} \lt \ce{S^{2-}} \lt \ce{P^{3-}} \]