Assigning Third-Period Elements from Periodic Trends
Three unlabeled samples of third-period elements are on a lab bench. The labels were lost, but the instructor confirms the samples are sodium, magnesium, and aluminum. A student measures some properties and records the following: Sample X has an atomic radius of 186 pm and a first ionization energy of 496 kJ/mol. Sample Y has a radius of 160 pm and a first ionization energy of 738 kJ/mol. Sample Z has a radius of 143 pm and a first ionization energy of 578 kJ/mol. All three are silvery solids that react with oxygen when heated. The student expected first ionization energy to increase from left to right across the period, so she is confused that Z has a lower value than Y even though Z is farther to the right. She also notices that the atomic radius decreases from X to Z. She must assign each sample to an element and explain both the general trend and the exception using electron configurations and shielding.
1. Assign samples X, Y, and Z to sodium, magnesium, and aluminum. Justify the assignments using the data and periodic trends.
Read the worked answer
Sample X = sodium. It has the largest radius (186 pm) and the lowest first ionization energy (496 kJ/mol). Sodium is the leftmost of the three, in Group 1, so its single valence electron is the easiest to remove.
Sample Y = magnesium. It has the next-largest radius (160 pm) and the highest first ionization energy (738 kJ/mol). Magnesium is to the right of sodium, so its valence electrons are held more tightly.
Sample Z = aluminum. It has the smallest radius (143 pm), consistent with being farthest to the right. Its first ionization energy is lower than magnesium's, which is the exception to the general trend. Aluminum follows magnesium in Period 3, so the radius should be smaller, and it is. The lower ionization energy is explained by aluminum's electron configuration, not by its position alone.
So the order by atomic number is X → Y → Z: Na → Mg → Al.
2. Explain why Sample Z (aluminum) has a lower first ionization energy than Sample Y (magnesium), even though Z has a greater nuclear charge. Use electron configurations and shielding in your answer.
Read the worked answer
The electron configurations are:
- Mg: [Ne] 3s²
- Al: [Ne] 3s² 3p¹
The electron removed from magnesium is a 3s electron. The electron removed from aluminum is a 3p electron. The 3p subshell is higher in energy than the 3s subshell and is shielded from the nucleus by the filled 3s electrons as well as the inner-core electrons.
Aluminum does have a greater nuclear charge than magnesium, which tends to hold electrons more tightly. However, the 3p electron is in a higher-energy orbital and experiences more shielding. That makes the 3p electron easier to remove than the 3s electron of magnesium. Removing an electron from magnesium also breaks a filled 3s subshell, which is relatively stable.
For these reasons, the first ionization energy of Al (578 kJ/mol) is lower than that of Mg (738 kJ/mol), despite the general left-to-right increase across Period 3.
3. A classmate says that because atomic radius decreases from X to Z, the first ionization energy must increase smoothly from X to Z. Identify the flaw in that reasoning using these data.
Read the worked answer
The flaw is treating atomic radius as the only factor that controls first ionization energy. Radius generally decreases across a period because the effective nuclear charge increases while electrons are added to the same principal energy level. That generally makes electrons harder to remove, so ionization energy tends to increase.
But the trend is not smooth because ionization energy also depends on the subshell of the electron being removed and on shielding. Aluminum's outermost electron is in 3p, which is higher in energy and more shielded than magnesium's 3s electron. As a result, Al has the smaller radius but a lower first ionization energy than Mg.
The data show exactly this: radius decreases from X (186 pm) to Y (160 pm) to Z (143 pm), but ionization energy goes 496 → 738 → 578 kJ/mol. The middle value is the highest, not the last. A smooth increase is a useful first approximation, not a rule that overrides electron configuration.