1. Atomic and Ionic Radii Across Period 3
Periodicity is the repeating pattern of properties of elements across different periods. As you move from left to right across Period 3 (Na to Ar), the physical properties change in a predictable way. Atomic radius is the first key trend. It consistently decreases across the period. This is because, from sodium to argon, a proton is added to the nucleus and an electron is added to the same outer shell (the third shell). The increasing nuclear charge pulls the electrons in this outer shell more strongly towards the nucleus, shrinking the atom. The shielding effect from the inner shells remains relatively constant, so it doesn't counteract the stronger pull. Ionic radius also shows clear trends. Metal atoms (Na, Mg, Al) lose their outer shell electrons to form positive ions (cations). These cations are much smaller than their parent atoms because they have lost an entire electron shell and the remaining electrons are pulled more tightly by the unchanged nuclear charge. Non-metal atoms (P, S, Cl) gain electrons to form negative ions (anions). These anions are larger than their parent atoms because the addition of electrons increases the repulsion between the electrons in the outer shell, causing it to expand.
Key term
Examiner insight
Common pitfall
Worked example 13 marks
Explain why the atomic radius of chlorine is smaller than that of sodium.
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Step 1: Identify the positions of Na and Cl. Both are in Period 3. Sodium is in Group 1 and Chlorine is in Group 17.
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Step 2: Compare their nuclear charge. Chlorine has 17 protons in its nucleus, while sodium has 11 protons. Therefore, chlorine has a greater nuclear charge.
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Step 3: Compare their electron arrangement. In both atoms, the outermost electrons are in the 3rd principal energy shell.
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Step 4: Relate nuclear charge and shielding to atomic radius. The increased nuclear charge in chlorine attracts the electrons in the 3rd shell more strongly than in sodium. The shielding from inner shell electrons is similar for both. This stronger attraction pulls the electron cloud closer to the nucleus, resulting in a smaller atomic radius for chlorine.
Worked example 23 marks
Explain why a magnesium ion (Mg²⁺) is significantly smaller than a magnesium atom (Mg).
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Step 1: Compare the number of protons and electrons. A Mg atom has 12 protons and 12 electrons (2.8.2). A Mg²⁺ ion has 12 protons and 10 electrons (2.8).
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Step 2: Identify the change in electron shells. The Mg atom has electrons in 3 shells, while the Mg²⁺ ion has electrons in only 2 shells. The entire outer shell has been removed.
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Step 3: Compare the electrostatic attraction. The 12 protons in the Mg²⁺ ion are now attracting only 10 electrons, leading to a much stronger effective pull on each remaining electron compared to the neutral atom. This pulls the remaining electron shells closer to the nucleus.
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Step 4: Conclude. The combination of losing the outermost shell and the increased attraction on the remaining electrons makes the Mg²⁺ ion much smaller than the Mg atom.
Recap
- Atomic radius decreases across Period 3.
- This is due to increasing nuclear charge attracting electrons in the same outer shell more strongly.
- Shielding effect from inner shells is relatively constant across the period.
- Positive ions (cations) are much smaller than their parent atoms.
- Negative ions (anions) are larger than their parent atoms.
Quick check
- Which particle is larger: a sulfur atom (S) or a sulfide ion (S²⁻)?1 mark
- Arrange Na⁺, Mg²⁺, and Al³⁺ in order of increasing ionic radius.1 mark