Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Suppose only the two electrons shield the outer electrons of period 2, so . Find for lithium () and fluorine (), and the ratio of their radii that the model predicts. Compare it with the real ratio, . What does the model leave out?
Atomic radius falls across a period, because the nuclear charge grows while new electrons enter the same shell. It grows down a group, because each period adds a shell. A cation is smaller than its atom, an anion larger. In isoelectronic species, more protons mean a smaller particle. The covalent radius is half the distance between two bonded nuclei, and the metallic radius half the distance between neighbouring nuclei in a metal. In a simple model, the radius is inversely proportional to the effective nuclear charge, .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Effective charges. Lithium: . Fluorine: .
Step 2. Model ratio. .
Step 3. Real ratio. .
Step 4. What is missing. Fluorine's seven outer electrons also shield each other a little, so is larger than and smaller than . The model exaggerates the shrinking.
Answer: the model predicts , but the real ratio is . Electrons in the same shell also shield, partly.
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