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Chapter 20: Problem 3

Explain why atomic radii decrease very gradually from scandium to copper.

Short Answer

Expert verified
Atomic radii decrease gradually from scandium to copper due to the increase in nuclear charge which pulls the electrons closer to the nucleus, contracting the atomic radius. However, this decrease is gradual because electron-electron repulsion within the 3d and 4s shells slightly counteracts the greater nuclear pull, preventing a dramatic decrease in atomic radius.

Step by step solution

01

Understand Atomic Radius

Atomic radius describes the size of an atom. It is typically measured from the center of the nucleus to the boundary of the surrounding electron cloud. Across a period in the periodic table, atomic radii usually decrease because electrons are added to the same energy level while the nucleus becomes more positively charged due to the addition of protons. The increased positive charge attracts the electrons more strongly, pulling them closer to the nucleus, which decreases the atomic radius.
02

Analyze the Electron Configuration from Scandium to Copper

Beginning with Scandium (Sc), which has the electron configuration [Ar] 3d1 4s2, we move across the period to Copper (Cu), which has the electron configuration [Ar] 3d10 4s1. The number of protons increases from 21 in Sc to 29 in Cu. These additional protons cause an increase in the nuclear charge, pulling the shell electrons towards the nucleus and slightly decreasing the atomic radius.
03

Explain the Gradual Decrease

The decrease is gradual because electron-electron repulsion in the same energy level (3d and 4s shells) counteracts the increased pull of the nucleus. This repulsion prevents the atomic radius from decreasing dramatically. The additional 3d electrons do not completely shield the increased nuclear charge, allowing the atomic radius to decrease, but at a slow rate.

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Most popular questions from this chapter

Complete these statements for the complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{CN}\right]^{2+}\) (a) The term "en" is the abbreviation for ____ (b) The oxidation number of \(\mathrm{Co}\) is _____(c) The coordination number of \(\mathrm{Co}\) is ____ (d) _____ is a bidentate ligand.

As we read across the first-row transition metals from left to right, the +2 oxidation state becomes more stable in comparison with the +3 state. Why is this so?

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