Question #20b55
1 Answer
(smallest atomic radii) H, Li, Na, K, Rb, Cs, Fr (largest atomic radii)
(smallest atomic radii) Ne, F, O, N, C, B, Be, Li (largest atomic radii)
Explanation:
The first group of elements increases as you move down on the periodic table due to the primary quantum number (the distance of the valence shell from the nucleus) increasing with each period. The effective nuclear charge (the attractive force felt by the valence shell) is the same for each atom, +1 charge from the nucleus (the difference between the core electrons and the nucleus) is the attractive force applied to the valence shell.
The second Period elements are interesting, the valence shell for each element has the same primary quantum number (2), but the effective nuclear charge appied to that shell increases as you move from left to right. Fluorine, despite having more protons and neutrons and electron than any other element in period 2, has the greatest attractive force being applied to those electrons due to the nucleus applying greater attractive force (+9 total, +7 net attraction of valence shell). As a result, the electrons are held closer to the nucleus making the atomic radii smaller when compared to lithium
