## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 8 - Section 8.6 - Periodic Trends in the Size of Atoms and Effective Nuclear Charge - For More Practice - Page 356: 8.5

#### Answer

The order of decreasing radius is: $Rb \gt Ca \gt Si \gt S \gt F$

#### Work Step by Step

We know, atomic radius increases down a column of the periodic table and decreases across a row of the periodic table. $Rb$ is in the first row and the fifth column of all the given elements, so it has the greatest atomic radius. After that, $Ca$ is in the 2nd row and 4th column, so its atomic radius is the next largest. Then, $Si$ and $S$ are both in the 3rd column, but $S$ is in the 16th row and $Si$ is in the 14th, so the radius of $Si$ is greater. Finally, $F$ is in the 2nd column and has the smallest atomic radius among these elements. So, the order of decreasing radius is: $Rb \gt Ca \gt Si \gt S \gt F$

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