Animated Solution for Chemistry - Periodicity in Properties: The increasing order of the atomic radii of the following elements is (A) C (B) O (C) F (D) Cl (E) Br

Mastering Atomic Radii

The periodic table is not just a list of elements; it is a beautifully organized map of atomic properties. One of the most fundamental properties you can read from this map is the atomic radius—the size of an atom. In this problem, we are tasked with arranging Carbon (C), Oxygen (O), Fluorine (F), Chlorine (Cl), and Bromine (Br) in increasing order of their atomic radii. To do this, we need to understand two competing forces: the pull of the nucleus and the expansion of electron shells.

Moving Across a Period

Let's first look at the elements in the second period: Carbon, Oxygen, and Fluorine. As we move from left to right across a period, we are adding protons to the nucleus and electrons to the same outermost shell.

Because the electrons are entering the same shell, they do not shield each other very effectively from the growing positive charge of the nucleus. This results in an increase in the Effective Nuclear Charge ($Z_{\text{eff}}$). The stronger nuclear pull draws the electron cloud closer, causing the atom to shrink. Therefore, across a period, atomic radius decreases.

Applying this logic, Carbon (Group 14) is the largest of the three, Oxygen (Group 16) is smaller, and Fluorine (Group 17) is the smallest. Mathematically, we can write this as:

Moving Down a Group

Now, let's shift our focus to the vertical columns, specifically Group 17 (the halogens), which contains Fluorine, Chlorine, and Bromine. As we move down a group, something dramatic happens: we add entirely new principal electron shells ($n$).

Even though the nuclear charge is increasing significantly as we go down, the addition of a new shell places the outermost electrons much further away from the nucleus. The inner shells also provide a strong shielding effect, canceling out much of the increased nuclear pull. The sheer physical space taken up by a new shell dominates the trend. Therefore, down a group, atomic radius increases.

Comparing our halogens, Fluorine (Period 2) is the smallest, Chlorine (Period 3) is larger, and Bromine (Period 4) is the largest.

Combining the Trends

To find the final order, we must synthesize our findings. The elements in Period 2 (C, O, F) only have two electron shells, making them inherently smaller than elements in Period 3 (Cl) and Period 4 (Br), which have three and four shells, respectively.

So, Fluorine is the absolute smallest element in our list. It is followed by Oxygen, and then Carbon. After the Period 2 elements, we move to the larger Period 3 element, Chlorine, and finally the largest Period 4 element, Bromine.

Our final increasing order of atomic radii is:

Mapping this back to the letters given in the question: (C) < (B) < (A) < (D) < (E). This perfectly matches option (b). By mastering the interplay between effective nuclear charge and principal quantum shells, you can confidently navigate any atomic radius problem!