Identify periodic trends

Atomic radius:

• decreases across a period from left to right due to increasing effective nuclear charge.
• increases down a group because new shells are added.

Compare the elements in Period 2

The elements are:

• $\text{C}$, $\text{O}$, $\text{F}$

In Period 2, moving left to right: $\text{C} > \text{O} > \text{F}$ So, increasing order among these three is: $\text{F} < \text{O} < \text{C}$

Compare the halogens down the group

The elements are:

• $\text{F}$, $\text{Cl}$, $\text{Br}$

Down Group 17, atomic radius increases: $\text{F} < \text{Cl} < \text{Br}$

Combine all comparisons

Since $\text{C}$ is in Period 2 and lies to the left of $\text{F}$ and $\text{O}$, it is larger than both $\text{F}$ and $\text{O}$. Also, atoms in lower periods are generally larger, so: $\text{C} < \text{Cl} < \text{Br}$

Therefore, the overall increasing order is: $\boxed{\text{F} < \text{O} < \text{C} < \text{Cl} < \text{Br}}$

Match with options

Given labels:

• $(a)=\text{C}$
• $(b)=\text{O}$
• $(c)=\text{F}$
• $(d)=\text{Cl}$
• $(e)=\text{Br}$

So the order becomes: $(c) < (b) < (a) < (d) < (e)$

This corresponds to Option B.