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# Select the smallest atom${A.{\text{ }}F} \\ {B.{\text{ }}Cl} \\ {C.{\text{ }}Br} \\ {D.{\text{ }}I}$

Last updated date: 20th Jun 2024
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Hint: We must know that halogens are non-metal in group 17. The size of atoms of the elements increases when we go down the group. Fluorine as a diatomic molecule has the weakest bond due to the repulsion of the small atoms between electrons. Also we must remember that the fluorine is at ${2^{nd}}$ period, chlorine is at ${3^{rd}}$, bromine is at $\;{4^{th}}$ and Iodine is at $\;{5^{th}}$.

While determining the size of an atom, its place on the periodic table is the most important factor that one should keep in mind.
Now when we move down a group in the periodic table, the atomic size gradually increases. This increase is because one shell is added as we move down each group. So an atom that is of a group that is lower on the periodic table will be larger than that is on a group above.
Next, while moving left to right, the size of an atom also tends to decrease. This decrease is mainly noticed in the elements that belong to the main group. While the number of shells is not affected in any way, the total electronegativity of the atom increases. Due to this, the electrons that are on the outer shells get pulled closer. This leads to an overall decrease in the size of the atom.
Now, applying this factor we can have a look at the position of the given atoms.
Comparing $F$ and $Cl$, as $Cl$ is in a group that is lower and hence it is larger than $F$.
Next we compare $Cl$ and $Br$, $Br$ is in a group that is lower than $Cl$. Thus it is larger than $Cl$.
When it comes to $Br$ and$I$, $I$ is in a group that is lower than $Br$. Thus, it is larger than Br.
We can now state that Fluorine is the smallest, followed by Chlorine, then Bromine, and finally Iodine.

So, the answer to this question is A.

Note: As the size of the atom increases its electron nucleus interaction/bonding reduces. Due to this, it becomes easier for an atom to lose its electron during bond formation. There has been a case where iodine forms bonds with fluorine and chlorine in which iodine acts as an electron donor and the other two act as an acceptor.
We must remember that the electron affinity is the energy which is released when electrons attach to the atom or nucleus. In the halogen group, the electron affinity order is $(At{\text{ }} < {\text{ }}I{\text{ }} < {\text{ }}Br{\text{ }} < \;F < \;Cl)$. This is because fluorine has small size compared to chlorine.