Question

Mark the smallest atom
(A) F
(B) Cl
(C) Br
(D) I

Answer 1

Hint: In the periodic table, along the period atomic radii of element decrease and effective nuclear charge increases because an electron is usually added to the same orbital, and down the group, size of element increases and effective nuclear charge decreases due to the addition of electron to new orbital.

Complete Step by Step Solution:
The given elements in the option are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). All the elements belong to the \[{{17}^{th}}\]group in the periodic table and are known as halogens. Fluorine is the first element of the \[{{17}^{th}}\], chlorine is the second, bromine is the third, and iodine is the last element.

As down the group, the nuclear charge increases equal to the added electron but in this electron is added to a new orbital. Thus, the nucleus attracts the outermost electron but not effectively (effective nuclear charge decreases). As the nucleus cannot hold the outermost electron tightly so the size of the atom increases down the group. Thus the decreasing trend of atomic radii down the group is given as\[\mathbf{F}\text{ }>\text{ }\mathbf{Cl}>\text{ }\mathbf{Br}\text{ }>\text{ }\mathbf{I}\]
From this, we can conclude that fluorine is the smallest of all other halogens (Cl, Br, and I).
Thus, the correct option is A.

Note: Also, it is easy to get to know the size of all the halogens by their atomic number. The atomic number of fluorine is 9, chlorine has 17 electrons, bromine has 35 electrons, and the atomic number of iodine is 53. As fluorine has less number of electrons in its atom than other halogens, thus its size is smaller than other halogens.