Mark the element which shows only one oxidation state in this compound.
(A) F
(B) Cl
(C) Br
(D) I

Hint: If any element has only one electron in its outermost orbital, thus, it can show only one oxidation state after donating that one electron or can say can form only one bond with another atom (valence will be one). But in some elements, the inner electron jumps to an empty orbital and increases its valency (bond with the number of elements), and show various oxidation states.

Complete Step by Step Solution:
To find the possible oxidation state of fluorine, chlorine, bromine, and iodine, their electronic configuration is needed. By electronic configuration, we can get to know if there is any empty outer orbital (in a subshell) present or not.

Atomic number of flour is 9 so its electronic configuration is \[2s{}^\text{2}\text{ }2{{p}^{5}}\]. As last shell of fluorine(\[{{2}^{nd}}\]shell) has no other subshell (except s and p in\[{{2}^{nd}}\]shell), thus, it’s outermost and inner electron do not have any other orbital to jump and so it can show only one oxidation state (\[+1\]) (three orbital in p and filling five electron as per Hund's principle, it is left with only one unpaired electron in its outermost subshell) such as

Whereas chlorine, bromine, and iodine belong to the group of fluorine (halogens), \[{{17}^{th}}\]. Fluorine is first element of the \[{{17}^{th}}\]group, chlorine is the second, bromine is the third, and iodine is the last element of group \[{{17}^{th}}\]. As down the group, the new orbital is introduced thus, the inner electron of chlorine, bromine, and iodine can jump to a new orbital and can exceed its oxidation state.
Thus, the correct option is A.

Note: It is important to note that in the first element of all groups there is no newly added orbital thus its electron can neither jump to a new orbital nor can exceed its oxidation number. While except for the first element of every group of the periodic table, all the other elements can exceed their oxidation state.