Which of the following cannot show variable oxidation state?

The correct option is Fluorine.

Step-by-step explanation:

1. Electronic Configuration and Valence Shell:
Fluorine (atomic number 9) has the electronic configuration:
$$1s^{2}2s^{2}2p^5$$
Because Fluorine belongs to the second period, its valence shell has the principal quantum number of n = 2. This shell consists of only the 2s and 2p subshells and lacks vacant d-orbitals (as there is no 2d subshell).

2. Absence of d-orbitals:
Other halogens such as Chlorine, Bromine, and Iodine belong to the third, fourth, and fifth periods respectively, and have vacant d-orbitals in their valence shells. These vacant d-orbitals allow them to excite paired electrons from their s and p orbitals to d-orbitals, enabling them to expand their octets and exhibit variable positive oxidation states (such as +1, +3, +5, and +7). Since Fluorine has no vacant d-orbitals, it cannot excite its electrons to expand its octet.

3. Extreme Electronegativity:
Fluorine is the most electronegative element in the periodic table. As a result, it always attracts shared electrons in a bond towards itself, exhibiting a constant oxidation state of -1 in all of its compounds (and 0 in its elemental diatomic state, F₂). It cannot exhibit any positive oxidation states because no other element is electronegative enough to share electrons away from Fluorine.

Therefore, due to the combined effect of its high electronegativity and the absence of vacant valence d-orbitals, Fluorine cannot show variable oxidation states.