Question

Which halogen does not show variable oxidation state?
A. ${F_2}$
B. $C{l_2}$
C. $B{r_2}$
D. ${I_2}$

Answer 1

Hint: An element's oxidation state is indicated by a number that is assigned to it to indicate how many electrons have been lost from it or consumed by it. Both positive and negative oxidation states are possible.

Complete Step by Step Solution:
Fluorine lacks d orbitals in its valence shell and is the most electro-negative element. It is exceedingly challenging to connect a more electronegative element with fluorine in order for it to display a positive oxidation state.

The lightest halogen, fluorine, is a very poisonous, pale yellow diatomic gas under normal conditions. It is the chemical element that is most reactive. It also has a strong affinity for electrons. With the exception of minute amounts of the free element in radium-irradiated fluorspar, fluorine is solely present in nature as chemical compounds.

Since F is the most electronegative element, it cannot possibly share an electron with another element that is more electronegative than it. Its great capacity to attract electrons (it is the most electronegative element) and the small size of its atoms are both responsible for its high level of chemical activity. Fluorine does not exhibit positive oxidation states due to this. Fluorine is the most electronegative atom so it accepts electrons very easily and shows only -1 oxidation state.
Hence the correct option is A.

Note: Excitation of outer s and p-orbitals into d-orbitals makes it possible to achieve positive oxidation states by making 3, 5, or 7 unpaired electrons readily available for bonding. Higher oxidation states are not capable of being displayed by elements without unoccupied d orbitals.