Question

Unlike other halogens, Fluoride, does not show higher oxidation states because
A. It is highly electronegative
B. It has no d orbital and only 1 unpaired electron
C. The atomic radius is small
D. Unstable electron and isoelectronic with neon

Answer 1

Hint: 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.

Complete Step by Step Solution:
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. It is a common element that makes up roughly 0.065% of the Earth's crust.

Covalent radii of fluorine atoms are relatively modest. 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.

Fluoride does not exhibit higher oxidation states than other halogens because it lacks d orbitals and only has one unpaired electron.
Therefore, choice B is the best one.

Note: Fluorine is a hazy yellow gas with an unpleasant odour when it is at room temperature. It is harmful to breathe in the gas. Fluorine turns into a yellow liquid when it cools. Fluorine is the most electronegative element, hence fluorine-rich atomic groups frequently have a negative charge.