Question

The element which never has positive oxidation state in any of its compounds is:
A. Chlorine
B. Bromine
C. Oxygen
D. Fluorine

Answer 1

Hint: Oxidation state is also known as oxidation number. It defines the ability to transfer electrons by an element for the formation of bonds between the atoms. Oxidation state is of two types-positive and negative.

Complete step by step answer:
When the oxidation number or state is positive, it indicates that the element loses its electrons. When the oxidation number or state is negative, it indicates that the element gains electrons. This is based on the electronegativity of the element. Electronegativity is the ability of an atom to pull electrons towards itself.
Now let’s focus on the options. Three of the given options are halogens. They are chlorine, bromine and fluorine. When the electronegativity of given elements are considered, fluorine has more electronegativity than chlorine, bromine and oxygen. When we move left to right of the periodic table, electronegativity is increased and when we move down the group, electronegativity is decreased. When certain elements are combined with more electronegative elements, then the element with low electronegativity has positive oxidation state and that with high electronegativity has negative oxidation state.
Generally, all of them have a negative oxidation state. But in certain cases, it possesses a positive oxidation state. For example, when chlorine is combined with a more electronegative element like fluorine, it possesses $ + 1$ oxidation state. This is similar to bromine and fluorine. But in the case of oxygen, generally, its oxidation state is $ - 2$. But when it is combined with fluorine to form ${{{F}}_2}{{O}}$, the oxidation state of oxygen is $ + 2$.
Thus the element which will not have positive oxidation state is fluorine.

Hence, option D is the correct answer.

Note: The electronic configuration of fluorine is $2,7$. It has the ability to accept one electron to form bonds. So its oxidation state is $ - 1$. While electronic configuration of oxygen is $2,6$. It can accept two electrons. Thus the oxidation state is $ - 2$. Fluorine is very small in size and thus has very high electronegativity. This is because it can accept electrons very easily and thereby attain octet.