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Of the following elements , which one has the same oxidation state in all of its compounds?
(A) Hydrogen
(B) Fluorine
(C) Carbon
(D) Oxygen

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Last updated date: 20th Jun 2024
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Answer
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Hint:Electronic shell configuration of elements play an important role in determining its oxidation state. Oxidation state of any element also depends on the compound to which it is being added.

Complete answer: Let us first learn about oxidation state.
Oxidation state: The oxidation state of an element in a compound can be defined as the charge acquired by its atom from other atoms on the basis of its electronegativity . An element can show multiple oxidation states. What value of oxidation will a compound show depends on the number of electrons in its valence shell as well as the compound in which it is being added.
The oxidation states shown by various elements-

Hydrogen: It is kept at first position in the periodic table. It shows an oxidation state of +1. Except in case of metal hydrides in which it shows an oxidation state of -1.

Florine: It belongs to group 17 of the periodic table. It has 7 valence electrons . It will receive one electron to receive noble gas configuration. It always shows an oxidation state of -1 irrespective of the compound to which it is added.

Carbon : It belongs to group 14 of the element. It can show multiple oxidation states most common being 4. It can also show +2,+4,-4;

Oxygen : This belongs to group 16 of the periodic table. It shows an oxidation state of -2 in common except for peroxides and superoxides.

Hence, the correct option is fluorine (B).

Note:We need to keep in mind that the most common oxidation state of any compound is dependent on its group number. Oxidation state depends upon the valence electrons.