Question

The oxidation number of hydrogen varies from:
(A) $ - 1$ to $ + 1$
(B) $0$ to$ + 1$
(C) $ - 1$ to $0$
(D) $0$ to $ + 2$

Answer 1

Hint: Oxidation number of an element is defined as the charge which an atom of the element has in its ion or appears to have when present in combined state with other atoms. Hydrogen shows both positive and negative oxidation numbers.

Complete step by step answer:
Let us first understand about oxidation numbers. Oxidation number of an element may be defined as the charge which an Atom of the element has in its ion or appears to have when present in the combined state with other atoms. Oxidation numbers are also called oxidation. There are few rules that have to be applied to determine the oxidation number of an atom in an ion or molecule.
(a) The oxidation state of all atoms of different elements in their respective elementary states and allotropic forms is taken to be zero. For example;\[{N_2},{\text{ }}C{l_2},{\text{ }}{H_2},{\text{ }}He,{\text{ }}{P_4},{\text{ }}B{r_2}\] etc; oxidation number of each atom is zero.
(b) The oxidation number of a monatomic ion is the same as the charge on it. Example, \[N{a^ + },{\text{ }}M{g^{2 + }}\] and \[A{l^{3 + }}\] has oxidation number \[ + 1,{\text{ }} + 2,{\text{ }} + 3\] respectively.
(c) The oxidation number of oxygen is \[ - 2\] in most of its compounds except in peroxides.
(d) In a compound formed by the union of metal with non metal, the metal atom has a positive oxidation number and the nonmetal atom will have a negative oxidation number.
(e) In neutral compounds, the sum of the oxidation number of all atoms is zero.
(f) In complex ions, the sum of the oxidation number of all the atoms in the ion is equal to the charge on the ion.
(g) Moreover, the oxidation number of hydrogen is \[ + 1\] when combined with non metals and is \[ - 1\] when combined with active metals.
Hence, we got to know that hydrogen shows the oxidation number of both $ - 1$ and $ + 1$. Moreover, it shows an oxidation number of 0 when present in their elementary state of $H_2$. So it’s oxidation number varies from $ - 1$ to $ + 1$.

So, the correct answer is Option A.

Note: Just as hydrogen, oxygen generally show the oxidation state of \[ - 2\] but in a peroxide compound, it shows the oxidation state of \[ - 1\]. Since being an electronegative element, oxygen did not casually show positive oxidation state but showed positive oxidation state when bonded with fluorine.