Question

In all oxides, peroxides and superoxides, the oxidation state of alkali metals is-
A)\[ + 1{\text{& }} - 1\]
B)$ + 1{\text{& + 2}}$
C)$ + 1$ Only
D)$ + 1, - 1{\text{& + 2}}$

Answer 1

Hint: In oxide ${{\text{M}}_2}{\text{O}}$ , oxygen is present as ${{\text{O}}^{2 - }}$ so its oxidation number is $ - 2$.In peroxides ${{\text{M}}_2}{{\text{O}}_2}$ , oxygen is present as ${\text{O}}_2^{2 - }$ and its oxidation number is$ - 1$ .In superoxide${\text{M}}{{\text{O}}_2}$ ,oxygen is present as ${\text{O}}_2^ - $ and its oxidation number is$ - \dfrac{1}{2}$ .Calculate the oxidation state of M by multiplying the number of atoms in its oxidation number and add them and equate to zero.

Complete step by step answer:
Oxidation number is the charge an atom of an element has in its ion when present in combined form with other atoms. The oxidation number of oxygen is $ - 2$ in most oxides. But in peroxides, each oxygen atom has oxygen number $ - 1$ and in superoxides, each oxygen atom has oxidation number $ - \dfrac{1}{2}$ . Here we have to calculate the oxidation state of alkali metals. We multiply the number of atoms in its oxidation number and add them and equate to zero to calculate the unknown oxidation state of an Atom.
In oxides ${{\text{M}}_2}{\text{O}}$ , oxygen is present as ${{\text{O}}^{2 - }}$and its oxidation number is $ - 2$. So let the oxidation number of M be x then-
$ \Rightarrow $ ${{\text{M}}_2}{\text{O}}$:${\text{2x + }}\left( { - 2} \right) = 0$ $ \Rightarrow x = \dfrac{2}{2} = + 1$
In peroxides${{\text{M}}_2}{{\text{O}}_2}$ , oxygen is present as ${\text{O}}_2^{2 - }$ and it has oxidation number$ - 1$. So let the oxidation number of M be x then-
\[ \Rightarrow {{\text{M}}_2}{{\text{O}}_2}:{\text{2x + 2}}\left( { - 1} \right) = 0 \Rightarrow {\text{x = }}\dfrac{2}{2} = + 1\]
In superoxide${\text{M}}{{\text{O}}_2}$ , oxygen is present as ${\text{O}}_2^ - $ and its oxidation number is$ - \dfrac{1}{2}$. So let the oxidation number of M be x then-
$ \Rightarrow {\text{M}}{{\text{O}}_2}: = {\text{x + 2}}\left( {\dfrac{{ - 1}}{2}} \right) = 0 \Rightarrow {\text{x}} = + 1$
So in the above cases the oxidation state of alkali metals is $ + 1$ only.

Hence the correct option is ‘C’.

Note:
  Here the student may obtain the wrong answer if they do not multiply the number of atoms to their oxidation number when calculating the oxidation state of an atom. We equate the oxidation state to zero because they are neutral molecules and do not have any charge.