Question

Oxidation state of Potassium in \[{{\rm{K}}_{\rm{2}}}{\rm{O,}}{{\rm{K}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{,K}}{{\rm{O}}_{\rm{2}}}\] are respectively:
(1) +1,+1,+1
(2) +1,+2, +4
(3) +1,+2,+2
(4) +1,+4,+2

Answer 1

Hint: To undergo the formation of a chemical bond, the number of electrons an atom gains or loses is termed the oxidation state. The oxidation state can be of positive or negative value. It can also be zero.

Complete Step by Step Solution:
Let's understand the various oxidation states of Oxygen. In most compounds, the oxidation state shown by Oxygen is -2, which means, it can accept two electrons. But, there are some exceptions as p-block elements exhibit variable oxidation states.
The exceptions are as follows:
1) Oxygen exhibits oxidation state value of -1 in peroxides, that is\[{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\] .
2) Oxygen exhibits oxidation state value of -1/2 in case of superoxides like\[{{\rm{K}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\] .
3) Oxygen exhibits oxidation state value of +1 in case of \[{{\rm{F}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\] .
Let’s find out oxidation state of element K in \[{{\rm{K}}_{\rm{2}}}{\rm{O}}\] . \[{{\rm{K}}_{\rm{2}}}{\rm{O}}\]is Potassium oxide. So, the oxidation state of O in \[{{\rm{K}}_{\rm{2}}}{\rm{O}}\]is -2. Let’s take x as the oxidation state of element K.
\[ \Rightarrow x \times 2 + \left( { - 2} \right) = 0\]
\[ \Rightarrow 2x - 2 = 0\]
\[ \Rightarrow 2x = 2\]
\[ \Rightarrow x = 1\]
Therefore, oxidation state of element K in \[{{\rm{K}}_{\rm{2}}}{\rm{O}}\]is +1.
Let’s find out the oxidation state of element K in \[{{\rm{K}}_{\rm{2}}}{{\rm{O}}_2}\] . \[{{\rm{K}}_{\rm{2}}}{{\rm{O}}_2}\]is Potassium peroxide. So, the oxidation state of O in \[{{\rm{K}}_{\rm{2}}}{{\rm{O}}_2}\]is -1. Let’s take x as the oxidation state of element K.
\[ \Rightarrow x \times 2 + 2 \times \left( { - 1} \right) = 0\]
\[ \Rightarrow 2x - 2 = 0\]
\[ \Rightarrow 2x = 2\]
\[ \Rightarrow x = 1\]
Therefore, oxidation state of element K in \[{{\rm{K}}_{\rm{2}}}{{\rm{O}}_2}\]is +1.
Let’s find out the oxidation state of element K in \[{\rm{K}}{{\rm{O}}_2}\] . \[{\rm{K}}{{\rm{O}}_2}\]is Potassium superoxide. So, the oxidation state of O in \[{\rm{K}}{{\rm{O}}_2}\]is -1/2. Let’s take x as the oxidation state of element K.
\[ \Rightarrow x + 2 \times \left( { - \dfrac{1}{2}} \right) = 0\]
\[ \Rightarrow x - 1 = 0\]
\[ \Rightarrow x = 1\]
Therefore, the oxidation state of element K in \[{\rm{K}}{{\rm{O}}_2}\]is +1.
Hence, option(1) is the right answer.

Note: The bonding of Oxygen with more electronegative atoms than it, exhibits a positive state of Oxygen. The only electronegative atom other than Oxygen is Fluorine.