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The oxidation state of potassium in ${K_2}O$ is _ _ _ _ _ _?
                    (A) +1
                    (B) -1
                    (C) +2
                    (D) -2

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Last updated date: 20th Jun 2024
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Answer
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Hint: The sum of the oxidation states of the constituent ions of any compound is equal to the oxidation state or charge of the entire compound. If not specified or written then the oxidation state of the compound is zero.

Complete answer:
-The oxidation state is defined as the total number of electrons that can be lost or gained by any atom to form bonds or to form any complex. If due to a reaction the oxidation state increases then it is known as oxidation and if the oxidation state decreases then it is known as reduction.
-The oxidation state or charge of the entire compound as a whole is equal to the sum of the oxidation states of its constituent ions.
So, now we will find out the oxidation state of K in ${K_2}O$.
Here the oxidation state of the entire compound is 0 and that of O atom is (-2). Let the oxidation state of K be ‘x’.
                                                    0 = 2(x) + (-2)
                                                    2 = 2x
                                                    x = 1
So, the oxidation state of K in ${K_2}O$ is (+1). Also K is an alkali metal who always shows (+1) oxidation state.
-Potassium oxide (${K_2}O$) is a pale yellow solid ionic compound and also highly reactive. It is a basic oxide. It is also deliquescent in nature and absorbs water from the atmosphere. It is highly useful in making fertilizers, cement and glassmaking. It is used in some medical processes as well.
-${K_2}O$ is formed by reacting potassium and oxygen and then treating the resulting potassium peroxide with metallic potassium. It can be shown in the form of following reaction:
                                              $2K + {K_2}{O_2} \to 2{K_2}O$
So, the correct answer is “Option A”.

Note: Potassium (K) belongs to the 1st group of the periodic tables, they are the alkali metals. They have an electronic configuration having 1 electron in the outermost shell ($n{s^1}$) and so they easily give up 1 electron to attain a noble gas configuration. So, they always show (+1) oxidation state in all compounds.