Question

Oxide, peroxide, and superoxide – all the three can be formed by:
A) \[Li,K\]
B) \[Na,K\]
C) \[K,Rb\]
D) \[Na,Cs\]

Answer 1

Hint: As we move down the group in the periodic table, the atomic radius increases. It is known that the superoxide ion \[\left( {{O_2}^ - } \right)\] generally remains stable only in the presence of large cations having high lattice energy.

Complete step by step answer:
The group 1 of the modern periodic table consists of alkali metals.
There are seven alkali metals known till now and from top to bottom they are represented as:
Lithium \[\left( {Li} \right)\], Sodium \[\left( {Na} \right)\], Potassium \[\left( K \right)\], Rubidium \[\left( {Rb} \right)\], Cesium \[\left( {Cs} \right)\], Francium \[\left( {Fr} \right)\].
The oxidation state of alkali metals is \[{\text{ + 1}}\].
 The order of ionic radii of the alkali metal is given as follows.
\[F{r^ + } > C{s^ + } > R{b^ + } > {K^ + } > N{a^ + } > L{i^ + }\]
We know that the alkali metal tarnishes in the air because of the formation of oxides.
We have seen alkali metal burning very fast in the presence of oxygen and this leads to the formation of different types of oxides.
Now, lithium with the reaction of oxygen only forms a monoxide product which can be represented as follows.
\[4Li + {O_2} \to 2L{i_2}O\]
In monoxide, the oxidation state of the oxygen atom is \[{O^{2 - }}\].
Now, sodium with the reaction of oxygen only forms a peroxide product which can be represented as follows.
\[2Na + {O_2} \to N{a_2}{O_2}\]
In peroxide, the oxidation state of the oxygen atom is \[{O^ - }\].
Now, Potassium \[\left( K \right)\], Rubidium \[\left( {Rb} \right)\], Cesium \[\left( {Cs} \right)\] with the reaction of oxygen leads to the formation of superoxide which can be represented as:
\[M + {O_2} \to M{O_2}\]
Here, \[M = K,Rb,Cs\].
In superoxide, the oxidation state of the oxygen molecule is \[{O_2}^ - \].
But Potassium \[\left( K \right)\] and Rubidium \[\left( {Rb} \right)\] can also form oxide and peroxide along with the formation of superoxide.

Therefore, we can conclude that the correct answer to this question is option C.

Note:
The Cesium \[\left( {Cs} \right)\] forms superoxide with the reaction of oxygen but it cannot form monoxide or peroxide to very high ionic radii and high lattice energy. In this type of question, we can eliminate options because lithium and sodium strictly form monoxide and peroxide respectively.