Which of the following is the correct order of stability for the given superoxides?

K O_{2} < R b O_{2} < C s O_{2}

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C s O_{2} < R b O_{2} < K O_{2}

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R b O_{2} < C s O_{2} < K O_{2}

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K O_{2} < C s O_{2} < R b O_{2}

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Solution

The correct option is A $K O_{2} < R b O_{2} < C s O_{2}$
With the progressive increase in the size of alkali metal ions, the stability of superoxides increases because the size of superoxide ion is large and large cation can be stabilised more by large anion.

Therefore, the order of stability is : $K O_{2} < R b O_{2} < C s O_{2}$ .

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Similar questions

Q. Assertion :The oxidation state of superoxide ion in

{K O}_{2}, {C s O}_{2}

and

{R b O}_{2}

- \frac{1}{2}

. Reason: Since the oxidation state of an alkali metal in any compound is always

+ 1

, the oxidation state of oxygen is

\frac{1}{2}

in the

O_{2}^{-}

ion.

Read the above assertion and reason and choose the correct option regarding it.

Q. The oxidation state of Na, K, Rb respectively in

N a_{2} O_{2}, K O_{2}, R b O_{2}

is :

$R b O_{2}$ is

The stability of $K_{2} O$ , $K_{2} O_{2}$ and $K O_{2}$ is in order $K_{2} O < K_{2} O_{2} < K O_{2}$ . The increasing stability at the size of metal ion increases is due to stabilization of:

Q. Potassium superoxide (

{K O}_{2}

) is used in rebreathing gas masks to generate oxygen:

{K O}_{2} (s) + H_{2} O (l) ⟶ K O H (s) + O_{2} (g)

If a reaction vessel contains 0.158 mol

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Which of the following is the correct order of stability for the given superoxides?

The correct option is A KO2<RbO2<CsO2With the progressive increase in the size of alkali metal ions, the stability of superoxides increases because the size of superoxide ion is large and large cation can be stabilised more by large anion.Therefore, the order of stability is : KO2<RbO2<CsO2.