A s-block element (M) reacts with oxygen to f | vedclass

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Question

(B) The reaction of alkali metals with excess oxygen leads to the formation of superoxides $(MO_{2})$.Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)

Vedclass · Accepted Answer

$K$

Vedclass · Answer

(B) The reaction of alkali metals with excess oxygen leads to the formation of superoxides $(MO_{2})$.Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)$ form superoxides ($KO_{2}$,$RbO_{2}$,$CsO_{2}$) which are paramagnetic due to the presence of one unpaired electron in the $\pi^* 2p$ molecular orbital of the superoxide ion $(O_{2}^{-})$.$KO_{2}$ is a pale yellow solid.Sodium $(Na)$ forms a peroxide $(Na_{2}O_{2})$,which is diamagnetic and white.Calcium $(Ca)$ and Magnesium $(Mg)$ form oxides ($CaO$,$MgO$) or peroxides $(CaO_{2})$,which are diamagnetic.Therefore,the element $(M)$ is $K$.

$A$ $s$-block element $(M)$ reacts with oxygen to form an oxide of the formula $MO_{2}$. The oxide is pale yellow in colour and paramagnetic. The element $(M)$ is:

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