Question

Oxidation number of rubidium in $Rb_{2}O,Rb_{2}O_{2}$ and $RbO_{2}$ , can be given respectively, as

• A. $+1,+4$ and $+2$
• B. $+2,+1$ and $+\frac{1}{2}$
• C. $+1,+1$ and $+1$
• D. $+1,+2$ and $+4$

Answer 1

Answer: (C)

Let oxidation number of Rb as x

$\underset{2 x - 2 = 0}{R b_{2} O} \rightarrow x=+1$

$\underset{2 x - 2 = 0}{R b_{2} O_{2}}\left(\right.peroxide\left.\right) \rightarrow x=+1$

In peroxide oxidation number of oxygen is taken as $-1$

$\underset{2 x - 2 = 0}{R b O_{2}}\left(\right.superoxide\left.\right) \rightarrow x=+1$

In superoxide oxidation number of oxygen is taken as $-\frac{1}{2}$