Oxidation number of rubidium in Rb2O,Rb2O2 and RbO2 , can be given respectively, as

Question

Oxidation number of rubidium in Rb2O,Rb2O2 and RbO2 , can be given respectively, as (A) +1,+4 and +2 (B) +2,+1 and +(1/2) (C) +1,+1 and +1 (D) +1,+2 and +4

Tardigrade Admin · Accepted Answer

Let oxidation number of Rb as x underset2 x - 2 = 0R b2 O arrow x=+1 underset2 x - 2 = 0R b2 O2(.peroxide.) arrow x=+1 In peroxide oxidation number of oxygen is taken as -1 underset2 x - 2 = 0R b O2(.superoxide.) arrow x=+1 In superoxide oxidation number of oxygen is taken as -(1/2)

Q. Oxidation number of rubidium in $R b_{2} O, R b_{2} O_{2}$ and $R b O_{2}$ , can be given respectively, as

$+ 1, + 4$ and $+ 2$

$+ 2, + 1$ and $+ \frac{1}{2}$

$+ 1, + 1$ and $+ 1$

$+ 1, + 2$ and $+ 4$

Q. Oxidation number of rubidium in Rb2​O,Rb2​O2​ and RbO2​ , can be given respectively, as

+1,+4 and +2

+2,+1 and +21​

+1,+1 and +1

+1,+2 and +4

Let oxidation number of Rb as x 2x−2=0Rb2​O​→x=+1 2x−2=0Rb2​O2​​(peroxide)→x=+1 In peroxide oxidation number of oxygen is taken as −1 2x−2=0RbO2​​(superoxide)→x=+1 In superoxide oxidation number of oxygen is taken as −21​

Q. Oxidation number of rubidium in $R b_{2} O, R b_{2} O_{2}$ and $R b O_{2}$ , can be given respectively, as

$+ 1, + 4$ and $+ 2$

$+ 2, + 1$ and $+ \frac{1}{2}$

$+ 1, + 1$ and $+ 1$

$+ 1, + 2$ and $+ 4$