Explanation of RbO2

Oxidation state of Rb
- The oxidation state of Rb is the charge that Rb would have if all bonds were ionic.
- In RbO2, oxygen is more electronegative than rubidium.
- Therefore, each oxygen atom would have a charge of -2.
- The overall charge of RbO2 is 0, so the sum of the charges of rubidium and oxygen must be 0.
- The oxidation state of Rb can be calculated as follows:
- (-2) x 2 = -4 (for the two oxygen atoms)
- Rb + (-4) = 0
- Rb = +4

RbO2 and its properties
- RbO2 is a dark blue solid at room temperature.
- It is a powerful oxidizing agent and reacts violently with water and other reducing agents.
- It is used in organic synthesis as an oxidant.
- It can also be used in the production of ceramics and as a catalyst.

Conclusion
- The oxidation state of Rb in RbO2 is +1.
- This is because oxygen is more electronegative than rubidium, and each oxygen atom has a charge of -2.
- The sum of the charges of rubidium and oxygen must be 0, so the oxidation state of Rb is calculated to be +1.

RbO2 is a superoxide.......A superoxide has 2 oxygen atoms with a negative charge on one oxygen atom. hence, that oxygen atom needs an e-¹ to fill its valence shell n this work is done by the s orbital of Rb. Hence, it has the O.S +1.hey, I think O2 does not satisfy the octet rule because as we know octet rule states that an atom has to have 8 e- in the outer shell. Oxygen has 6 valence electrons, the bonds should be 8-6=2 bonds. So we need 2 more covalent bonds to form an octet.