Calculate the change in oxidation state of the oxygen, on reaction with the following alkali methals on heating: (a) Li, (b) Na and (c ) Rb.

Explanation

a. `Li underset((excess))+O_(2) overset(Delta)rarr underset("Lithium monoxide")(Li_(2)O)`
Let the oxidation state of O in `Li_(2)O=x`
Since the oxidation state of `Li = + 1`
`:.2(+1)+ x = 0`
`x= -2`
The change in the oxidation state of oxygen is form 0 to -2.
b. `Na_((s))+underset(("excess"))O_(2)overset(Delta)rarr Na_(2)O_(2(s))`
Let the oxidation state of 'O' in `Na_(2)O_(2)` is x.
since the oxidation state of Na is + 1.
`:. 2(+1)+2x = 0`
`x= -1`
The Change in oxidation state of oxygen is form 0 to - 1.
c. `Rd_((s))+ underset((excess))(O_(2))overset(Delta)rarr RbO_(2)(s)`
Let the oxidation state of oxygen in `RbO_(2)` is x. Since the oxidation state of Rb is + 1.
`:. (+1)+2x = 0`
`x= -(1)/(2)`
The oxidation state of O in `RbO_(2)` is `-1//2`. The change in oxidation state of O is form 0 to `-1//2`.