Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Explanation

The given statement can be verified by three examples given below:
(I) (i) `underset(underset("Excess")(("reducing agent")))(4Na(s))+ underset(("oxidising agent"))(O_(2)(g)) rarr underset(("Compound of lower O.S"))(Na_(2)overset(-2)(O)(s))`
(ii) `underset(("reducing agent"))(2Na(s)) + underset(underset("Excess")(("oxidising agent")))(2O_(2)(g)) rarr underset(("Compound of higher O.S"))(Na_(2)overset(-1)O_(2) (g))`
In the above example (II), in reaction (i), Na is the reducing agent which is in excess, while in reaction (ii), `O_(2)` is the oxidizing agent which is in excess. Reaction (i) leads to the formation of `Na_(2)O` (O oxidation state= -2,) while reaction (ii) leads to the formation of `Na_(2)O_(2)` (O oxidation state = -1).
(III) (i) `underset(underset("Excess")(("reducing agent")))(P_(4)(s))+ underset(("oxidising agent"))(6Cl_(2)(g)) rarr underset(("Compound of lower O.S"))(overset(+3)(4PCl_(3)(I)))`
(ii) `underset(("reducing agent"))(P_(4)(s)) + underset(underset("Excess")(("oxidising agent"))) (10Cl_(2)(g))rarr underset(("Compound of higher O.S"))(overset(+5)(4PCl_(5)(s)))`
In the above example (III), in reaction (i), P is the reducing agent which is in excess, while in reaction (ii), Cl is the oxidizing agent which is in excess. Reaction (i) leads to the formation of `PCl_(3)` (P oxidation state = +3) while reaction (ii) leads to the formation of `PCl_(5)` (P oxidation state = +5)