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 the following illustrations :
(a) `underset(("reducing agent Excess"))(2C(s))+underset(("oxidising agent"))(O_(2)(g))tounderset(("Compound of lower O.S."))overset(+2)(2CO(g))" "......(i)`
`underset(("reducing agent"))(C(s))+underset(("oxidising angent Excess"))(O_(2)(g))tounderset(("Compound of higher O.S."))overset(+4)(CO_(2)(g))" "......(ii)`
In reaction (i), the reducing agent, i.e., carbon is in excess, whereas in reaction (ii) the oxidising agent, i.e., `O_(2)` is in excess. Reaction (i) leads to the formation of CO (O.S. of C = + 2), while reaction (ii) leads to the formation of `CO_(2)` (O.S. of C=+4)
(b) `underset(("reducing agent Excess"))(4Na(s))+underset(("oxidising agent"))(O_(2)(g))tounderset(("Compound of lower O.S."))(Na_(2)overset(-2)(O)(s))" "......(i)`
`underset(("reducing agent"))(2Na(s))+underset(("oxidising agent Excess"))(2O_(2)(g))tounderset(("Compound of higher O.S."))(Na_(2)overset(-1)(O_(2))(g))" ".......(ii)`
(c ) `underset(("reducing agent Excess"))(P_(4)(s))+underset(("oxidising agent"))(6Cl_(2)(g))tounderset("Compound of lower O.S.")(4overset(+3)(P)Cl_(3)(l))" ".......(ii)`
`underset(("reducing agent"))(P_(4)(s))+underset(("oxidising agent Excess"))(10Cl_(2)(g))tounderset("Compound or higher O.S.")(4overset(+5)(P)Cl_(5)(s))" "........(ii)`