Question

If \[C{{l}_{2}}\], is passed through hot aqueous NaOH, the products formed have Cl in
different oxidation states. These are indicated as:
A.) -l and +1
B.) -l and +5
C.) +l and +5
D.) -l and +3

Answer 1

Hint: Sodium hydroxide and chlorine gas react in different ways according to the temperature and concentration of the solution and it gives different products. The reaction undergoes disproportionation but in presence of hot alkali the halogen that is oxidized goes to a higher oxidation state.

Complete step by step solution:

Oxidation - reduction reaction occurs when hot aqueous NaOH and \[C{{l}_{2}}\] react together and these reactions are more special because they are disproportionation reactions. It results in the formation of sodium chloride, sodium chlorate(iii) and water as products.
The chemical equation is given as:

\[\overset{hot\,aqueous}{\mathop{6NaOH}}\,(aq)\,+\,3C{{l}_{2}}(g)\,\to \,5NaCl(aq)\,+\,NaCl{{O}_{3}}(aq)\,+\,3{{H}_{2}}O(l)\]

We can see that the only thing to have changed is the chlorine. Chlorine gas is oxidized and reduced to \[Cl{{O}_{3}}^{-}\] ion (in the \[NaCl{{O}_{3}}\]) and \[C{{l}^{-}}\]ion (in the NaCl) respectively. It goes from an oxidation 0 in the chlorine molecules on the left-hand side to -1 (in the NaCl) and +5 (in the \[NaCl{{O}_{3}}\]).

Thus, the obtained compounds have Cl in -1 and +5 oxidation states and the correct answer is (b).

Note: The same reaction will give different products when temperature and concentrations are changed. Cold dilute sodium hydroxide will give sodium chloride, sodium chlorate(i) and water as products.