Question
Question: Consider the following reaction: \( {\text{6NaOH + 3C}}{{\text{l}}_{\text{2}}} \to {\text{5NaCl +...
Consider the following reaction:
6NaOH + 3Cl2→5NaCl + A + 3H2O
What is the oxidation number of chlorine in A?
(A) +5
(B) −1
(C) +3
(D) +1
Solution
The oxidation number of an element is equal to the total number of electrons which have been removed from an element or which have added to an element to reach its present state. It can be calculated by adding the oxidation number of each element in the compound. Neutral molecules have total charge/oxidation state equal to zero.
Complete stepwise solution:
There are certain to determine the oxidation number of an element that can be delineated as follows:
The oxidation state of an element in the uncombined state or Free State is equal to zero. This applies regardless to the structure of the element.
The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
The most electronegative element is assigned the negative oxidation state. The less electronegative element is assigned a positive oxidation state.
The oxidation state of chlorine is zero in the reactant side as it exists in the Free State, however in the product side, the chlorine reacts with sodium hydroxide to form sodium chloride and sodium chlorate or NaClO3 . Hence the oxidation state of chlorine in ‘A’ which is sodium chlorate is [1 + x + ( - 2 \times3)] = 0
⇒x=+5 .
So the correct answer is option A, the oxidation state of chlorine in the compound “A” is +5 .
Note:
The oxidation number of an element should not be confused with the valence of the element. Oxidation number can be even zero and of fractional order, however, valance can never be equal to zero. But for certain elements the valence is equal to the oxidation number of the elements, such as the alkali metals and the alkaline earth metals.