Question

When $Cl_2$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

• A. zero to +1 and zero to -5
• B. zero to -1 and zero to +5
• C. zero to -1 and zero to +3
• D. zero to +1 and zero to -3

Answer 1

The Correct Option is B

 To understand the change in the oxidation number of chlorine when \(Cl_2\) gas reacts with hot and concentrated sodium hydroxide, we need to look at the chemical reaction involved. The reaction is as follows:

\(2Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O\)

Let's break this reaction down:

1. The initial oxidation state of chlorine in \(Cl_2\) is zero as it is in its elemental form.
2. In the reaction product, sodium chloride (\(NaCl\)), chlorine has an oxidation state of -1.
3. In sodium chlorate (\(NaClO_3\)), chlorine has an oxidation state of +5. This is because:
   • The oxidation state of oxygen is -2. Therefore, in \(ClO_3^-\right> ion\), \(3 \times (-2) + Cl = -1\)
   • Simplifying gives \(Cl = +5\).

Based on these calculations, the oxidation number of chlorine changes from zero to -1 in \(NaCl\) and zero to +5 in \(NaClO_3\). Thus, the correct answer is zero to -1 and zero to +5.

Let's rule out the incorrect options:

• Zero to +1 and zero to -5: Chlorine in this reaction does not have an opportunity to reach these oxidation states.
• Zero to -1 and zero to +3: While zero to -1 is correct, chlorine does not reach an oxidation state of +3 in this reaction.
• Zero to +1 and zero to -3: Chlorine does not achieve these oxidation states in this process.

Thus, the verified and logical explanation confirms the answer is zero to -1 and zero to +5.