Question:
Chlorine gas reacts with cold, dilute NaOH to produce NaCl, H\(_2\)O and (X). With hot, concentrated NaOH, it produces NaCl, H\(_2\)O and (Y). Identify the correct statements regarding the oxidation states of chlorine in X and Y.
• The oxidation state of chlorine in Y is same as that of nitrogen in nitric acid
• The oxidation state of chlorine in X is same as that of phosphorus in phosphinic acid
• The sum of the oxidation states of chlorine in X and Y is same as that of iodine in iodic acid
Chlorine gas reacts with cold, dilute NaOH to produce NaCl, H\(_2\)O and (X). With hot, concentrated NaOH, it produces NaCl, H\(_2\)O and (Y). Identify the correct statements regarding the oxidation states of chlorine in X and Y.
• The oxidation state of chlorine in Y is same as that of nitrogen in nitric acid
• The oxidation state of chlorine in X is same as that of phosphorus in phosphinic acid
• The sum of the oxidation states of chlorine in X and Y is same as that of iodine in iodic acid
• The oxidation state of chlorine in Y is same as that of nitrogen in nitric acid
• The oxidation state of chlorine in X is same as that of phosphorus in phosphinic acid
• The sum of the oxidation states of chlorine in X and Y is same as that of iodine in iodic acid
Show Hint
Cold dilute NaOH forms hypochlorite (\(+1\)), whereas hot concentrated NaOH forms chlorate (\(+5\)).
Updated On: Jun 8, 2026
- I, II only
- II, III only
- I, III only
- I, II, III
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Chlorine undergoes disproportionation reaction with sodium hydroxide. The products depend upon temperature and concentration conditions.
Step 1: Reaction with cold, dilute NaOH.
\[ Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O \] Thus: \[ X = NaOCl \] Oxidation state of chlorine in \(NaOCl\): \[ x + (-2) = -1 \] \[ x = +1 \]
Step 2: Reaction with hot, concentrated NaOH.
\[ 3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O \] Thus: \[ Y = NaClO_3 \] Oxidation state of chlorine in \(NaClO_3\): \[ x + 3(-2) = -1 \] \[ x = +5 \]
Step 3: Evaluate the statements.
The oxidation states obtained are: \[ X : +1,\qquad Y : +5 \] Therefore, statements I and III are correct. Hence, the correct option is: \[ \boxed{(C)} \]
Step 1: Reaction with cold, dilute NaOH.
\[ Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O \] Thus: \[ X = NaOCl \] Oxidation state of chlorine in \(NaOCl\): \[ x + (-2) = -1 \] \[ x = +1 \]
Step 2: Reaction with hot, concentrated NaOH.
\[ 3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O \] Thus: \[ Y = NaClO_3 \] Oxidation state of chlorine in \(NaClO_3\): \[ x + 3(-2) = -1 \] \[ x = +5 \]
Step 3: Evaluate the statements.
The oxidation states obtained are: \[ X : +1,\qquad Y : +5 \] Therefore, statements I and III are correct. Hence, the correct option is: \[ \boxed{(C)} \]
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