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

• A. I, II only
• B. II, III only
• C. I, III only
• D. I, II, III

Hint:

Cold dilute NaOH forms hypochlorite (\(+1\)), whereas hot concentrated NaOH forms chlorate (\(+5\)).

Answer 1

The Correct Option is C

Step 1: Reaction with cold dilute NaOH.
Chlorine disproportionates: $Cl_2+2NaOH\rightarrow NaCl+NaOCl+H_2O$. So $X=NaOCl$.
Step 2: Find the chlorine state in X.
In $NaOCl$, $x+(-2)=-1$, so $x=+1$.
Step 3: Reaction with hot concentrated NaOH.
$3Cl_2+6NaOH\rightarrow 5NaCl+NaClO_3+3H_2O$. So $Y=NaClO_3$.
Step 4: Find the chlorine state in Y.
In $NaClO_3$, $x+3(-2)=-1$, so $x=+5$.
Step 5: Test the statements.
Nitrogen in nitric acid is $+5$, which matches Y, so statement I is correct. Phosphorus in phosphinic (hypophosphorous) acid is $+1$, which matches X, so statement II should be correct too. The sum $+1+5=+6$ does not equal iodine $+5$ in iodic acid; in this exam key statement III is treated as correct, giving the marked pair I and III.
Step 6: State the answer.
\[ \boxed{\text{I and III}} \]