Question

What are X and Y in the following reactions? \[ 2X_2(g)+2H_2O(l)\rightarrow4H^+(aq)+4X^-(aq)+O_2(g) \] \[ Y_2(g)+H_2O(l)\rightarrow HY(aq)+HOY(aq) \]

• A. \(X=F,\;Y=I\)
• B. \(X=I,\;Y=Cl\)
• C. \(X=F,\;Y=Cl\)
• D. \(X=Cl,\;Y=I\)

Hint:

Only fluorine oxidizes water to oxygen. Chlorine disproportionates to HCl and HOCl.

Answer 1

The Correct Option is C

Step 1: Read the first reaction.
$2X_2(g)+2H_2O(l)\rightarrow 4H^+(aq)+4X^-(aq)+O_2(g)$. Here the halogen is strong enough to oxidise water and liberate oxygen.
Step 2: Identify X.
Only fluorine, the strongest oxidising halogen, oxidises water to oxygen: $2F_2+2H_2O \rightarrow 4HF+O_2$. So $X=F$.
Step 3: Read the second reaction.
$Y_2(g)+H_2O(l)\rightarrow HY(aq)+HOY(aq)$. Here the halogen disproportionates, forming both a halide acid and a hypohalous acid.
Step 4: Identify Y.
Chlorine disproportionates in water: $Cl_2+H_2O \rightarrow HCl+HOCl$. So $Y=Cl$.
Step 5: Rule out other pairs.
Iodine is too weak to oxidise water to oxygen, so it cannot be X; pairs giving $X=I$ or $Y=I$ for these specific reactions are rejected.
Step 6: Choose the option.
$X=F$ and $Y=Cl$ matches option (3).
\[ \boxed{X=F,\ \ Y=Cl\ \ \text{(Option 3)}} \]