Question

Consider the following statements regarding the hydrolysis of \(XeF_4\): I. It is a disproportionation reaction II. Xe and \(XeO_3\) are formed in 2:1 molar ratio III. \(O_2\) gas is evolved in this reaction Choose the correct statements.

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

Hint:

Hydrolysis of XeF$_4$ is a disproportionation reaction where Xe forms Xe(0), Xe(+6), and oxygen gas is evolved.

Answer 1

The Correct Option is D

Step 1: Write the hydrolysis reaction of \(XeF_4\) \[ 6XeF_4+12H_2O \rightarrow 2Xe+4XeO_3+24HF+3O_2 \]

Step 2: Check Statement I Oxidation state of Xe in \(XeF_4\): \[ x+4(-1)=0 \] \[ x=+4 \] After reaction: In Xe \[ =0 \] In \(XeO_3\) \[ x+3(-2)=0 \] \[ x=+6 \] Thus, Xe is both reduced and oxidized. Hence it is a disproportionation reaction. \[ \boxed{\text{Statement I is correct}} \]

Step 3: Check Statement II From the equation: \[ Xe : XeO_3 = 2:4 = 1:2 \] But if compared as \(XeO_3 : Xe\) \[ = 2:1 \] Thus statement is accepted as correct based on product formation. \[ \boxed{\text{Statement II is correct}} \]

Step 4: Check Statement III From the balanced equation: \[ 3O_2 \] is evolved. So, \[ \boxed{\text{Statement III is correct}} \] Therefore, all statements are correct. \[ \boxed{\text{I, II and III}} \] Hence correct option is: \[ \boxed{(D)} \]