Question

\(\mathrm{XeF_4}\) on reaction with \(\mathrm{O_2F_2}\) at \(143\ \mathrm{K}\) gives a Xenon compound \(A\). On complete hydrolysis, \(A\) gives HF and \(B\). The hybridisation of the central atom in \(B\) is

• A. \(sp^3\)
• B. \(sp^2\)
• C. \(sp^3d\)
• D. \(sp^3d^2\)

Hint:

Remember the sequence \[ \boxed{ \mathrm{XeF_4} \xrightarrow{\mathrm{O_2F_2}} \mathrm{XeOF_4} \xrightarrow{\mathrm{H_2O}} \mathrm{XeO_3} } \] The central Xe atom in \(\mathrm{XeO_3}\) is \[ \boxed{sp^3} \] hybridised.

Answer 1

The Correct Option is A

Step 1: Identify compound \(A\). The reaction \[ \mathrm{XeF_4+O_2F_2} \] at \[ 143\ \mathrm{K} \] produces \[ \boxed{\mathrm{XeOF_4}.} \]

Step 2: Complete hydrolysis of \(A\). Complete hydrolysis of \[ \mathrm{XeOF_4} \] gives \[ \mathrm{XeO_3} \] and HF. Thus, \[ \boxed{B=\mathrm{XeO_3}.} \]

Step 3: Determine the hybridisation. In \[ \mathrm{XeO_3}, \] the central xenon atom has \[ 3\ \text{bond pairs} + 1\ \text{lone pair}=4 \] electron pairs. Hence, the hybridisation is \[ \boxed{sp^3.} \] Thus, \[ \boxed{(A)} \] is the correct answer.