Question

Observe the following reactions:
XeF$_6$ + H$_2$O $\rightarrow$ A + 2HF
XeF$_6$ + 2H$_2$O $\rightarrow$ B + 4HF
Hybridization of central atom in A, B respectively is:

• A. sp$^3$d, sp$^3$d
• B. sp$^3$d$^2$, sp$^3$d
• C. sp$^3$d, sp$^3$
• D. sp$^3$d$^2$, sp$^3$

Hint:

Hydrolysis pattern of XeF$_6$:
• XeF$_6$ + H$_2$O $\rightarrow$ XeOF$_4$ (SN 6)
• XeF$_6$ + 2H$_2$O $\rightarrow$ XeO$_2$F$_2$ (SN 5)
• XeF$_6$ + 3H$_2$O $\rightarrow$ XeO$_3$ (SN 4) Each step reduces steric number by 1.

Answer 1

The Correct Option is B

Concept: This question deals with the partial hydrolysis of Xenon hexafluoride (XeF$_6$) and the subsequent determination of the steric numbers and molecular geometries using the VSEPR theory. Xenon (Xe) is a noble gas belonging to Group 18 and carries 8 valence electrons in its outermost shell. When computing hybridization, we calculate the steric number: \[ SN = (\text{number of } \sigma\text{-bonds}) + (\text{number of lone pairs}) \]
• SN = 4 $\Rightarrow$ sp$^3$
• SN = 5 $\Rightarrow$ sp$^3$d
• SN = 6 $\Rightarrow$ sp$^3$d$^2$

Step 1: Compound A (XeOF$_4$). XeF$_6$ + H$_2$O $\rightarrow$ XeOF$_4$ + 2HF
• Xe forms 4 Xe–F bonds and 1 Xe=O bond (1 $\sigma$ contribution).
• Total $\sigma$ bonds = 5
• Lone pairs on Xe = 1 \[ SN = 5 + 1 = 6 \Rightarrow sp^3d^2 \] So A is sp$^3$d$^2$ hybridized.

Step 2: Compound B (XeO$_2$F$_2$). XeF$_6$ + 2H$_2$O $\rightarrow$ XeO$_2$F$_2$ + 4HF
• 2 Xe–F bonds + 2 Xe=O bonds (2 $\sigma$)
• Total $\sigma$ bonds = 4
• Lone pairs on Xe = 1 \[ SN = 4 + 1 = 5 \Rightarrow sp^3d \] So B is sp$^3$d hybridized.

Step 3: Conclusion Hybridization of A and B are: \[ sp^3d^2 \;,\; sp^3d \] Hence Option (2) is correct.