TS EAMCET · Chemistry · Chemical Bonding and Molecular Structure
\(\mathrm{XeF}_4\) is square planar while \(\mathrm{XeF}_6\) has a distorted octahedral structure. What is the correct explanation for this observation?
- A Both molecules have one lone pair of electrons
- B Both molecules have two lone pairs of electrons
- C \(\mathrm{XeF}_4\) does not have any lone pair of electrons; \(\mathrm{XeF}_6\) has one lone pair of electrons on \(\mathrm{Xe}\)
- D \(\mathrm{XeF}_4\) has two lone pairs of electrons on \(\mathrm{Xe}^2 \mathrm{XeF}_6\) has one lone pair of electrons on Xe
Answer & Solution
Correct Answer
(D) \(\mathrm{XeF}_4\) has two lone pairs of electrons on \(\mathrm{Xe}^2 \mathrm{XeF}_6\) has one lone pair of electrons on Xe
Step-by-step Solution
Detailed explanation
\(\mathrm{XeF}_4\) is square planar and \(s p^3 d^2\)-hybridised. \(4 \sigma+2 l p=6\)-hybrid orbital \(s p^3 d^2\)-hybridisation \(\mathrm{XeF}_6\) has distorted octahedral structure and \[s p^3 d^3 \text {-hybridised. } \] \(6 \sigma+1 l p=7\)-hybrid orbital…
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