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NEET · Chemistry · STD 11 - 4. Chemical bonding and molecular structure
The correct geometry and hybridization for \(XeF_4\) are
- A octahedral, \(sp^3 d^2\)
- B trigonal bipyramidal, \(sp^3 d\)
- C planar triangle, \(sp^3d^3\)
- D square planar, \(sp^3d^2\)
Answer & Solution
Correct Answer
(A) octahedral, \(sp^3 d^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{XeF}_{4},\) has octahedral geometry where hybridization of \(\mathrm{Xe}\) is \(s p^{3} d^{2}\)
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