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NEET · Chemistry · STD 11 - 4. Chemical bonding and molecular structure

The correct geometry and hybridization for \(XeF_4\) are

  1. A octahedral, \(sp^3 d^2\)
  2. B trigonal bipyramidal, \(sp^3 d\)
  3. C planar triangle, \(sp^3d^3\)
  4. D square planar, \(sp^3d^2\)
Verified Solution

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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