Q.

The number of orbitals involved in the hybridisation of XeF2 is :

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answer is 5.

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Detailed Solution

H=1/2[V+M−C+A]where H = Number of orbitals involved in hybridisation           V = Valence electrons of central atom           M = Number of monovalent atoms linked with central atom          C = Charge on the cation          A = Charge on the anion          H=1/2[8+2−0+0]=5
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