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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