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Bro4 - Valance electrons = 7 +6(4) + 1 = 32 o U Lewis structure. Bx ö- Bro Formal charge - valance a Nonbonding valedectrons - bonding e- 2 Br F.C - 7-0 - 14 :00 double bond 0-) F.C. - 4 - 420 single bond o -> F.C - 6 - 6 - 22-1 6 molecular geometry = tetrahedral.
(since there is no lone pairs in 1 Bv atomz and 4 bonds present ). c) Hybridization: Number of a bonds in Br = 4. number of band lone pairs in Br - o Hybrik dization = number of sbonds & number of lone pair= 4 + 0 = 4 Hence hybridization is - Sp3
Xe & Number of valance electrons - 8 + 6 (3) = 26. formal charge 0 = 6 -4 -472 30 xe - 8 -2 – 17/₂ = 0 is 3 bond (6) electron geometry = tetrahedral (since there groups and one lone pair present in xe) molecular geometry = Triagonal pyramidal I lone pari 3 o bonds and 3 t. 41 c) The central atom xe has is sp3 Hence hybridization
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