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Draw the structure of BrF_(5). |
Answer» Solution :The electronic configuration of Br (Z = 35) is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4S^(2) 4p_(x)^(2) 4p_(y)^(2)4p_(z)^(1)`. It has only one half-filled orbital. But to form five Br-F bonds, we need five half-filled ORBITALS. To achieve this, one electron each of `4p_(x) and 4p_(y)` gets excited to 4d-orbitals. The resulting six orbitals of the fourth shell of Br in the 2ND  excited state undergo `SP^(3)d^(2)`-hydridization. As a result, `BeF_(5)` has octahedral (or square pyramidal) geometry with one position occupied by a lone pair. 
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