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Asked by Ashwin Last Modified
Answer 
Bromine Pentafluoride, 
BrF5
Draw the 3D molecular structure using VSEPR rules, The molecular geometry of BrF5 is square pyramidal with asymmetric charge distribution on the central atom.
Bromine has 35 electrons in atomic structure he can share his 5 electron with fluorine n makes brf5 fluorine has 7 electrone in his outer orbital so he can share 1 eleceone with bromine n complete his orbital by taking 1 electron from bromine but bromine is more powerful than fluorine so he can not gain his electron but he shares so bromine shares his 5 electron with 5 fluorine n makes a structure of square pyramidal it is bcuz of the dis-attraction of fluorine ion it makes a angel of 90° so it can be situated in a form of square pyramidal form. Thus it is a structure of square pyramidal in brf5 with 90° each angel.
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has total of
valence electrons, and that's why it's geometry is octahedral. Now as there can be only 5 bond pairs possible, one pair is left as lone pair. The question is, where to place it? Normally we would place the lone pair in equitorial position, but in this case, that increases the repulsionm between the lone pairs and bond-pairs, alongside that the bond-pairs repulse each other too much after getting close because of lp-bp repulsion. The structure with lone pair at axial position only has the 4 l.p.-b.p. repulsions. So, The shape is similar to a pyramid with a square base. It's not ideal, as the
bond angle is 84.8°, instead of the ideal 90°.
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