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Question
Based on VSEPR theory, the number of 90 degree F−Br−F angles in BrF5 are :
Based on VSEPR theory, the number of 90 degree F−Br−F angles in BrF5 are :
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Solution
The correct option is D 0
According to VSEPR, the valence electron pairs surrounding an atom tend to repel each other, and will, therefore, adopt an arrangement that minimizes this repulsion, thus, determining the molecule's geometry. All four planar bonds (F−Br−F) will reduce from 90o to 84.8o after lone pair−bond pair repulsion.
So, there are no 90-degree F−Br−F angles in BrF5.
According to VSEPR, the valence electron pairs surrounding an atom tend to repel each other, and will, therefore, adopt an arrangement that minimizes this repulsion, thus, determining the molecule's geometry. All four planar bonds (F−Br−F) will reduce from 90o to 84.8o after lone pair−bond pair repulsion.
So, there are no 90-degree F−Br−F angles in BrF5.
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