Based on rigorous VSEPR theory incorporating lone-pair repulsions, what is the exact number of perfe β Chemical Bonding Chemistry Question
Question
Based on rigorous VSEPR theory incorporating lone-pair repulsions, what is the exact number of perfect $90^\circ$ $F-Br-F$ bond angles present in the $BrF_5$ molecule?
Answer: 0
π‘ Solution & Explanation
$BrF_5$ is $sp^3d^2$ hybridized with 5 bond pairs and 1 lone pair, adopting a square pyramidal geometry. Because the single lone pair situated below the basal plane exerts greater repulsion ($lp-bp > bp-bp$) on the four equatorial $Br-F$ bonds, they are pushed upward toward the axial fluorine. Consequently, all adjacent $F-Br-F$ angles shrink to slightly less than $90^\circ$ (approx $84.8^\circ$), leaving exactly 0 angles at a perfect $90^\circ$.
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