For the reaction Br_2 + F_2 (excess) rightarr | vedclass

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(A) The reaction of $Br_2$ with excess $F_2$ is given by: $Br_2 + 5F_2 \rightarrow 2BrF_5$ Thus, the product $P$ is $BrF_5$. In $BrF_5$, the central $

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$BrF_5$, Square pyramidal

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(A) The reaction of $Br_2$ with excess $F_2$ is given by: $Br_2 + 5F_2 \rightarrow 2BrF_5$ Thus, the product $P$ is $BrF_5$. In $BrF_5$, the central $Br$ atom has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. According to $VSEPR$ theory, the steric number is $5 + 1 = 6$, which corresponds to $sp^3d^2$ hybridization. The geometry is octahedral, but due to the presence of one lone pair, the shape is square pyramidal.

For the reaction $Br_2 + F_2$ (excess) $\rightarrow P$, the molecular formula and structure of $P$, respectively, are

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