The molecule/ion with square pyramidal shape | vedclass

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(C) To determine the shape,we calculate the steric number using the formula: $Steric \ Number = \frac{1}{2} 	imes (V + M - C + A)$,where $V$ is the n

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$BrF_5$

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(C) To determine the shape,we calculate the steric number using the formula: $Steric \ Number = \frac{1}{2} 	imes (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $BrF_5$: $V = 7$ (for $Br$),$M = 5$ (for $F$). $Steric \ Number = \frac{1}{2} 	imes (7 + 5) = 6$.$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization with an octahedral electron geometry.Since there are $5$ bond pairs and $1$ lone pair,the molecular geometry is square pyramidal.

The molecule/ion with square pyramidal shape is:

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