The structure of IF_{7} is | vedclass

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(D) The central iodine atom $(I)$ in $IF_{7}$ has $7$ valence electrons. It forms $7$ bonds with $7$ fluorine atoms. The steric number is calculated a

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Pentagonal bipyramidal

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(D) The central iodine atom $(I)$ in $IF_{7}$ has $7$ valence electrons. It forms $7$ bonds with $7$ fluorine atoms. The steric number is calculated as: $	ext{Steric Number} = \frac{1}{2} 	imes (V + M - C + A) = \frac{1}{2} 	imes (7 + 7) = 7$. $A$ steric number of $7$ corresponds to $sp^{3}d^{3}$ hybridization,which results in a pentagonal bipyramidal geometry. In this structure,five fluorine atoms lie in a pentagonal plane with $I-F$ bond angles of $72^{\circ}$,and two fluorine atoms are positioned above and below the plane at $90^{\circ}$ to the equatorial plane.

The structure of $IF_{7}$ is

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