The hybridisation of the central atom I in IF | vedclass

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Question

(D) In the molecule $IF_7$,the central atom $I$ (Iodine) has $7$ valence electrons. All $7$ valence electrons are involved in forming $7$ sigma bonds

Vedclass · Accepted Answer

$sp^3d^3$

Vedclass · Answer

(D) In the molecule $IF_7$,the central atom $I$ (Iodine) has $7$ valence electrons. All $7$ valence electrons are involved in forming $7$ sigma bonds with $7$ fluorine atoms. The steric number is calculated as: $	ext{Steric Number} = 	ext{Number of sigma bonds} + 	ext{Number of lone pairs} = 7 + 0 = 7$. $A$ steric number of $7$ corresponds to $sp^3d^3$ hybridization. The geometry and shape of $IF_7$ is pentagonal bipyramidal,where $5$ $F$ atoms are in the equatorial plane and $2$ $F$ atoms are in the axial positions.

The hybridisation of the central atom $I$ in $IF_7$ is:

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Which of the underlined atoms in the molecules shown below have $sp$-hybridization?
$(u) H_2\underline{C}=CH-CH_3$
$(v) CH_2=\underline{C}=CHCl$
$(w) CH_3-\underline{C}H_2^+$
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$(y) CH_3-\underline{C} \equiv N$
$(z) (CH_3)_2C=\underline{N}-NH_2$

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