Among the following, the CORRECT combinations | vedclass

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(A) $1$. $IF_3$: Iodine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. Steric number = $3 + 2 = 5$ ($sp^3d$ hybr

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$A$, $B$ and $C$ only

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(A) $1$. $IF_3$: Iodine has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. Steric number = $3 + 2 = 5$ ($sp^3d$ hybridization). Due to $2$ lone pairs, it is $T$-shaped. (Correct)$2$. $IF_5$: Iodine has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. Steric number = $5 + 1 = 6$ ($sp^3d^2$ hybridization). Due to $1$ lone pair, it is square pyramidal. (Correct)$3$. $IF_7$: Iodine has $7$ valence electrons. It forms $7$ bonds with $F$ atoms and has $0$ lone pairs. Steric number = $7 + 0 = 7$ ($sp^3d^3$ hybridization). It is pentagonal bipyramidal. (Correct)$4$. $ClO_4^-$: Chlorine has $7$ valence electrons. It forms $4$ bonds with $O$ atoms and has $0$ lone pairs. Steric number = $4 + 0 = 4$ ($sp^3$ hybridization). It is tetrahedral, not square planar. (Incorrect)Thus, $A$, $B$, and $C$ are correct.

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