In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three?

Question

Accepted Answer

To determine the hybridization of the central atom in each species, we can use the steric number method (number of sigma bonds + number of lone pairs).

Step 1: Determine hybridization for SF₄

In SF₄, Sulfur (S) is the central atom. It has 6 valence electrons. It forms 4 sigma bonds with 4 Fluorine atoms. The remaining 2 electrons form 1 lone pair.
Steric number = 4 (sigma bonds) + 1 (lone pair) = 5.
A steric number of 5 corresponds to sp³d hybridization.

Step 2: Determine hybridization for I₃^-

In I₃^-, the central Iodine atom has 7 valence electrons + 1 (for the negative charge) = 8 electrons. It forms 2 sigma bonds with the other two Iodine atoms. The remaining 6 electrons form 3 lone pairs.
Steric number = 2 (sigma bonds) + 3 (lone pairs) = 5.
A steric number of 5 corresponds to sp³d hybridization.

Step 3: Determine hybridization for SbCl₅^2-

In SbCl₅^2-, Antimony (Sb) is the central atom. It has 5 valence electrons + 2 (for the negative charge) = 7 electrons. It forms 5 sigma bonds with 5 Chlorine atoms. The remaining 2 electrons form 1 lone pair.
Steric number = 5 (sigma bonds) + 1 (lone pair) = 6.
A steric number of 6 corresponds to sp³d² hybridization.

Step 4: Determine hybridization for PCl₅

In PCl₅, Phosphorus (P) is the central atom. It has 5 valence electrons. It forms 5 sigma bonds with 5 Chlorine atoms. There are no lone pairs.
Steric number = 5 (sigma bonds) + 0 (lone pairs) = 5.
A steric number of 5 corresponds to sp³d hybridization.

Option Analysis:

A) SF₄: sp³d hybridization.
B) I₃^-: sp³d hybridization.
C) SbCl₅^2-: sp³d² hybridization. This is different from the others.
D) PCl₅: sp³d hybridization.

Therefore, the central atom in SbCl₅^2- has a different type of hybridization compared to the other three species.

Correct Answer: (C)

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