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Chapter 11: Problem 77

The anion \(I_{4}^{2-}\) is linear, and the anion \(I_{5}^{-}\) is V-shaped, with a \(95^{\circ}\) angle between the two arms of the V. For the central atoms in these ions, propose hybridization schemes that are consistent with these observations.

Short Answer

Expert verified
The hybridization of iodine in the \(I_{4}^{2-}\) anion is sp3d, and in the \(I_{5}^{-}\) anion is sp3d2.

Step by step solution

01

Analysis of \(I_{4}^{2-}\) Anion

In the \(I_{4}^{2-}\) anion, the central iodine atom is bonded to two other iodine atoms and also has three lone pairs. The geometry is linear, meaning all atoms and lone pairs are in one line.
02

Determine Hybridization of \(I_{4}^{2-}\)

The number of electron pairs around the central atom is the sum of bonded atoms and lone pairs, which is 2 (bonds) + 3 (lone pairs) = 5. According to the hybridization theory, this corresponds to sp3d hybridization.
03

Analysis of \(I_{5}^{-}\) Anion

In the \(I_{5}^{-}\) anion, the central iodine atom is bonded to five other iodine atoms and has no lone pairs. The molecule is V-shaped with a \(95^{\circ}\) angle between the two arms of the V.
04

Determine Hybridization of \(I_{5}^{-}\)

The number of electron pairs around the central atom is 5, all of which are bonding pairs. This corresponds to sp3d2 hybridization.

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