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Chapter 10: Problem 38

Give the formula of an anion comprised of iodine and fluorine in which the iodine atom is \(s p^{3}\) \(d^{2}\) -hybridized.

Short Answer

Expert verified
The formula of the anion is \[IF_{6}^{-}\]

Step by step solution

01

Understand the Hybridization

From the given exercise, the iodine atom is \(sp^{3}d^{2}\) hybridized. This means that there are total of 6 hybrid orbitals formed from 1 \(s\), 3 \(p\), and 2 \(d\) orbitals. Each of these orbitals can make one bond. Therefore, the iodine atom can make 6 bonds.
02

Construct the Molecule

In this case, the molecule is composed of iodine and fluorine. Fluorine forms one bond, so it can occupy the hybrid orbitals.
03

Formulate the Anion

The negative charge in an anion indicates that there are extra electrons. Since iodine forms 6 bonds with fluorines, one electron must remain unpaired. This electron gives the molecule a negative charge. Thus, the formula of this anion is \[IF_{6}^{-}\] .

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Most popular questions from this chapter

Describe the bonding scheme of the AsH \(_{3}\) molecule in terms of hybridization.

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Predict the geometry of the following molecules using the VSEPR method: (a) \(\mathrm{HgBr}_{2}\), (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{ar}-\) rangement of atoms is \(\mathrm{NNO}\) ), (c) \(\mathrm{SCN}^{-}\) (arrangement of atoms is \(\mathrm{SCN}\) ).
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