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Chapter 10: Problem 32

What is the hybridization state of \(\mathrm{Si}\) in \(\mathrm{SiH}_{4}\) and in \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3} ?\)

Short Answer

Expert verified
In both molecules, \(\mathrm{SiH}_{4}\) and \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3}\), the silicon atom is in the \(sp^{3}\) hybridization state.

Step by step solution

01

Analyze \(\mathrm{SiH}_{4}\)

Look at the molecule \(\mathrm{SiH}_{4}\). In this molecule, silicon (Si) is the central atom and is bonded with four hydrogen (H) atoms. As silicon is making 4 sigma bonds and has no lone pair of electrons, it undergoes \(sp^{3}\) hybridization to form tetrahedral structure.
02

Analyze \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3}\)

Next, let's find the hybridization state of silicon (Si) in \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3}\). Here, we can consider the molecule as two \(\mathrm{SiH}_{3}\) units joined together by a sigma bond formed by two silicon (Si) atoms. Each silicon atom is involved with three hydrogen (H) atoms and one silicon (Si) atom, meaning there are four sigma bonds. There are no lone pairs of electrons on any silicon atom. This also leads to \(sp^{3}\) hybridization, resulting in a tetrahedral structure.

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Most popular questions from this chapter

The compound 1,2 -dichloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\right)\) is nonpolar, while \(c i s\) -dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has a dipole moment: The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in 1,2 -dichloroethane but not in \(c i s\) -dichloroethylene.

Use molecular orbital theory to compare the relative stabilities of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{+}\)

List the following molecules in order of increasing dipole moment: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CBr}_{4}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{HF}, \mathrm{NH}_{3}, \mathrm{CO}_{2}\)

The formation of \(\mathrm{H}_{2}\) from two \(\mathrm{H}\) atoms is an energetically favorable process. Yet statistically there is less than a 100 percent chance that any two \(\mathrm{H}\) atoms will undergo the reaction. Apart from energy considerations, how would you account for this observation based on the electron spins in the two \(\mathrm{H}\) atoms?

Write the electron configuration of the cyanide ion \(\left(\mathrm{CN}^{-}\right) .\) Name a stable molecule that is isoelectronic with the ion.
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