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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
The hybridization state of Si in SiH4 and in H3Si-SiH3 is sp3.

Step by step solution

01

Identify the Steric Number of Si in SiH4

Steric number is given by the formula \[SN = Number\ of\ sigma-bonds + Number\ of\ lone\ pairs + Number\ of\ single\ electrons\]. In SiH4, Si is bonded to four H atoms through sigma bonds. It has no lone pairs of electrons or unpaired electrons. So, the steric number of Si in SiH4 would be \(4+0+0=4\).
02

Determine the Hybridization of Si in SiH4

The hybridization is determined by the steric number. If the steric number(SN) is 4, the hybridization is sp3. So, the hybridization of Si in SiH4 is sp3.
03

Identify the Steric Number of Si in H3Si-SiH3

In H3Si-SiH3, each Si atom is bonded to three H atoms and one Si atom, all through sigma bonds. There are no lone pairs of electrons or unpaired electrons. So, the steric number of Si in H3Si-SiH3 would be \(4+0+0=4\).
04

Determine the Hybridization of Si in H3Si-SiH3

As highlighted before, if the steric number(SN) is 4, the hybridization is sp3. So, the hybridization of Si in H3Si-SiH3 is sp3.

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Most popular questions from this chapter

Predict the bond angles for these molecules: (a) \(\mathrm{BeCl}_{2},\) (b) \(\mathrm{BCl}_{3},\) (c) \(\mathrm{CCl}_{4},\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}\) (e) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (arrangement of atoms: \(\mathrm{ClHgHgCl}\) ), (f) \(\mathrm{SnCl}_{2}\) (g) \(\mathrm{H}_{2} \mathrm{O}_{2},\) (h) \(\mathrm{SnH}_{4}\).

Sketch the shape of a linear triatomic molecule, a trigonal planar molecule containing four atoms, a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule. Give the bond angles in each case.

Which of these species is not likely to have a tetrahedral shape (a) \(\operatorname{SiBr}_{4}\) (b) \(\mathrm{NF}_{4}^{+},(\mathrm {c}) \mathrm{SF}_{4}\) (d) \(\mathrm{BeCl}_{4}^{2-}\) (e) \(\mathrm{BF}_{4}^{-},\) (f) \(\mathrm{AlCl}_{4}^{-}\)

The compound 1,2 -dichloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\right)\) is nonpolar, while cis-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.

Define these terms: bonding molecular orbital, antibonding molecular orbital, pi molecular orbital, sigma molecular orbital.
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