Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 56

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) HCN, (d) \(\mathrm{TeCl}_{2}\) (c) \(\mathrm{SO}_{3}\)

Short Answer

Expert verified
The hybridization of the central atoms in the given compounds can be summarized as follows: \( (a) \mathrm{Si} \) in \(\mathrm{SiCl}_{4}\) is sp3 hybridized, \( (b) \mathrm{C} \) in HCN is sp hybridized, \( (c) \mathrm{Te} \) in \(\mathrm{TeCl}_{2}\) is sp3 hybridized, and \( (d) \mathrm{S} \) in \(\mathrm{SO}_{3}\) is sp2 hybridized.

Step by step solution

01

(a) Determine hybridization of Si in SiCl4

First, we need to identify the central atom, which is Si in SiCl4. Next, we must find the total number of valence electrons by looking at the periodic table. Si has 4 valence electrons and Cl has 7, so in SiCl4 there are a total of 4 Si electrons + 4 Cl electrons * 7 Cl electrons = 32 electrons. Since all four chlorines are bonded to the silicon, no lone pairs are on the silicon atom. Therefore, silicon has 4 electron domains (4 single bonds to 4 chlorine atoms), which corresponds to a tetrahedral electron domain geometry. The hybridization that matches tetrahedral geometry is sp3.
02

(b) Determine hybridization of C in HCN

In HCN, the central atom is carbon. Carbon has 4 valence electrons, nitrogen has 5, and hydrogen has 1, making the total number of valence electrons = 4 + 5 + 1 = 10 electrons. Carbon is bonded to one hydrogen with a single bond and one nitrogen with a triple bond. So, the carbon has 2 electron domains (since the triple bond is considered as one domain). This corresponds to a linear electron domain geometry. The hybridization that matches linear geometry is sp.
03

(c) Determine hybridization of Te in TeCl2

The central atom in TeCl2 is tellurium (Te), which has 6 valence electrons. Each chlorine atom has 7 valence electrons, making the total number of valence electrons = 6 + 2 * 7 = 20 electrons. Tellurium forms two single bonds with the two chlorine atoms. Given there are 20 valence electrons, this leaves us with 16 electrons unaccounted for, which form 8 lone pairs. Tellurium has 2 lone pairs and 2 single bonds, giving it 4 electron domains (2 bonding + 2 lone pairs). This corresponds to a tetrahedral electron domain geometry. The hybridization that matches tetrahedral geometry is sp3.
04

(d) Determine hybridization of S in SO3

The central atom in SO3 is sulfur. Sulfur has 6 valence electrons and oxygen has 6 valence electrons, making the total number of valence electrons = 6 + 3 * 6 = 24 electrons. Sulfur forms a double bond with each of the three oxygen atoms. So, sulfur has 3 electron domains (3 double bonds). This corresponds to a trigonal planar electron domain geometry. The hybridization that matches trigonal planar geometry is sp2. In summary: (a) Si in SiCl4 is sp3 hybridized (b) C in HCN is sp hybridized (c) Te in TeCl2 is sp3 hybridized (d) S in SO3 is sp2 hybridized

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$\mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6}\), you run into a problem. What is it? (c) What could you do to resolve the difficulty in part \((\mathrm{b}) ?(\mathbf{d})\) The molecule \(\mathrm{IF}_{7}\) has a pentagonalbipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\).

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same? Explain.

(a) If you combine two atomic orbitals on two different atoms to make a new orbital, is this a hybrid orbital or a molecular orbital? (b) If you combine two atomic orbitals on one atom to make a new orbital, is this a hybrid orbital or a molecular orbital? (c) Does the Pauli exclusion principle (Section 6.7\()\) apply to MOs? Explain.

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN},\) and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N},\) and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O} ?\) (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free