Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 52

What is the hybridization of the central atom in (a) \(\mathrm{PBr}_{5}\), (b) \(\mathrm{CH}_{2} \mathrm{O},\) (c) \(\mathrm{O}_{3},(\mathbf{d}) \mathrm{NO}_{2} ?\)

Short Answer

Expert verified
The hybridization of the central atoms in the given molecules are as follows: (a) \( PBr_5 \): sp3d (b) \( CH_2O \): sp2 (c) \( O_3 \): sp3 (d) \( NO_2 \): sp2

Step by step solution

01

Determining the central atom

For each of the molecules, we need to identify the central atom which is the atom that is the least electronegative and has the highest connectivity to other atoms: (a) PBr5: Phosphorus (P) is the central atom (b) CH2O: Carbon (C) is the central atom (c) O3: Oxygen (O) is the central atom (d) NO2: Nitrogen (N) is the central atom
02

Calculate the electron domains

For each central atom, calculate the number of electron domains which is the sum of its bonds and lone pairs of electrons. (a) PBr5: Phosphorus (P) has 5 bonds (one with each Br atom) and no lone pairs. So, electron domains = 5. (b) CH2O: Carbon (C) has 2 bonds (one with each H atom) and 1 double bond (with the O atom). So, electron domains = 3. (c) O3: Oxygen (O) has 4 electron domains. Two of them are bonds with Z type atoms (the other two O atoms) and two lone pairs. (d) NO2: Nitrogen (N) has three electron domains: 1 double bond (with one O atom), 1 single bond (with the other O atom), and 1 unpaired electron (odd electron species). So, electron domains = 3.
03

Determine the Hybridization

Based on the number of electron domains, we can determine the hybridization of each central atom. (a) PBr5: 5 electron domains correspond to sp3d hybridization. (b) CH2O: 3 electron domains correspond to sp2 hybridization. (c) O3: 4 electron domains correspond to sp3 hybridization. (d) NO2: 3 electron domains correspond to sp2 hybridization. To conclude, the hybridization of the central atoms in the given molecules are as follows: (a) PBr5: sp3d (b) CH2O: sp2 (c) O3: sp3 (d) NO2: sp2

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

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

(a) The \(\mathrm{PH}_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3}\), has a small dipole moment. How is this possible, given that all the atoms are the same?

(a) Write a single Lewis structure for \(\mathrm{N}_{2} \mathrm{O},\) and determine the hybridization of the central \(\mathrm{N}\) atom. (b) Are there other possible Lewis structures for the molecule? (c) Would you expect \(\mathrm{N}_{2} \mathrm{O}\) to exhibit delocalized \(\pi\) bonding?

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: $(\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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