Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 42

Among the following molecules, \(\mathrm{p}-\mathrm{p}\) overlap takes place in (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{BeCl}_{2}\) (c) \(\mathrm{F}_{2}\) (d) HF

Short Answer

Expert verified
Question: Identify the molecule in which p-p orbital overlap takes place among the following molecules: (a) H2, (b) BeCl2, (c) F2, and (d) HF. Answer: (c) F2

Step by step solution

01

Analyze Hydrogen Molecule (H2)

In H2 molecule, each hydrogen atom has one electron in its 1s orbital. For both hydrogen atoms to share their electrons and form a bond, the 1s orbitals of the two hydrogen atoms will overlap. Thus, the bond formation in H2 is due to the overlap of two 1s orbitals (s-s overlap).
02

Analyze Beryllium Chloride Molecule (BeCl2)

In BeCl2 molecule, the central atom, beryllium, has two valence electrons in its 2s orbital. Each chlorine atom has seven valence electrons, with three in 2p orbitals and two in 2s orbitals. Beryllium forms two bonds with two chlorine atoms by sharing its two 2s electrons. In this case, the bond formation is due to the overlap of one 2s orbital from beryllium and one 3p orbital from each chlorine atom (s-p overlap).
03

Analyze Fluorine Molecule (F2)

In F2 molecule, each fluorine atom has seven valence electrons, with three electrons in 2p orbitals and two pairs of electrons in the 2s orbitals. For both fluorine atoms to share a pair of electrons and form a bond, the 2p orbitals of two fluorine atoms will overlap. Thus, the bond formation in F2 is due to the overlap of two 2p orbitals (p-p overlap).
04

Analyze Hydrogen Fluoride Molecule (HF)

In the HF molecule, hydrogen has one electron in its 1s orbital, and fluorine has seven valence electrons. Hydrogen shares its 1s electron with one unpaired electron in the 2p orbital of fluorine to form a bond. The bond formation in HF is due to the overlap of one 1s orbital of hydrogen and one 2p orbital of fluorine (s-p overlap).
05

Identify the Molecule with p-p Overlap

Among the molecules, the F2 molecule has a bond formed by the overlap of two 2p orbitals (p-p overlap). Thus, the correct answer is (c) F2.

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 valence electronic configurations of two elements are \(4 \mathrm{~s}^{1}\) and \(3 \mathrm{~s}^{2} 3 \mathrm{p}^{5}\), respectively. The type of bond expected to be present between them is (a) polar covalent bond (b) nonpolar covalent bond (c) metallic bond (d) ionic bond

A solid compound "A" consists of simple ions. Choose corresponding answers from the choices given below. Nature of the bond present in \(A\) will be (a) ionic (b) covalent (c) molecular (d) hydrogen bond

For each of the questions, four choices have been provided. Select the correct alternative. Expanded octet occurs in (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{PF}_{5}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{O}_{2}\)

Identify the compound that exhibits maximum ionic character among the following: (a) \(\mathrm{MgCl}_{2}\) (b) \(\mathrm{BaCl}_{2}\) (c) \(\mathrm{NaCl}\) (d) \(\mathrm{CsCl}\)

The bond formed by end on overlap is than the bond formed by side on overlap.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic 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