   Chapter 20, Problem 62E

Chapter
Section
Textbook Problem

# Give the Lewis structure, molecular structure, and hybridization of the oxygen atom for OF2. Would you expect OF2 to be a strong oxidizing agent like O2F2 discussed in Exercise 61?

Interpretation Introduction

Interpretation: The Lewis structure, molecular structure and hybridization of central atom in OF2 is to be stated. If OF2 is strong oxidizing agent like O2F2 is to be stated.

Concept introduction: The following steps are to be followed to determine the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• The value 1e is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by 2 to find the number of electron pairs.
• This further gives the hybridization of the given compound.

The substance that gains electron is known as oxidizing agent.

To determine: The Lewis structure, molecular structure and hybridization of the central atom in OF2 ; if OF2 is strong oxidizing agent like O2F2 .

Explanation

Explanation

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of oxygen (O) is 8 and its electronic configuration is,

1s22s22p4

The valence electron of oxygen is 6 .

The atomic number of fluorine (F) is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule OF2 is made of two fluorine atoms and single oxygen atom; hence, the total number of valence electrons is,

O+2F=(6+2×7)=20

The formula of number of electron pairs is,

X=V+M±C2

Where,

• X is the number of electron pairs.
• V is the valence electrons of central atom.
• M is the number of monovalent atoms.
• C is the charge on compound

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