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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
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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