Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 10, Problem 10.93QE

(a)

Interpretation Introduction

Interpretation:

The species, which have stronger bonds between, B2andB2- has to be predicted.

Concept Introduction:

Bond energy:

The bond order is the number of electrons pairs shared between two atoms in the formation of the bond.  The amount of energy required to break a bond is called bond dissociation energy.  Thus greater the bond order greater is the bon energy.

Bond order =(Nb)-(Na)2

Where, Nb is number of electrons in bonding orbital.

Na is number of electrons in antibonding orbital

(a)

Expert Solution
Check Mark

Answer to Problem 10.93QE

The bond in B2 molecule is stronger than the B2 ion.

Explanation of Solution

The molecular electronic configuration of boron molecule is,

  B2=σ1s2σ*1s2σ2s2σ*2s2π2px1π2py1

The number of electrons in the bonding orbital=6

The number of electrons in the antibonding orbital=4

The bond order of boron molecule can be calculated as,

Bond order =(6)-(4)2=1

The molecular electronic configurations of B2 ion is,

  B2=σ1s2σ*1s2σ2s2σ*2s2π2px1

The number of electrons in the bonding orbital is five.

The number of electrons in the antibonding orbital is four.

Bond order of B2 ion can be calculated as

Bond order =(5)-(4)2=0.5

The bond order of B2 is greater B2 ion which means that the bond in B2 molecule is stronger than the B2 ion.

(b)

Interpretation Introduction

Interpretation:

The species, which have stronger bonds between, C2andC2+ has to be predicted.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 10.93QE

The bond in C2 ion molecule is stronger than the C2+ ion.

Explanation of Solution

The molecular electronic configuration of C2 molecule is,

  σ1s2σ*1s2σ2s2σ*2s2π2Px2π2py2σ2pz1

Bond order of C2 can be calculated as,

  Bond order =(Nb)-(Na)2

There are six electrons in bonding and four electrons in antibonding orbitals.

  Bond order=(9)-(4)2                  =2.5

The molecular electronic configuration of C2+ ion is,

  σ1s2σ*1s2σ2s2σ*2s2σ2px1

The number of electrons in the bonding orbital is five and the number of electrons in antibonding orbital is 4.

Bond order C2+ can be calculated as,

  Bond order=(5)-(4)2                  =0.5

The bond order of the C2 is greater C2+ ion which means that the bonds in C2 is stronger than the C2+ ion.

(c)

Interpretation Introduction

Interpretation:

The species that have stronger bonds between O2andO22+ has to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 10.93QE

The bond in O22+ ion molecule is stronger than the O2 ion.

Explanation of Solution

The molecular configuration of O2=σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2π2py2π*2px1π*2py1

There are ten electrons in bonding orbital and six electrons antibonding orbitals.

  Bond order=(10)-(6)2                  =42=2

Bond order of O2 is two.

The molecular configuration of O22+=σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2π2py2

Bond order of O22+ ion is,

  Bond order=(10)-(4)2                  =62=3

Bond order of O22+ ion is three.

The bond order of the O22+ ion is greater than the oxygen molecule that means that bonds in O22+ ion are stronger than the oxygen molecule.

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Chapter 10 Solutions

Chemistry: Principles and Practice

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