Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 9, Problem 9.91QP

(a)

Interpretation Introduction

Interpretation: The electronic configuration, bond order and the ionic species that can exist is to be predicted from given molecular ions.

Concept introduction: When two atomic orbitals come close to each other they lose their identity and form new pair of orbitals knows as molecular orbitals. Among the two molecular orbitals formed one has energy lower than the atomic orbitals is known as bonding molecular orbital and the other has energy higher than the atomic orbitals and is known as antibonding molecular orbital.

To determine: The orbital electronic configuration of the given ionic species.

(a)

Expert Solution
Check Mark

Answer to Problem 9.91QP

Solution

The electronic configuration of molecular species is given as follows.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

Charge on N2+ is +1 .

The total number of valence electrons in N2+ is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofNatoms×Valenceelectrons innitrogen±Charge)=2×51=9

The total number of valence electrons in N2+ is 9 .

According to the molecular orbital theory the electronic configuration of N2+ is

N2+=2s)2*2s)22p)42p)1

Oxygen has six valence electrons.

Charge on O2+ is +1 .

The total number of valence electrons in O2+ is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofOatoms×Valenceelectronsofoxygen ±Charge)=2×61=11

The total number of valence electrons in O2+ is 11 .

According to the molecular orbital theory the electronic configuration of O2+ is

O2+=2s)2*2s)22p)22p)4*2p)1

Carbon has four valence electrons.

Charge on C2+ is +1

The total number of valence electrons in C2+ is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofCatoms×Valenceelectronsincarbon±Charge)=2×41=7

The total number of valence electrons in C2+ is 7 .

According to the molecular orbital theory the electronic configuration of C2+ is

C2+=2s)2*2s)22p)3

Bromine has seven valence electrons.

Charge on Br22 is 2

The total number of valence electrons in Br22 is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofBratoms×Valenceelectronsinbromine±Charge)=2×7(2)=16

The total number of valence electrons in Br22 is 16 .

According to the molecular orbital theory the electronic configuration of Br22- is

Br22-=4s)2*4s)24p)22p)4*2p)4*4p)2

(b)

Interpretation Introduction

To determine: The bond order of given molecular ionic species.

(b)

Expert Solution
Check Mark

Answer to Problem 9.91QP

Solution

All the given molecular ionic species except Br22 have nonzero bo9mnd order.

Explanation of Solution

Explanation

The electronic configuration of N2+ is,

N2+=2s)2*2s)22p)42p)1

The bond order of N2+ is calculate by formula,

Bondorder=12(NumberofelectronsinbondingmolecularorbitalsNumberofelectroninantibondingmolecularorbitals)

The number of bonding electrons in N2+=7 .

The number of antibonding electrons in N2+=2 .

Substitute the value of number of electrons in bonding and antibonding molecular orbitals of N2+ in the above equation.

Bondorder=12(72)=2.5

The bond order of N2+ is 2.5 .

The electronic configuration of O2+ is,

O2+=2s)2*2s)22p)22p)4*2p)1

The bond order of O2+ is calculate by formula,

Bondorder=12(NumberofelectronsinbondingmolecularorbitalsNumberofelectroninantibondingmolecularorbitals)

The number of bonding electrons in O2+=8 .

The number of antibonding electrons in O2+=3 .

Substitute the value of number of electrons in bonding and antibonding molecular orbitals of O2+ in the above equation.

Bondorder=12(83)=2.5

The bond order of O2+ is 2.5 .

The electronic configuration of C2+ is,

C2+=2s)2*2s)22p)3

The bond order of C2+ is calculate by formula,

Bondorder=12(NumberofelectronsinbondingmolecularorbitalsNumberofelectroninantibondingmolecularorbitals)

The number of bonding electrons in C2+=5 .

The number of antibonding electrons in C2+=3 .

Substitute the value of number of electrons in bonding and antibonding molecular orbitals of C2+ in the above equation.

Bondorder=12(53)=1.5

The bond order of C2+ is 1.5 .

The electronic configuration of Br22- is,

Br22-=4s)2*4s)24p)22p)4*2p)4*4p)2

The bond order of Br22- is calculate by formula,

Bondorder=12(NumberofelectronsinbondingmolecularorbitalsNumberofelectroninantibondingmolecularorbitals)

The number of bonding electrons in Br22-=8 .

The number of antibonding electrons in Br22-=8 .

Substitute the value of number of electrons in bonding and antibonding molecular orbitals of Br22- in the above equation.

Bondorder=12(88)=0

The bond order of Br22- is 0 .

(c)

Interpretation Introduction

To determine: The molecular ion species that are expected to exist

(c)

Expert Solution
Check Mark

Answer to Problem 9.91QP

Solution

All the molecular species except Br22- are expected to exist.

Explanation of Solution

Explanation

The bond order of a molecule is in direct proportion with its stability. Higher the bond order higher is the stability of the molecule.

Molecular species with nonzero bond order are expected to exist. All the given molecular ions except Br22- have nonzero bond order. Therefore, all the molecular species except Br22- are expected to exist.

Conclusion

The molecular species with nonzero bond order are expected to exist.

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Chapter 9 Solutions

Chemistry

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