CHEMISTRY W/WRKBK AND SMARTWORK (LL)
CHEMISTRY W/WRKBK AND SMARTWORK (LL)
5th Edition
ISBN: 9780393693447
Author: Gilbert
Publisher: NORTON
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Chapter 9, Problem 9.106QP

(a)

Interpretation Introduction

Interpretation: The molecular ions which have electrons in π antibonding molecular orbitals from the given molecules is to be predicted.

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. The filling electrons in molecular orbitals follow Aufbau’s principle and hund’s rule.

To determine: If the molecular ion O2 has electrons in π antibonding molecular orbital.

(a)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion O2 has three electrons in π antibonding molecular orbital.

Explanation of Solution

Explanation

Oxygen has six valence electrons.

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

Totalnumberofvalenceelectrons=(NumberofOatoms×Valenceelectronsinoxygen±Charge)=2×6+1=13

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

O2=2s)2*2s)22p)22p)4*2p)3

Three electrons are present in the π antibonding orbital in O2 .

(b)

Interpretation Introduction

To determine: If the molecular ion O22 has electrons in π antibonding molecular orbital.

(b)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion O22 has four electrons in π antibonding molecular orbitals

Explanation of Solution

Explanation

Oxygen has six valence electrons.

The total number of valence electrons in O22 is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofOatoms×Valenceelectronsinoxygen±Charge)=2×6(2)=14

According to the molecular orbital theory the electronic configuration of O22 is,

O22=2s)2*2s)22p)22p)4*2p)4

Three electrons are present in the π antibonding orbital in O22 .

(c)

Interpretation Introduction

To determine: If the molecular ion N22 has electrons in π antibonding molecular orbital.

(c)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion N22 has two electrons in π antibonding molecular orbitals

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

The total number of valence electrons in N22 is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofNatoms×Valenceelectronsinnitrogen±Charge)=2×5(2)=12

According to the molecular orbital theory the electronic configuration of N22 is,

N22=2s)2*2s)22p)12p)4(π2py*1=π2pz*1)

Two electrons are present in the π antibonding orbital in N22 .

(d)

Interpretation Introduction

To determine: The molecular ion F2+ has π antibonding molecular orbital.

(d)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion F2+ has three electrons in π antibonding molecular orbitals

Explanation of Solution

Explanation

Fluorine has seven valence electrons.

The total number of valence electrons in F2+ is calculated by formula,

Totalnumberofvalenceelectrons=(NumberofFatoms×Valenceelectronsinfluorine±Charge)=2×71=13

According to the molecular orbital theory the electronic configuration of F2+ is,

F2+=2s)2*2s)22p)22p)4(π2py*22pz*1)

Three electrons are present in the π antibonding orbital in F2+ .

(e)

Interpretation Introduction

To determine: The molecular ion N2+ has π antibonding molecular orbital.

(e)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion N2+ has no electron in π antibonding molecular orbitals.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

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

Totalnumberofvalenceelectrons=(NumberofNatoms×Valenceelectronsinnitrogen±Charge)=2×51=9

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

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

No electron is present in the π antibonding molecular orbital in N2+ .

(f)

Interpretation Introduction

To determine: If the molecular ion O2+ has electrons in π antibonding molecular orbital.

(f)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion O2+ has one electron in π antibonding molecular orbitals

Explanation of Solution

Explanation

Oxygen has six valence electrons.

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

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

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

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

One unpaired electron is present in π antibonding molecular orbital in O2+ .

(g)

Interpretation Introduction

To determine: If the molecular ion C2+ has π antibonding molecular orbital.

(g)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion C2+ has no electron in π antibonding molecular orbitals

Explanation of Solution

Explanation

Carbon has four valence electrons.

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

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

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

C2+=2s)2*2s)22px22py1)

No electron is present in π antibonding molecular orbital in C2+ .

(g)

Interpretation Introduction

To determine: If the molecular ion Br22 has electrons in π antibonding molecular orbital.

(g)

Expert Solution
Check Mark

Answer to Problem 9.106QP

Solution

The molecular ion Br22 has four electrons in π antibonding molecular orbitals

Explanation of Solution

Explanation

Bromine has seven valence electrons.

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

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

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

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

Four electrons are present in π antibonding molecular orbitals in Br22- .

Conclusion

Electrons in antibonding molecular orbitals of the molecule decrease the strength of the bond. All the given molecular ion species except N2+ and C2+ have electrons in π antibonding molecular orbitals.

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Chapter 9 Solutions

CHEMISTRY W/WRKBK AND SMARTWORK (LL)

Ch. 9.7 - Prob. 11PECh. 9.7 - Prob. 12PECh. 9.7 - Prob. 13PECh. 9 - Prob. 9.1VPCh. 9 - Prob. 9.2VPCh. 9 - Prob. 9.3VPCh. 9 - Prob. 9.4VPCh. 9 - Prob. 9.5VPCh. 9 - Prob. 9.6VPCh. 9 - Prob. 9.7VPCh. 9 - Prob. 9.8VPCh. 9 - Prob. 9.9QPCh. 9 - Prob. 9.10QPCh. 9 - Prob. 9.11QPCh. 9 - Prob. 9.12QPCh. 9 - Prob. 9.13QPCh. 9 - Prob. 9.14QPCh. 9 - Prob. 9.15QPCh. 9 - Prob. 9.16QPCh. 9 - Prob. 9.17QPCh. 9 - Prob. 9.18QPCh. 9 - Prob. 9.19QPCh. 9 - Prob. 9.20QPCh. 9 - Prob. 9.21QPCh. 9 - Prob. 9.22QPCh. 9 - Prob. 9.23QPCh. 9 - Prob. 9.24QPCh. 9 - Prob. 9.25QPCh. 9 - Prob. 9.26QPCh. 9 - Prob. 9.27QPCh. 9 - Prob. 9.28QPCh. 9 - Prob. 9.29QPCh. 9 - Prob. 9.30QPCh. 9 - Prob. 9.31QPCh. 9 - Prob. 9.32QPCh. 9 - Prob. 9.33QPCh. 9 - Prob. 9.34QPCh. 9 - Prob. 9.35QPCh. 9 - Prob. 9.36QPCh. 9 - Prob. 9.37QPCh. 9 - Prob. 9.38QPCh. 9 - Prob. 9.39QPCh. 9 - Prob. 9.40QPCh. 9 - Prob. 9.41QPCh. 9 - Prob. 9.42QPCh. 9 - Prob. 9.43QPCh. 9 - Prob. 9.44QPCh. 9 - Prob. 9.45QPCh. 9 - Prob. 9.46QPCh. 9 - Prob. 9.47QPCh. 9 - Prob. 9.48QPCh. 9 - Prob. 9.49QPCh. 9 - Prob. 9.50QPCh. 9 - Prob. 9.51QPCh. 9 - Prob. 9.52QPCh. 9 - Prob. 9.53QPCh. 9 - Prob. 9.54QPCh. 9 - Prob. 9.55QPCh. 9 - Prob. 9.56QPCh. 9 - Prob. 9.57QPCh. 9 - Prob. 9.58QPCh. 9 - Prob. 9.59QPCh. 9 - Prob. 9.60QPCh. 9 - Prob. 9.61QPCh. 9 - Prob. 9.62QPCh. 9 - Prob. 9.63QPCh. 9 - Prob. 9.64QPCh. 9 - Prob. 9.65QPCh. 9 - Prob. 9.66QPCh. 9 - Prob. 9.67QPCh. 9 - Prob. 9.68QPCh. 9 - Prob. 9.69QPCh. 9 - Prob. 9.70QPCh. 9 - Prob. 9.71QPCh. 9 - Prob. 9.72QPCh. 9 - Prob. 9.73QPCh. 9 - Prob. 9.74QPCh. 9 - Prob. 9.75QPCh. 9 - Prob. 9.76QPCh. 9 - Prob. 9.77QPCh. 9 - Prob. 9.78QPCh. 9 - Prob. 9.79QPCh. 9 - Prob. 9.80QPCh. 9 - Prob. 9.81QPCh. 9 - Prob. 9.82QPCh. 9 - Prob. 9.83QPCh. 9 - Prob. 9.84QPCh. 9 - Prob. 9.85QPCh. 9 - Prob. 9.86QPCh. 9 - Prob. 9.87QPCh. 9 - Prob. 9.88QPCh. 9 - Prob. 9.89QPCh. 9 - Prob. 9.90QPCh. 9 - Prob. 9.91QPCh. 9 - Prob. 9.92QPCh. 9 - Prob. 9.93QPCh. 9 - Prob. 9.94QPCh. 9 - Prob. 9.95QPCh. 9 - Prob. 9.96QPCh. 9 - Prob. 9.97QPCh. 9 - Prob. 9.98QPCh. 9 - Prob. 9.99QPCh. 9 - Prob. 9.100QPCh. 9 - Prob. 9.101QPCh. 9 - Prob. 9.102QPCh. 9 - Prob. 9.103QPCh. 9 - Prob. 9.104QPCh. 9 - Prob. 9.105QPCh. 9 - Prob. 9.106QPCh. 9 - Prob. 9.107QPCh. 9 - Prob. 9.108QPCh. 9 - Prob. 9.109QPCh. 9 - Prob. 9.110QPCh. 9 - Prob. 9.111QPCh. 9 - Prob. 9.112QPCh. 9 - Prob. 9.113QPCh. 9 - Prob. 9.114QPCh. 9 - Prob. 9.115APCh. 9 - Prob. 9.116APCh. 9 - Prob. 9.117APCh. 9 - Prob. 9.118APCh. 9 - Prob. 9.119APCh. 9 - Prob. 9.120APCh. 9 - Prob. 9.121APCh. 9 - Prob. 9.122APCh. 9 - Prob. 9.123APCh. 9 - Prob. 9.124APCh. 9 - Prob. 9.125APCh. 9 - Prob. 9.126APCh. 9 - Prob. 9.127APCh. 9 - Prob. 9.128APCh. 9 - Prob. 9.129APCh. 9 - Prob. 9.130APCh. 9 - Prob. 9.131APCh. 9 - Prob. 9.132APCh. 9 - Prob. 9.133APCh. 9 - Prob. 9.134APCh. 9 - Prob. 9.135APCh. 9 - Prob. 9.136APCh. 9 - Prob. 9.137APCh. 9 - Prob. 9.138APCh. 9 - Prob. 9.139APCh. 9 - Prob. 9.140APCh. 9 - Prob. 9.141APCh. 9 - Prob. 9.142APCh. 9 - Prob. 9.143AP
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