Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
5th Edition
ISBN: 9780393615296
Author: Rein V. Kirss (Author), Natalie Foster (Author), Geoffrey Davies (Author) Thomas R. Gilbert (Author)
Publisher: W. W. Norton
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Chapter 9, Problem 9.105QP

(a)

Interpretation Introduction

Interpretation: The molecular ions with one or more unpaired electron from the given molecules are to be identified.

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 N2+ has unpaired electrons.

(a)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The N2+ ion has one unpaired electron.

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

One unpaired electron is present in σ2p orbital in N2+ .

(b)

Interpretation Introduction

To determine: If the molecular ion O2+ has unpaired electrons.

(b)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The O2+ ion has one unpaired electron.

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 the π2p* orbital in O2+ .

(c)

Interpretation Introduction

To determine: If the molecular ion C2+ has unpaired electrons.

(c)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The C2+ ion has one unpaired electron.

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)

One unpaired electron is present in the π2py orbital in C2+ .

(d)

Interpretation Introduction

To determine: If the molecular ion Br22 has unpaired electron.

(d)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The Br22 ion has no unpaired electron.

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*4p)4*4p)2

No unpaired electron is present in any orbital of Br22- .

(e)

Interpretation Introduction

To determine: If the molecular ion O2 has unpaired electron.

(e)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The O2 ion has one unpaired electron.

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

One unpaired electron is present in the π2p* orbital in O2 .

(f)

Interpretation Introduction

To determine: If the molecular ion O22 has unpaired electron.

(f)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The O22 ion has no unpaired electron.

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

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

No unpaired electron is present in any orbital of O22 .

(g)

Interpretation Introduction

To determine: If the molecular ion N22 has unpaired electron.

(g)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The N22 ion has two unpaired electrons.

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

N2+=2s)2*2s)22p)22p)4(π2py*1=π2pz*1)

Two unpaired electrons are present in (π2py*1=π2pz*1) orbitals of N22 .

(h)

Interpretation Introduction

To determine: If the molecular ion F2+ has unpaired electron.

(h)

Expert Solution
Check Mark

Answer to Problem 9.105QP

Solution

The F2+ ion has one unpaired electron.

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(π2p*)3

One unpaired electron is present in π2p* orbital in F2+ .

Conclusion

The molecules which contain unpaired electrons in their molecular orbitals are paramagnetic in nature. All the molecular ion species except Br22-andO22- , have unpaired electrons.

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Chapter 9 Solutions

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)

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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