Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 8, Problem 8.171AP

(a)

Interpretation Introduction

Interpretation: The justification for the isoelectronic behaviour of N2F+ and N2O ; the Lewis structure of N2F+ ; the atom that has the +1 formal charge in the Lewis structure; the possibility of N2F+ to have resonance forms and the possibility of the central atom of N2F+ to be fluorine atom is to be stated.

Concept introduction: The Lewis symbols of the atoms indicate the total number of valence electrons surrounding the atom or the ions.

To determine: The justification for the isoelectronic behaviour of N2F+ and N2O .

(a)

Expert Solution
Check Mark

Answer to Problem 8.171AP

Solution

The total valence electrons of N2F+ is same as that of N2O . Therefore, N2F+ and N2O are isoelectronic.

Explanation of Solution

Explanation

The atomic number of nitrogen is 7 .

The electronic configuration of nitrogen is [He]2s22p3 . Therefore, nitrogen has five valence electrons.

The atomic number of fluorine is 9 .

The electronic configuration of fluorine is [He]2s22p5 . Therefore, fluorine has seven valence electrons.

The atomic number of oxygen is 6 .

The electronic configuration of oxygen is [He]2s22p4 . Therefore, oxygen has six valence electrons.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, N2F+ has an electron less with that of the valence electrons of two nitrogen and fluorine atoms.

Thus, the total number of valence electrons of N2F+ is 2×5+71=16 and the total valence electrons of N2O is 2×5+6=16 .

Hence, the total number of valence electrons of N2F+ and N2O is same. Therefore, N2F+ and N2O are isoelectronic.

(b)

Interpretation Introduction

To determine: The Lewis structure of N2F+ .

(b)

Expert Solution
Check Mark

Answer to Problem 8.171AP

Solution

The Lewis structure of N2F+ shows 16 valence electrons.

Explanation of Solution

Explanation

The atomic number of nitrogen is 7 .

The electronic configuration of nitrogen is [He]2s22p3 . Therefore, nitrogen has five valence electrons.

The atomic number of fluorine is 9 .

The electronic configuration of fluorine is [He]2s22p5 . Therefore, fluorine has seven valence electrons.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, N2F+ has an electron less with that of the valence electrons of two nitrogen and fluorine atoms.

Thus, the total number of valence electrons of N2F+ is 2×5+71=16 .

Therefore, the Lewis structure of N2F+ is,

Chemistry, Chapter 8, Problem 8.171AP , additional homework tip  1

Figure 1

(c)

Interpretation Introduction

To determine: The atom that has the +1 formal charge in the Lewis structure.

(c)

Expert Solution
Check Mark

Answer to Problem 8.171AP

Solution

The atom that has +1 formal charge in the Lewis structure is the central nitrogen atom.

Explanation of Solution

Explanation

Formal charge is calculated by the formula,

Formalcharge=(Valenceelectrons(Numberofnonbondingelectrons+Numberofbondingelectrons2))

Therefore, the formal charge of nitrogen containing six bonding electrons and two nonbonding electrons is,

Formalcharge=(5(2+62))=55=0

The formal charge of nitrogen containing eight bonding electrons is,

Formalcharge=(5(0+82))=54=+1

The formal charge of nitrogen containing four bonding electrons and four nonbonding electrons is,

Formalcharge=(5(4+42))=56=1

(d)

Interpretation Introduction

To determine: The possibility of N2F+ to have resonance forms.

(d)

Expert Solution
Check Mark

Answer to Problem 8.171AP

Solution

There are three resonance forms of N2F+ .

Explanation of Solution

Explanation

The Lewis structure of N2F+ is,

Chemistry, Chapter 8, Problem 8.171AP , additional homework tip  2

Figure 1

Therefore, the possible resonance forms of N2F+ are,

  • Nitrogen bonded to nitrogen with triple bond an fluorine bonded to nitrogen with single bond.
  • Nitrogen bonded to nitrogen with double bond an fluorine bonded to nitrogen with double bond.
  • Nitrogen bonded to nitrogen with single bond an fluorine bonded to nitrogen with triple bond.

Thus, the resonating structures of N2F+ are,

Chemistry, Chapter 8, Problem 8.171AP , additional homework tip  3

Figure 2

Therefore, there are three resonance forms

(e)

Interpretation Introduction

To determine: The possibility of the central atom of N2F+ to be fluorine.

(e)

Expert Solution
Check Mark

Answer to Problem 8.171AP

Solution

It is possible for the central atom of N2F+ to be fluorine.

Explanation of Solution

Explanation

Principle quantum number of fluorine is 2 . Therefore, fluorine has one s and three p orbitals to form total four bonds. Therefore, being the central atom of N2F+ fluorine forms minimum two bond and maximum four bonds with the two nitrogen atoms.

Conclusion

  1. a. The total valence electrons of N2F+ is same as that of N2O . Therefore, N2F+ and N2O are isoelectronic.
  2. b. The Lewis structure of N2F+ shows 16 valence electrons.
  3. c. The atom that has +1 formal charge in the Lewis structure is the central nitrogen atom.
  4. d. There are three resonance forms of N2F+ .
  5. e. It is possible for the central atom of N2F+ to be fluorine.

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Chapter 8 Solutions

Chemistry

Ch. 8.7 - Prob. 11PECh. 8.8 - Prob. 12PECh. 8 - Prob. 8.1VPCh. 8 - Prob. 8.2VPCh. 8 - Prob. 8.3VPCh. 8 - Prob. 8.4VPCh. 8 - Prob. 8.5VPCh. 8 - Prob. 8.6VPCh. 8 - Prob. 8.7VPCh. 8 - Prob. 8.8VPCh. 8 - Prob. 8.9VPCh. 8 - Prob. 8.10VPCh. 8 - Prob. 8.11VPCh. 8 - Prob. 8.12VPCh. 8 - Prob. 8.13VPCh. 8 - Prob. 8.14VPCh. 8 - Prob. 8.15VPCh. 8 - Prob. 8.16VPCh. 8 - Prob. 8.17VPCh. 8 - Prob. 8.18VPCh. 8 - Prob. 8.19QPCh. 8 - Prob. 8.20QPCh. 8 - Prob. 8.21QPCh. 8 - Prob. 8.22QPCh. 8 - Prob. 8.23QPCh. 8 - Prob. 8.24QPCh. 8 - Prob. 8.25QPCh. 8 - Prob. 8.26QPCh. 8 - Prob. 8.27QPCh. 8 - Prob. 8.28QPCh. 8 - Prob. 8.29QPCh. 8 - Prob. 8.30QPCh. 8 - Prob. 8.31QPCh. 8 - Prob. 8.32QPCh. 8 - Prob. 8.33QPCh. 8 - Prob. 8.34QPCh. 8 - Prob. 8.35QPCh. 8 - Prob. 8.36QPCh. 8 - Prob. 8.37QPCh. 8 - Prob. 8.38QPCh. 8 - Prob. 8.39QPCh. 8 - Prob. 8.40QPCh. 8 - Prob. 8.41QPCh. 8 - Prob. 8.42QPCh. 8 - Prob. 8.43QPCh. 8 - Prob. 8.44QPCh. 8 - Prob. 8.45QPCh. 8 - Prob. 8.46QPCh. 8 - Prob. 8.47QPCh. 8 - Prob. 8.48QPCh. 8 - Prob. 8.49QPCh. 8 - Prob. 8.50QPCh. 8 - Prob. 8.51QPCh. 8 - Prob. 8.52QPCh. 8 - Prob. 8.53QPCh. 8 - Prob. 8.54QPCh. 8 - Prob. 8.55QPCh. 8 - Prob. 8.56QPCh. 8 - Prob. 8.57QPCh. 8 - Prob. 8.58QPCh. 8 - Prob. 8.59QPCh. 8 - Prob. 8.60QPCh. 8 - Prob. 8.61QPCh. 8 - Prob. 8.62QPCh. 8 - Prob. 8.63QPCh. 8 - Prob. 8.64QPCh. 8 - Prob. 8.65QPCh. 8 - Prob. 8.66QPCh. 8 - Prob. 8.67QPCh. 8 - Prob. 8.68QPCh. 8 - Prob. 8.69QPCh. 8 - Prob. 8.70QPCh. 8 - Prob. 8.71QPCh. 8 - Prob. 8.72QPCh. 8 - Prob. 8.73QPCh. 8 - Prob. 8.74QPCh. 8 - Prob. 8.75QPCh. 8 - Prob. 8.76QPCh. 8 - Prob. 8.77QPCh. 8 - Prob. 8.78QPCh. 8 - Prob. 8.79QPCh. 8 - Prob. 8.80QPCh. 8 - Prob. 8.81QPCh. 8 - Prob. 8.82QPCh. 8 - Prob. 8.83QPCh. 8 - Prob. 8.84QPCh. 8 - Prob. 8.85QPCh. 8 - Prob. 8.86QPCh. 8 - Prob. 8.87QPCh. 8 - Prob. 8.88QPCh. 8 - Prob. 8.89QPCh. 8 - Prob. 8.90QPCh. 8 - Prob. 8.91QPCh. 8 - Prob. 8.92QPCh. 8 - Prob. 8.93QPCh. 8 - Prob. 8.94QPCh. 8 - Prob. 8.95QPCh. 8 - Prob. 8.96QPCh. 8 - Prob. 8.97QPCh. 8 - Prob. 8.98QPCh. 8 - Prob. 8.99QPCh. 8 - Prob. 8.100QPCh. 8 - Prob. 8.101QPCh. 8 - Prob. 8.102QPCh. 8 - Prob. 8.103QPCh. 8 - Prob. 8.104QPCh. 8 - Prob. 8.105QPCh. 8 - Prob. 8.106QPCh. 8 - Prob. 8.107QPCh. 8 - Prob. 8.108QPCh. 8 - Prob. 8.109QPCh. 8 - Prob. 8.110QPCh. 8 - Prob. 8.111QPCh. 8 - Prob. 8.112QPCh. 8 - Prob. 8.113QPCh. 8 - Prob. 8.114QPCh. 8 - Prob. 8.115QPCh. 8 - Prob. 8.116QPCh. 8 - Prob. 8.117QPCh. 8 - Prob. 8.118QPCh. 8 - Prob. 8.119QPCh. 8 - Prob. 8.120QPCh. 8 - Prob. 8.121QPCh. 8 - Prob. 8.122QPCh. 8 - Prob. 8.123QPCh. 8 - Prob. 8.124QPCh. 8 - Prob. 8.125QPCh. 8 - Prob. 8.126QPCh. 8 - Prob. 8.127QPCh. 8 - Prob. 8.128QPCh. 8 - Prob. 8.129QPCh. 8 - Prob. 8.130QPCh. 8 - Prob. 8.131QPCh. 8 - Prob. 8.132QPCh. 8 - Prob. 8.133QPCh. 8 - Prob. 8.134QPCh. 8 - Prob. 8.135QPCh. 8 - Prob. 8.136QPCh. 8 - Prob. 8.137QPCh. 8 - Prob. 8.138QPCh. 8 - Prob. 8.139APCh. 8 - Prob. 8.140APCh. 8 - Prob. 8.141APCh. 8 - Prob. 8.142APCh. 8 - Prob. 8.143APCh. 8 - Prob. 8.144APCh. 8 - Prob. 8.145APCh. 8 - Prob. 8.146APCh. 8 - Prob. 8.147APCh. 8 - Prob. 8.148APCh. 8 - Prob. 8.149APCh. 8 - Prob. 8.150APCh. 8 - Prob. 8.151APCh. 8 - Prob. 8.152APCh. 8 - Prob. 8.153APCh. 8 - Prob. 8.154APCh. 8 - Prob. 8.155APCh. 8 - Prob. 8.156APCh. 8 - Prob. 8.157APCh. 8 - Prob. 8.158APCh. 8 - Prob. 8.159APCh. 8 - Prob. 8.160APCh. 8 - Prob. 8.161APCh. 8 - Prob. 8.162APCh. 8 - Prob. 8.163APCh. 8 - Prob. 8.164APCh. 8 - Prob. 8.165APCh. 8 - Prob. 8.166APCh. 8 - Prob. 8.167APCh. 8 - Prob. 8.168APCh. 8 - Prob. 8.169APCh. 8 - Prob. 8.170APCh. 8 - Prob. 8.171APCh. 8 - Prob. 8.172APCh. 8 - Prob. 8.173APCh. 8 - Prob. 8.174APCh. 8 - Prob. 8.175APCh. 8 - Prob. 8.176AP
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