CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<
CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<
7th Edition
ISBN: 9781319090876
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 2, Problem 2G.3E

(a)

Interpretation Introduction

Interpretation:

Valence shell electronic configuration of F2, F2+ and F22 species has to be determined.

Concept Introduction:

Molecular orbital diagram is a linear combination of atomic orbitals of similar energy and similar symmetry. It is formed by the proper overlap of the atomic orbitals.

There are 3 types of molecular orbitals as follows:

1. Bonding molecular orbital: They are formed by the constructive interference of atomic orbitals and electrons in it stabilize the molecule and are of lesser in energy.

2. Antibonding molecular orbital: This type of orbitals increases the energy of molecule and destabilizes it and weakens the bond between the atoms.

3. Non-bonding molecular orbital: These types of orbitals have energy similar to atomic orbitals that is addition or removal of electron does not change the energy of molecule.

The order of energy in molecular orbital follows two rules as follows:

1. For atomic number less than or equal to 14 order of energy is,

  σ1s<σ1s<σ2s<σ2s<(π2px=π2py)<σ2pz<(π2px=π2py)<σ2pz

2. For atomic number more than 14 order of energy is,

  σ1s<σ1s<σ2s<σ2s<σ2pz<(π2px=π2py)<(π2px=π2py)<σ2pz

(a)

Expert Solution
Check Mark

Explanation of Solution

For F2 molecule:

The symbol for fluorine is F with atomic number 9. Its electronic configuration is [He]2s22p5. Thus, it possesses 7 valence electrons.

One negative charge is added up in total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in F2. It is calculated as follows:

  Total valence electrons=7(2)+1=15

Hence, 15 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in F2 is more than 14 so electronic configuration is (σ2s)2(σ2s*)2(σ2p)2( π2p)4( π2p*)4(σ2p)1.

Hence, the electronic configuration of F2 is (σ2s)2(σ2s*)2(σ2p)2( π2p)4( π2p*)4(σ2p)1.

For F2+ molecule:

The symbol for fluorine is F with atomic number 9. Its electronic configuration is [He]2s22p5. Thus, it possesses 7 valence electrons.

One positive charge is subtracted from the total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in F2+. It is calculated as follows:

  Total valence electrons=7(2)1=13

Hence, 13 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in F2+ is more than 14 so electronic configuration is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3.

Hence, the electronic configuration of F2+ is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3.

For F22 molecule:

The symbol for fluorine is F with atomic number 9. Its electronic configuration is [He]2s22p5. Thus, it possesses 7 valence electrons.

Two negative charges are added up in total valence count.

Thus total valence electrons is sum of the valence electrons for each atom in F22. It is calculated as follows:

  Total valence electrons=7(2)+2=16

Hence, 16 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in F22 is more than 14 so electronic configuration is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p)2.

Hence, the electronic configuration of F22 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p)2.

(b)

Interpretation Introduction

Interpretation:

Bond order of F2, F2+ and F22 species has to be determined.

Concept Introduction:

Bond order b is defined as total number of bonds between the atoms. It is calculated as follows:

  b=12[number of elctrons in bonding orbitalsnumber of electrons in antibonding orbitals]        (1)

(b)

Expert Solution
Check Mark

Explanation of Solution

The electronic configuration of F2 is (σ2s)2(σ2s*)2(σ2p)2( π2p)4( π2p*)4(σ2p)1.

Substitute 8 for number of electrons in bonding orbitals and 7 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  b=12(87)=12=0.5

Hence, the bond order of the molecule F2 is 0.5.

The electronic configuration of F2+ is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3.

Substitute 8 for number of electrons in bonding orbitals and 5 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  b=12(85)=32=1.5

Hence, the bond order of the molecule F2+ is 1.5.

The electronic configuration of F22 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p)2.

Substitute 8 for number of electrons in bonding orbitals and 8 for number of electrons in antibonding orbitals in equation (1) to calculate bond order of F22.

  b=12(88)=0

Hence, the bond order of the molecule F22 is 0.

(c)

Interpretation Introduction

Interpretation:

Molecular orbital diagram of F2, F2+ and F22 species have to be drawn.

Concept Introduction:

Refer to part (a)

(c)

Expert Solution
Check Mark

Explanation of Solution

For F2 molecule:

The electronic configuration of F2 is (σ2s)2(σ2s*)2(σ2p)2( π2p)4( π2p*)4(σ2p)1. The molecular orbital diagram of F2 molecule is as follows:

CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<, Chapter 2, Problem 2G.3E , additional homework tip  1

Since F2 molecule has one unpaired electron, therefore it is paramagnetic.

For F2+ molecule:

The electronic configuration of F2+ is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3. The molecular orbital diagram of F2+ molecule is as follows:

CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<, Chapter 2, Problem 2G.3E , additional homework tip  2

Since F2+ molecule has one unpaired electron, therefore it is paramagnetic.

For F22 molecule:

The electronic configuration of F22 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p)2.

The molecular orbital diagram of F22 molecule is as follows:

CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<, Chapter 2, Problem 2G.3E , additional homework tip  3

Since F22 molecule has no unpaired electron, therefore it is diamagnetic.

(d)

Interpretation Introduction

Interpretation:

Character of highest energy orbital of F2 F2+ and F22 species has to be determined.

Concept Introduction:

Refer to part (a)

(d)

Expert Solution
Check Mark

Explanation of Solution

For F2 molecule:

The electronic configuration of F2 is (σ2s)2(σ2s*)2(σ2p)2( π2p)4( π2p*)4(σ2p)1. The highest energy orbital electron in F2 is present in (σ2p)1. Therefore, it contains σ character.

For F2+ molecule:

The electronic configuration of F2+ is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3. The highest energy orbital electron in F2+ is present in (π2p*)3. Therefore, it contains π character.

For F22 molecule:

The electronic configuration of F22 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p)2. The highest energy orbital electron in F22 is present in (σ2p)2. Therefore, it contains σ character.

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Chapter 2 Solutions

CHEMICAL PRIN. (LL)+ 2 SEM SAPLING >BI<

Ch. 2 - Prob. 2A.3ECh. 2 - Prob. 2A.4ECh. 2 - Prob. 2A.5ECh. 2 - Prob. 2A.6ECh. 2 - Prob. 2A.7ECh. 2 - Prob. 2A.8ECh. 2 - Prob. 2A.9ECh. 2 - Prob. 2A.10ECh. 2 - Prob. 2A.11ECh. 2 - Prob. 2A.12ECh. 2 - Prob. 2A.13ECh. 2 - Prob. 2A.14ECh. 2 - Prob. 2A.15ECh. 2 - Prob. 2A.16ECh. 2 - Prob. 2A.17ECh. 2 - Prob. 2A.18ECh. 2 - Prob. 2A.19ECh. 2 - Prob. 2A.20ECh. 2 - Prob. 2A.21ECh. 2 - Prob. 2A.22ECh. 2 - Prob. 2A.23ECh. 2 - Prob. 2A.24ECh. 2 - Prob. 2A.25ECh. 2 - Prob. 2A.26ECh. 2 - Prob. 2A.27ECh. 2 - Prob. 2A.28ECh. 2 - Prob. 2A.29ECh. 2 - Prob. 2A.30ECh. 2 - Prob. 2B.1ASTCh. 2 - Prob. 2B.1BSTCh. 2 - Prob. 2B.2ASTCh. 2 - Prob. 2B.2BSTCh. 2 - Prob. 2B.3ASTCh. 2 - Prob. 2B.3BSTCh. 2 - Prob. 2B.4ASTCh. 2 - Prob. 2B.4BSTCh. 2 - Prob. 2B.5ASTCh. 2 - Prob. 2B.5BSTCh. 2 - Prob. 2B.1ECh. 2 - Prob. 2B.2ECh. 2 - Prob. 2B.3ECh. 2 - Prob. 2B.4ECh. 2 - Prob. 2B.5ECh. 2 - Prob. 2B.6ECh. 2 - Prob. 2B.7ECh. 2 - Prob. 2B.8ECh. 2 - Prob. 2B.9ECh. 2 - Prob. 2B.10ECh. 2 - Prob. 2B.11ECh. 2 - Prob. 2B.12ECh. 2 - Prob. 2B.13ECh. 2 - Prob. 2B.14ECh. 2 - Prob. 2B.15ECh. 2 - Prob. 2B.16ECh. 2 - Prob. 2B.17ECh. 2 - Prob. 2B.18ECh. 2 - Prob. 2B.19ECh. 2 - Prob. 2B.20ECh. 2 - Prob. 2B.21ECh. 2 - Prob. 2B.22ECh. 2 - Prob. 2B.23ECh. 2 - Prob. 2B.24ECh. 2 - Prob. 2C.1ASTCh. 2 - Prob. 2C.1BSTCh. 2 - Prob. 2C.2ASTCh. 2 - Prob. 2C.2BSTCh. 2 - Prob. 2C.3ASTCh. 2 - Prob. 2C.3BSTCh. 2 - Prob. 2C.1ECh. 2 - Prob. 2C.2ECh. 2 - Prob. 2C.3ECh. 2 - Prob. 2C.4ECh. 2 - Prob. 2C.5ECh. 2 - Prob. 2C.6ECh. 2 - Prob. 2C.7ECh. 2 - Prob. 2C.8ECh. 2 - Prob. 2C.9ECh. 2 - Prob. 2C.10ECh. 2 - Prob. 2C.11ECh. 2 - Prob. 2C.12ECh. 2 - Prob. 2C.13ECh. 2 - Prob. 2C.14ECh. 2 - Prob. 2C.15ECh. 2 - Prob. 2C.16ECh. 2 - Prob. 2C.17ECh. 2 - Prob. 2C.18ECh. 2 - Prob. 2D.1ASTCh. 2 - Prob. 2D.1BSTCh. 2 - Prob. 2D.2ASTCh. 2 - Prob. 2D.2BSTCh. 2 - Prob. 2D.1ECh. 2 - Prob. 2D.2ECh. 2 - Prob. 2D.3ECh. 2 - Prob. 2D.4ECh. 2 - Prob. 2D.5ECh. 2 - Prob. 2D.6ECh. 2 - Prob. 2D.7ECh. 2 - Prob. 2D.8ECh. 2 - Prob. 2D.9ECh. 2 - Prob. 2D.10ECh. 2 - Prob. 2D.11ECh. 2 - Prob. 2D.12ECh. 2 - Prob. 2D.13ECh. 2 - Prob. 2D.14ECh. 2 - Prob. 2D.15ECh. 2 - Prob. 2D.16ECh. 2 - Prob. 2D.17ECh. 2 - Prob. 2D.18ECh. 2 - Prob. 2D.19ECh. 2 - Prob. 2D.20ECh. 2 - Prob. 2E.1ASTCh. 2 - Prob. 2E.1BSTCh. 2 - Prob. 2E.2ASTCh. 2 - Prob. 2E.2BSTCh. 2 - Prob. 2E.3ASTCh. 2 - Prob. 2E.3BSTCh. 2 - Prob. 2E.4ASTCh. 2 - Prob. 2E.4BSTCh. 2 - Prob. 2E.5ASTCh. 2 - Prob. 2E.5BSTCh. 2 - Prob. 2E.1ECh. 2 - Prob. 2E.2ECh. 2 - Prob. 2E.3ECh. 2 - Prob. 2E.4ECh. 2 - Prob. 2E.5ECh. 2 - Prob. 2E.6ECh. 2 - Prob. 2E.7ECh. 2 - Prob. 2E.8ECh. 2 - Prob. 2E.9ECh. 2 - Prob. 2E.10ECh. 2 - Prob. 2E.11ECh. 2 - Prob. 2E.12ECh. 2 - Prob. 2E.13ECh. 2 - Prob. 2E.14ECh. 2 - Prob. 2E.15ECh. 2 - Prob. 2E.16ECh. 2 - Prob. 2E.17ECh. 2 - Prob. 2E.18ECh. 2 - Prob. 2E.19ECh. 2 - Prob. 2E.20ECh. 2 - Prob. 2E.21ECh. 2 - Prob. 2E.22ECh. 2 - Prob. 2E.23ECh. 2 - Prob. 2E.24ECh. 2 - Prob. 2E.25ECh. 2 - Prob. 2E.26ECh. 2 - Prob. 2E.27ECh. 2 - Prob. 2E.28ECh. 2 - Prob. 2E.29ECh. 2 - Prob. 2E.30ECh. 2 - Prob. 2F.1ASTCh. 2 - Prob. 2F.1BSTCh. 2 - Prob. 2F.2ASTCh. 2 - Prob. 2F.2BSTCh. 2 - Prob. 2F.3ASTCh. 2 - Prob. 2F.3BSTCh. 2 - Prob. 2F.4ASTCh. 2 - Prob. 2F.4BSTCh. 2 - Prob. 2F.1ECh. 2 - Prob. 2F.2ECh. 2 - Prob. 2F.3ECh. 2 - Prob. 2F.4ECh. 2 - Prob. 2F.5ECh. 2 - Prob. 2F.6ECh. 2 - Prob. 2F.7ECh. 2 - Prob. 2F.8ECh. 2 - Prob. 2F.9ECh. 2 - Prob. 2F.10ECh. 2 - Prob. 2F.11ECh. 2 - Prob. 2F.12ECh. 2 - Prob. 2F.13ECh. 2 - Prob. 2F.14ECh. 2 - Prob. 2F.15ECh. 2 - Prob. 2F.16ECh. 2 - Prob. 2F.17ECh. 2 - Prob. 2F.18ECh. 2 - Prob. 2F.19ECh. 2 - Prob. 2F.20ECh. 2 - Prob. 2F.21ECh. 2 - Prob. 2G.1ASTCh. 2 - Prob. 2G.1BSTCh. 2 - Prob. 2G.2ASTCh. 2 - Prob. 2G.2BSTCh. 2 - Prob. 2G.1ECh. 2 - Prob. 2G.2ECh. 2 - Prob. 2G.3ECh. 2 - Prob. 2G.4ECh. 2 - Prob. 2G.5ECh. 2 - Prob. 2G.6ECh. 2 - Prob. 2G.7ECh. 2 - Prob. 2G.8ECh. 2 - Prob. 2G.9ECh. 2 - Prob. 2G.11ECh. 2 - Prob. 2G.12ECh. 2 - Prob. 2G.13ECh. 2 - Prob. 2G.14ECh. 2 - Prob. 2G.15ECh. 2 - Prob. 2G.16ECh. 2 - Prob. 2G.17ECh. 2 - Prob. 2G.18ECh. 2 - Prob. 2G.19ECh. 2 - Prob. 2G.20ECh. 2 - Prob. 2G.21ECh. 2 - Prob. 2G.22ECh. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 2.4ECh. 2 - Prob. 2.5ECh. 2 - Prob. 2.6ECh. 2 - Prob. 2.7ECh. 2 - Prob. 2.8ECh. 2 - Prob. 2.9ECh. 2 - Prob. 2.10ECh. 2 - Prob. 2.11ECh. 2 - Prob. 2.12ECh. 2 - Prob. 2.13ECh. 2 - Prob. 2.14ECh. 2 - Prob. 2.17ECh. 2 - Prob. 2.19ECh. 2 - Prob. 2.22ECh. 2 - Prob. 2.23ECh. 2 - Prob. 2.24ECh. 2 - Prob. 2.25ECh. 2 - Prob. 2.26ECh. 2 - Prob. 2.27ECh. 2 - Prob. 2.28ECh. 2 - Prob. 2.29ECh. 2 - Prob. 2.30ECh. 2 - Prob. 2.31ECh. 2 - Prob. 2.32ECh. 2 - Prob. 2.33ECh. 2 - Prob. 2.34ECh. 2 - Prob. 2.35ECh. 2 - Prob. 2.36ECh. 2 - Prob. 2.37ECh. 2 - Prob. 2.39ECh. 2 - Prob. 2.40ECh. 2 - Prob. 2.41ECh. 2 - Prob. 2.42ECh. 2 - Prob. 2.43ECh. 2 - Prob. 2.44ECh. 2 - Prob. 2.45ECh. 2 - Prob. 2.46ECh. 2 - Prob. 2.47ECh. 2 - Prob. 2.48ECh. 2 - Prob. 2.49ECh. 2 - Prob. 2.50ECh. 2 - Prob. 2.51ECh. 2 - Prob. 2.52ECh. 2 - Prob. 2.53ECh. 2 - Prob. 2.54ECh. 2 - Prob. 2.55ECh. 2 - Prob. 2.56ECh. 2 - Prob. 2.57ECh. 2 - Prob. 2.58ECh. 2 - Prob. 2.59ECh. 2 - Prob. 2.60ECh. 2 - Prob. 2.61ECh. 2 - Prob. 2.62ECh. 2 - Prob. 2.63ECh. 2 - Prob. 2.64E
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