Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified. Concept Introduction: Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. According to this theory there are two types of orbitals, (1) Bonding orbitals (2) Antibonding orbitals Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals. The electronic configuration of oxygen molecule O 2 can be represented as follows, ( σ 1s ) 2 ( σ* 1s ) 2 ( σ 2s ) 2 ( σ* 2s ) 2 ( σ 2p ) 2 ( π 2p ) 4 ( π * 2p ) 2 The symbol * represent the antibonding orbital Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis. Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule. Bond order: It is the measure of number of electron pairs shared between two atoms. Bond order = 1 2 ( Number of electrons in bondoing MOs- Number of electrons in antibonding MOs ) Bond length is inversely proportional to the bond order. Atoms with unpaired electrons are called Paramagnetic . Paramagnetic atoms are attracted to a magnet. Atoms with paired electrons are called diamagnetic . Diamagnetic atoms are repelled by a magnet

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Answer 1

Question

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

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Students have asked these similar questions

(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³

fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer

34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10

Answer 2

Question

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

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Answer 3

Question

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

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Answer 4

Question

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

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Answer 5

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

Answer 6

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$

$F2+Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-72=1.5$

Therefore,

The bond order decreases as follows,

$Bond order: F2+ > F2 > F2− 1.5 > 1 > 0.5$

Bond order is inversely proportional to bond length.

Thus, increasing bond length is,

$Bond length: F2+ < F2 < F2−$

Answer 7

Chapter 9, Problem 9B.2E

Interpretation Introduction

Interpretation:

Increasing order of bond length of the given species has to be arranged and also the bond order of each has to be identified.

Concept Introduction:

Molecular orbital (MO) theory is a method of determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

(1) Bonding orbitals

(2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule $O2$ can be represented as follows,

$(σ1s)2(σ1s)2(σ2s)2(σ2s)2 (σ2p)2( π2p)4( π2p)2$

The symbol represent the antibonding orbital

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared un-hybridized orbital’s (p, d, etc.) electron density are concentrated in above and below of the plane of the molecule.

Bond order: It is the measure of number of electron pairs shared between two atoms.

$Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)$

Bond length is inversely proportional to the bond order.

Atoms with unpaired electrons are called Paramagnetic. Paramagnetic atoms are attracted to a magnet.

Atoms with paired electrons are called diamagnetic. Diamagnetic atoms are repelled by a magnet

Answer 8

Expert Solution & Answer

Explanation of Solution

Given species: $F2−$

There are $19$ electrons in $F2−$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2 (σ*2pz)1$

MO diagram for $F2−$ ion can be drawn as follows,

$(σ*2pz) ↑ → (Anti bonding electrons) ↑↓ − ↑↓ (π*2px) (π*2py) −−↑↓−↑↓−↑−− −−↑↓−↑↓−↑↓−− 2px 2py 2pz 2px 2py 2pz ↑↓ −↑↓ ( π2px) (π2py) ↑↓→ ( Bonding electrons) ( σ2pz)$

$−−↑↓−− → (Anti bonding electrons) (σ2s)* −−↑↓−− −−↑↓−− 2s 2s −−↑↓−− →(Bonding electrons) (σ2s)2$

$−−↑↓−− → (Anti bonding electrons) (σ1s)* −−↑↓−− −−↑↓−− 1s 1s −−↑↓−− →(Bonding electrons) (σ1s)2$

$F2−Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-92=0.5$

Given species: $F2$

There are $18$ electrons in $F2$ . According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)2$

$F2Bond order = 12(Number of electrons in bondoing MOs-Number of electrons in antibonding MOs)= 10-82=1$

Given species: $F2+$

There are $17$ electrons in $F2+$ one electron is added to the anti-bonding orbital. According to the MO theory its electron configuration can be written as,

$(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2 (σ2p)2( π2px)2( π2py)2( π*2px)2 ( π*2py)1$