The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from F 2 to I 2 . Two reasons are to be stated for high reactivity of F 2 among other halogens. The reason for the fact, the boiling point of HF is much higher than the boiling points of HCl, HBr , and HI . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have − 1, +1, +3, +5 and +7 oxidation states are to be stated. Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen. To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from F 2 to I 2 ; two reasons for high reactivity of F 2 among other halogens; the reason for the fact, the boiling point of HF is much higher than the boiling points of HCl, HBr , and HI ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have − 1, +1, +3, +5 and +7 oxidation states.

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Answer 1

Question

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

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(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

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Answer 2

Question

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

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Answer 3

Question

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

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Answer 4

Question

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

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Answer 5

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

Answer 6

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

Answer 7

Chapter 20, Problem 9RQ

Interpretation Introduction

Interpretation: The valence electron configuration for halogens is to be stated. The reason for the increase in the melting and boiling point from $F2$ to $I2$ . Two reasons are to be stated for high reactivity of $F2$ among other halogens. The reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ . The definition of halide ions and explanation for the stability of halide ions is to be stated. The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are to be stated.

Concept introduction: Halogen elements are highly reactive. All are nonmetals. They form hydrides when react with hydrogen.

To determine: The valence electron configuration for halogens; the reason for the increase in the melting and boiling point from $F2$ to $I2$ ; two reasons for high reactivity of $F2$ among other halogens; the reason for the fact, the boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ ; the definition of halide ions and explanation for the stability of halide ions; the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states.

Answer 8

Expert Solution & Answer

Answer to Problem 9RQ

Answer

Solutions are as follows.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Explanation

The valence electron configuration for halogens is $ns2np5$ .

The common halogen elements are,

Fluorine
Chlorine
Bromine
Iodine

The valence electron configuration for halogens is $ns2np5$ .

Explanation

Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength.

The melting and boiling point of halogens depend on bond strength $X−X$ (where $X$ is halogen atoms). Higher the bond strength, higher will be the melting and boiling point. Since, bond strength increases down the group. Therefore, melting and boiling point increases from $F2$ to $I2$ .

Explanation

The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low.

The molecule $F2$ is most reactive due to the following reasons,

The size of $F2$ is very small as compared to other halogens.
The energy of $F−F$ bond is low due to non-bonding electrons in this molecule.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

The boiling point of compound is said to be high if it can form hydrogen bonds.

The electronegativity of halogen atoms decreases down the group. Fluorine molecule is most electronegative. Due to high electronegativity, $F2$ molecule easily forms hydrogen bonds. On the other hand, electronegativity of other halogen groups is very small that makes them unable to form hydrogen bonds.

The boiling point of $HF$ is much higher than the boiling points of $HCl, HBr$ , and $HI$ because Fluorine has the ability to form hydrogen bond.

Addition of an electron to hydride yields halide ion $(X−)$ . Halide salts are stable because they obtain inert gas electron configuration.

The combination of hydrogen $(H)$ with halogens $(X)$ forms hydrides. The reaction is,

$X2+H2→2HX$

Addition of an electron to hydride yields halide ion $(X−)$ . The reaction is,

$X+e−→X−$

Halide salts are stable because they obtain inert gas electron configuration.

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

The compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

The compound of Chlorine that has $−1$ oxidation state is $HCl$ , that is,

In $HCl$ , the oxidation state of Hydrogen atom $(H)$ is $+1$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Hydrogen and Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$1+x=0x=−1$

Therefore, the oxidation state of Chlorine in $HCl$ is $−1$ .

The compound of Chlorine that has $+1$ oxidation state is $Cl2O$ , that is,

In $Cl2O$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains single Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$(−2)+2x=0x=+1$

Therefore, the oxidation state of Chlorine in $Cl2O$ is $+1$ .

The compound of Chlorine that has $+3$ oxidation state is $Cl2O3$ , that is,

In $Cl2O3$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains three Oxygen and two Chlorine atom. Since, overall charge on stable compound is zero. Therefore,

$3(−2)+2x=0x=+3$

Therefore, the oxidation state of Chlorine in $Cl2O3$ is $+3$ .

The compound of Chlorine that has $+5$ oxidation state is $Cl2O5$ , that is,

In $Cl2O5$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains five Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$5(−2)+2x=0x=+5$

Therefore, the oxidation state of Chlorine in $Cl2O5$ is $+5$ .

The compound of Chlorine that has $+7$ oxidation state is $Cl2O7$ , that is,

In $Cl2O7$ , the oxidation state of Oxygen atom $(O)$ is $−2$ .

It is assumed that oxidation state of Chlorine $(Cl)$ is $x$ .

The given compound contains seven Oxygen and two Chlorine atoms. Since, overall charge on stable compound is zero. Therefore,

$7(−2)+2x=0x=+7$

Therefore, the oxidation state of Chlorine in $Cl2O7$ is $+7$ .

Hence, the compounds of chlorine that have $−1, +1, +3, +5$ and $+7$ oxidation states are,

$HCl, Cl2O, Cl2O3, Cl2O5$ and $Cl2O7$

Answer 12

Conclusion

Conclusion

Halogen elements are highly reactive. The valence electron configuration for the halogen elements is $ns2np5$ . Melting and boiling point increases from $F2$ to $I2$ due to increase in the bond strength. The molecule $F2$ is most reactive because the size of $F2$ is very small and the energy of $F−F$ bond is low

Answer 13

Ch. 20 - What are the two most abundant elements by mass in...Ch. 20 - Prob. 2RQ Ch. 20 - Prob. 3RQ Ch. 20 - What is the valence electron configuration for the...Ch. 20 - Prob. 5RQ Ch. 20 - Prob. 6RQ Ch. 20 - Table 19-14 lists some common nitrogen compounds...Ch. 20 - Prob. 8RQ Ch. 20 - Prob. 9RQ Ch. 20 - Prob. 10RQ

Answer to Problem 9RQ

Explanation of Solution

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