For each of the following groups, place the atoms and/or ions in order of decreasing size. a. V, V 2+ , V 3+ , V 5+ b. Na + , K + , Rb + , Cs + c. Te 2− , I − , Cs + , Ba 2+ d. P, P − , P 2− , P 3− . e. O 2− , S 2− , Se 2− , Te 2−

Question

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Answer 1

Textbook Question

Chapter 8, Problem 54E

For each of the following groups, place the atoms and/or ions in order of decreasing size.

a. V, V²⁺, V³⁺, V⁵⁺

b. Na⁺, K⁺, Rb⁺, Cs⁺

c. Te²⁻, I⁻, Cs⁺, Ba²⁺

d. P, P⁻, P²⁻, P³⁻.

e. O²⁻, S²⁻, Se²⁻, Te²⁻

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

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Answer 2

Textbook Question

Chapter 8, Problem 54E

For each of the following groups, place the atoms and/or ions in order of decreasing size.

a. V, V²⁺, V³⁺, V⁵⁺

b. Na⁺, K⁺, Rb⁺, Cs⁺

c. Te²⁻, I⁻, Cs⁺, Ba²⁺

d. P, P⁻, P²⁻, P³⁻.

e. O²⁻, S²⁻, Se²⁻, Te²⁻

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

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Answer 3

Textbook Question

Chapter 8, Problem 54E

For each of the following groups, place the atoms and/or ions in order of decreasing size.

a. V, V²⁺, V³⁺, V⁵⁺

b. Na⁺, K⁺, Rb⁺, Cs⁺

c. Te²⁻, I⁻, Cs⁺, Ba²⁺

d. P, P⁻, P²⁻, P³⁻.

e. O²⁻, S²⁻, Se²⁻, Te²⁻

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

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Answer 4

Textbook Question

Chapter 8, Problem 54E

For each of the following groups, place the atoms and/or ions in order of decreasing size.

a. V, V²⁺, V³⁺, V⁵⁺

b. Na⁺, K⁺, Rb⁺, Cs⁺

c. Te²⁻, I⁻, Cs⁺, Ba²⁺

d. P, P⁻, P²⁻, P³⁻.

e. O²⁻, S²⁻, Se²⁻, Te²⁻

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $V ,V2+,V3+,V5+$ .

Answer 13

Answer to Problem 54E

The size of ions

$V ,V2+,V3+,V5+$ decreases in the order

$V>V2+>V3+>V5+$ .

Answer 14

Explanation of Solution

In a periodic table the size of ions depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

The ions $V2+,V3+and V5+$ are positive ions where $V5+$ is smallest because it contains large positive charge and have high nuclear attraction.

Hence, the decreasing order is $V>V2+>V3+>V5+$ .

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Na+ ,K+,Rb+ and Cs+$ .

Answer 20

Answer to Problem 54E

The size of ions

$Na+ ,K+,Rb+ and Cs+$ decreases in the order

$Cs+>Rb+ >K+>Na+$ .

Answer 21

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $Na , K , Rb and Cs$ lie in 1^st group. Hence, the decreasing order is $Cs>Rb >K>Na$ .

The size of ions also depends on the nuclear attraction on the valence electrons. Positive ions are formed by removing electrons from outer shell. Hence, formation of positive ions or cation not only vacant the orbitals but also decrease the electron-electron repulsion. As a result size of cations is smaller than neutral atom.

Hence, the decreasing order of ions is $Cs+>Rb+ >K+>Na+$ .

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $Te2− ,I−,Cs+,Ba2+$ .

Answer 27

Answer to Problem 54E

The size of ions

$Te2− ,I−,Cs+,Ba2+$ decreases in the order

$Te2−>I−>Cs+>Ba2+$ .

Answer 28

Explanation of Solution

In a periodic table, the size of elements decreases from left to right along the period. Also if the periodic number increases then atomic size also increases. The elements $Cs and Ba$ lie in 6^th period and elements $Te and I$ lie in 5^th period. Hence, the decreasing order of atomic size is $Cs>Ba>Te>I$ .

The electronic configuration of ions $Te2−,I−,Cs+,Ba2+$ is,

$Te2−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$I−$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Cs+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

$Ba2+$ : $1s22s22p63s23p63d104s24p64d105s25p6$

The number of electrons present in $Te2−,I−,Cs+,Ba2+$ ions is 54. Hence these ions are isoelectronic. Since the atomic size is increased with increase of anion charge and decrease with increase of cation charge for isoelectronic configuration, the order of decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ ,

Answer 29

Conclusion

For isoelectronic ions, the size of atom decreases with increase in cation charge and increase with anion charge. Hence, the decreasing size of ions is $Te2−>I−>Cs+>Ba2+$ .

Answer 30

(d)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

Answer 31

(d)

Expert Solution

Answer 32

(d)

Expert Solution

Answer 33

Expert Solution

Answer 34

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $P , P−, P2−,P3−$ .

Answer 35

Answer to Problem 54E

The size of ions

$P , P−, P2−,P3−$ decreases in the order

$P3−>P2−>P−>P$ .

Answer 36

Explanation of Solution

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $P3−>P2−>P−>P$ .

Answer 37

Conclusion

The size of ions depends on the gaining or losing of electrons. The size of ions increases by gaining electrons whereas size decreases by losing electrons. The decreasing order of ions is $P3−>P2−>P−>P$

Answer 38

(e)

Expert Solution

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .

Answer 39

(e)

Expert Solution

Answer 40

(e)

Expert Solution

Answer 41

Expert Solution

Answer 42

Interpretation Introduction

Interpretation: The given ions are to be placed in decreasing order of size.

Concept introduction: The ionic size depends on the number of electrons transferred. The atomic size of cations is smaller than anions. Also the size of ions increases by gaining electrons whereas size decreases by accepting electrons.

To determine: The decreasing order of size of ions $O2−, S−, Se2−,Te2−$ .

Answer 43

Answer to Problem 54E

The size of ions

$O2−, S−, Se2−,Te2−$ decreases in the order

$Te2−>Se2−>S−>O2−$ .

Answer 44

Explanation of Solution

In a periodic table, the size of elements increases down the group. The elements $O , S , Se ,Te$ lie in 16^th group. Hence, the decreasing order is $Te > Se>S > O$ .

Addition of electrons leads to increase in electron-electron repulsion which causes the electrons to spread out more in space. As a result, size of anion is larger than neutral atom.

Hence, the decreasing order of ions is $Te2−>Se2−>S−>O2−$ .