Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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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
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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+andV5+ 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
Check Mark
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+andCs+.

Answer to Problem 54E

The size of ions Na+,K+,Rb+andCs+ 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,RbandCs lie in 1st 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
Check Mark
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 CsandBa lie in 6th period and elements TeandI lie in 5th 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
Check Mark
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
Check Mark
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 16th 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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Chapter 8 Solutions

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