Chapter 13, Problem 29E

(a)

Interpretation Introduction

Interpretation:

The following atoms and ions should be placed in order of decreasing size.

  $\text{Cu},{\text{ Cu}}^{+},{\text{Cu}}^{2+}$

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

The outermost electrons in an atom are known as valence electrons. The size of atoms or ions depends on the attraction between nucleus and valence electrons. When the electron(s) is(are) removed from an atom results in the formation of cation and increase in nuclear charge attraction. When the electron(s) is(are) added in an atom results in the formation of anion and decrease in nuclear charge.

Now, if the nuclear charge of an atom or ion is lower then, the size of atom or ion is large, if whereas the nuclear charge of an atom or ion is greater then, the size of atom or ion is small.

Answer to Problem 29E

The order of decreasing size is ${\text{Cu > Cu}}^{+}{\text{> Cu}}^{2+}$ .

Explanation of Solution

Anions are formed when a neutral atom or element gain electron(s) whereas cations are formed when a neutral atom or element lose electron(s).

The given atoms and ions are: $\text{Cu},{\text{ Cu}}^{+},{\text{Cu}}^{2+}$

Now, according to the rules, when positive charge on atom increases then, the nuclear attraction on valence electrons also increases and this results in decrease in size.

Hence, size of cation is smaller than the size of neutral atom as removal of electrons occurs.

Thus, the order of decreasing size is ${\text{Cu > Cu}}^{+}{\text{> Cu}}^{2+}$ .

(b)

Interpretation Introduction

Interpretation:

The following atoms and ions should be placed in order of decreasing size.

  ${\text{Ni}}^{2+},{\text{ Pd}}^{2+},{\text{Pt}}^{2+}$

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

The outermost electrons in an atom are known as valence electrons. The size of atoms or ions depends on the attraction between nucleus and valence electrons. When the electron(s) is(are) removed from an atom results in the formation of cation and increase in nuclear charge attraction. When the electron(s) is(are) added in an atom results in the formation of anion and decrease in nuclear charge.

Now, if the nuclear charge of an atom or ion is lower then, the size of atom or ion is large, if whereas the nuclear charge of an atom or ion is greater then, the size of atom or ion is small.

Answer to Problem 29E

The order of decreasing size is ${\text{Pt}}^{2+}>{\text{Pd}}^{2+}{\text{> Ni}}^{2+}$ .

Explanation of Solution

Anions are formed when a neutral atom or element gain electron(s) whereas cations are formed when a neutral atom or element lose electron(s).

The given atoms and ions are: ${\text{Ni}}^{2+},{\text{ Pd}}^{2+},{\text{Pt}}^{2+}$

Now, $\text{Ni},\text{ Pd},\text{Pt}$ belongs to group 10 of the periodic table.

In periodic table, down the group atomic size increases whereas the left to right atomic size decreases. Therefore, increasing order is $\text{Ni}<\text{Pd}<\text{Pt}$

Thus, the order of decreasing size is $\text{Pt}>\text{Pd> Ni}$ .

Similarly, order of decreasing size for given ions is ${\text{Pt}}^{2+}>{\text{Pd}}^{2+}{\text{> Ni}}^{2+}$

(c)

Interpretation Introduction

Interpretation:

The following atoms and ions should be placed in order of decreasing size.

  $\text{O},{\text{ O}}^{-},{\text{O}}^{2-}$

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

The outermost electrons in an atom are known as valence electrons. The size of atoms or ions depends on the attraction between nucleus and valence electrons. When the electron(s) is(are) removed from an atom results in the formation of cation and increase in nuclear charge attraction. When the electron(s) is(are) added in an atom results in the formation of anion and decrease in nuclear charge.

Now, if the nuclear charge of an atom or ion is lower then, the size of atom or ion is large, if whereas the nuclear charge of an atom or ion is greater then, the size of atom or ion is small.

Answer to Problem 29E

The order of decreasing size is ${\text{O}}^{2-}{\text{> O}}^{-}\text{> O}$ .

Explanation of Solution

Anions are formed when a neutral atom or element gain electron(s) whereas cations are formed when a neutral atom or element lose electron(s).

The given atoms and ions are: $\text{O},{\text{ O}}^{-},{\text{O}}^{2-}$

Now, according to the rules, $\mathrm{O}$ is formed when one electron is added in neutral atom (O) and $\mathrm{O}$ is formed when two electrons are added in neutral atom (O).

Hence, size of anion is larger than the size of neutral atom as addition of electrons occurs.

Thus, the order of decreasing size is ${\text{O}}^{2-}{\text{> O}}^{-}\text{> O}$ .

(d)

Interpretation Introduction

Interpretation:

The following atoms and ions should be placed in order of decreasing size.

  ${\text{La}}^{3+},{\text{ Eu}}^{3+},{\text{Gd}}^{\text{3+}}{\text{,Yb}}^{\text{3+}}$

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

The outermost electrons in an atom are known as valence electrons. The size of atoms or ions depends on the attraction between nucleus and valence electrons. When the electron(s) is(are) removed from an atom results in the formation of cation and increase in nuclear charge attraction. When the electron(s) is(are) added in an atom results in the formation of anion and decrease in nuclear charge.

Now, if the nuclear charge of an atom or ion is lower then, the size of atom or ion is large, if whereas the nuclear charge of an atom or ion is greater then, the size of atom or ion is small.

Answer to Problem 29E

The order of decreasing size is ${\text{La}}^{3+}>{\text{Eu}}^{3+}>{\text{Gd}}^{3+}{\text{>Yb}}^{3+}$ .

Explanation of Solution

Anions are formed when a neutral atom or element gain electron(s) whereas cations are formed when a neutral atom or element lose electron(s).

The given atoms and ions are: ${\text{La}}^{3+},{\text{ Eu}}^{3+},{\text{Gd}}^{\text{3+}}{\text{,Yb}}^{\text{3+}}$

Now, $\text{La},\text{ Eu},\text{Gd, Yb}$ belongs to same period of the periodic table.

In periodic table, down the group atomic size increases whereas the left to right atomic size decreases. Therefore, decreasing order of size is $\text{La}>\text{Eu}>\text{Gd>Yb}$

Similarly, order of decreasing size for given ions is ${\text{La}}^{3+}>{\text{Eu}}^{3+}>{\text{Gd}}^{3+}{\text{>Yb}}^{3+}$

(e)

Interpretation Introduction

Interpretation:

The following atoms and ions should be placed in order of decreasing size.

  ${\text{Te}}^{2-},{\text{ I}}^{-},{\text{Cs}}^{\text{+}}{\text{, Ba}}^{2+},{\mathrm{La}}^{3+}$

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

The outermost electrons in an atom are known as valence electrons. The size of atoms or ions depends on the attraction between nucleus and valence electrons. When the electron(s) is(are) removed from an atom results in the formation of cation and increase in nuclear charge attraction. When the electron(s) is(are) added in an atom results in the formation of anion and decrease in nuclear charge.

Now, if the nuclear charge of an atom or ion is lower then, the size of atom or ion is large, if whereas the nuclear charge of an atom or ion is greater then, the size of atom or ion is small.

Answer to Problem 29E

The order of decreasing size is ${\text{Te}}^{2-}>{\text{I}}^{-}>{\mathrm{Cs}}^{+}>{\text{Ba}}^{2-}>{\text{La}}^{3+}$ .

Explanation of Solution

Anions are formed when a neutral atom or element gain electron(s) whereas cations are formed when a neutral atom or element lose electron(s).

The given atoms and ions are: ${\text{Te}}^{2-},{\text{ I}}^{-},{\text{Cs}}^{\text{+}}{\text{, Ba}}^{2+},{\mathrm{La}}^{3+}$

Now, $\text{Cs},\text{ Ba},\text{ and La}$ belongs to sixth period and $\text{Te},\text{ I}$ belongs to fifth period of the periodic table.

According to trends of atoms in periodic table, if the periodic number increases then, the atomic size increases. Also, in periodic table, down the group atomic size increases whereas the left to right atomic size decreases.

Thus, order of decreasing size of neutral atoms is $\mathrm{Cs}>\text{Ba}>\text{La}>\text{Te}>\text{I}$ .

Now, the given atoms after losing or gaining of electrons acquire inert gas configuration.

For isoelectronic species, the size of ion increases with increase in anionic charge and deceases with increase in cationic charge.

Hence, order of decreasing size is ${\text{Te}}^{2-}>{\text{I}}^{-}>{\mathrm{Cs}}^{+}>{\text{Ba}}^{2-}>{\text{La}}^{3+}$