Chapter 8, Problem 51E

Which of the following ions have noble gas electron configurations?

a. Fe²⁺, Fe³⁺, Sc³⁺, Co³⁺

b. Tl⁺, Te²⁻, Cr³⁺

c. Pu⁴⁺, Ce⁴⁺, Ti⁴⁺

d. Ba²⁺, Pt²⁺, Mn²⁺

Interpretation Introduction

Interpretation: The ions that shows noble gas configuration is to be identified.

Concept introduction: Nobel gases are those gases which are stable in nature. The electronic configuration gives the representation of distribution of electrons of atoms and molecules. The outer-most shell of these gases is completely filled.

To determine: The ions that shows the noble gas electronic configuration from ${\text{Fe}}^{2+},{\text{Fe}}^{3+},{\text{Sc}}^{3+}$ and ${\text{Co}}^{3+}$.

Answer to Problem 51E

${\text{Sc}}^{3+}$ will show noble gas configuration.

Explanation of Solution

For ${\text{Fe}}^{2+}$ ion,

The atomic number of iron $\left(\text{Fe}\right)$ is 26. The electronic configuration of iron is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{6}4{\text{s}}^2$

It loses two electrons to form ${\text{Fe}}^{2+}$. The electronic configuration of ${\text{Fe}}^{2+}$ is

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^6$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

For ${\text{Fe}}^{3+}$ ion,

The atomic number of iron $\left(\text{Fe}\right)$ is 26. The electronic configuration of iron is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{6}4{\text{s}}^2$

It loses three electrons to form ${\text{Fe}}^{3+}$. The electronic configuration of ${\text{Fe}}^{3+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^5$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

For ${\text{Sc}}^{3+}$ ion,

The atomic number of scandium $\left(\text{Sc}\right)$ is 21. The electronic configuration of scandium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{1}}{\text{4s}}^{\text{2}}$

It loses three electrons to form ${\text{Sc}}^{3+}$. The electronic configuration of ${\text{Sc}}^{3+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

The outer most shell is completely filled. Hence, it will show noble gas configuration.

For ${\text{Co}}^{3+}$ ion,

The atomic number of cobalt $\left(\text{Co}\right)$ is 27. The electronic configuration of cobalt is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{7}}{\text{4s}}^{\text{2}}$

It loses three electrons to form ${\text{Co}}^{3+}$. The electronic configuration of ${\text{Co}}^{3+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^6$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

Interpretation Introduction

Interpretation: The ions that shows noble gas configuration is to be identified.

Concept introduction: Nobel gases are those gases which are stable in nature. The electronic configuration gives the representation of distribution of electrons of atoms and molecules. The outer-most shell of these gases is completely filled.

To determine: The ions that shows the noble gas electronic configuration from ${\text{Tl}}^{+},{\text{Te}}^{2-}$ and ${\text{Cr}}^{3+}$.

Answer to Problem 51E

${\text{Te}}^{2-}$ will show noble gas configuration.

Explanation of Solution

For ${\text{Tl}}^{+}$ ion,

The atomic number of thallium $\left(\text{Tl}\right)$ is 81. The electronic configuration of thallium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{1}}$

It loses an electron to form ${\text{Tl}}^{+}$. The electronic configuration of ${\text{Tl}}^{+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

For ${\text{Te}}^{2-}$ ion,

The atomic number of tellurium $\left(\text{Te}\right)$ is 52. The electronic configuration of tellurium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{4}}$

It gains two electrons to to form ${\text{Te}}^{2-}$. The electronic configuration of ${\text{Te}}^{2-}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^6$

The outer most shell is completely filled. Hence, it will show noble gas configuration.

For ${\text{Cr}}^{3+}$ ion,

The atomic number of chromium $\left(\text{Cr}\right)$ is 24. The electronic configuration of chromium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}{\text{4s}}^{\text{1}}$

It loses three electrons to to form ${\text{Cr}}^{3+}$. The electronic configuration of ${\text{Cr}}^{3+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^3$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

Interpretation Introduction

Interpretation: The ions that shows noble gas configuration is to be identified.

Concept introduction: Nobel gases are those gases which are stable in nature. The electronic configuration gives the representation of distribution of electrons of atoms and molecules. The outer-most shell of these gases is completely filled.

To determine: The ions that shows the noble gas electronic configuration from ${\text{Pu}}^{4+},{\text{Ce}}^{4+}$ and ${\text{Ti}}^{4+}$.

Answer to Problem 51E

${\text{Ti}}^{4+}$ and ${\text{Ce}}^{4+}$ will show noble gas configuration.

Explanation of Solution

For ${\text{Pu}}^{4+}$ ion,

The atomic number of plutonium $\left(\text{Pu}\right)$ is 94. The electronic configuration of plutonium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}{\text{5f}}^{\text{6}}{\text{6d}}^{\text{0}}{\text{7s}}^{\text{2}}$

It loses four electrons to form ${\text{Pu}}^{4+}$. The electronic configuration of ${\text{Pu}}^{4+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}{\text{5f}}^4$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

For ${\text{Ce}}^{4+}$ ion,

The atomic number of cerium $\left(\text{Ce}\right)$ is 58. The electronic configuration of cerium is,

$\begin{array}{l}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{1}}{\text{5d}}^{\text{1}}{\text{6s}}^{\text{2}}\hfill \end{array}$

It loses four electrons to to form ${\text{Ce}}^{4+}$. The electronic configuration of ${\text{Ce}}^{4+}$ is

$\begin{array}{l}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}\hfill \end{array}$

The outer most shell is completely filled. Hence, it will show noble gas configuration.

For ${\text{Ti}}^{4+}$ ion,

The atomic number of titanium $\left(\text{Ti}\right)$ is 22. The electronic configuration of titanium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}$

It loses four electrons to to form ${\text{Ti}}^{4+}$. The electronic configuration of ${\text{Ti}}^{4+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

The outer most shell is completely filled. Hence, it will show noble gas configuration.

Interpretation Introduction

Interpretation: The ions that shows noble gas configuration is to be identified.

Concept introduction: Nobel gases are those gases which are stable in nature. The electronic configuration gives the representation of distribution of electrons of atoms and molecules. The outer-most shell of these gases is completely filled.

To determine: The ions that shows the noble gas electronic configuration from ${\text{Ba}}^{2+},{\text{Pt}}^{2+}$ and ${\text{Mn}}^{2+}$.

Answer to Problem 51E

${\text{Ba}}^{2+}$ will show noble gas configuration.

Explanation of Solution

For ${\text{Ba}}^{2+}$ ion,

The atomic number of barium $\left(\text{Ba}\right)$ is 56. The electronic configuration of barium is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{2}}$

It loses two electrons to form ${\text{Ba}}^{2+}$. The electronic configuration of ${\text{Ba}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$

The outer most shell is completely filled. Hence, it will show noble gas configuration.

For ${\text{Pt}}^{2+}$ ion,

The atomic number of platinum $\left(\text{Pt}\right)$ is 79. The electronic configuration of platinum is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{9}}{\text{6s}}^{\text{1}}$

It loses two electrons to to form ${\text{Pt}}^{2+}$. The electronic configuration of ${\text{Pt}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^8$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.

For ${\text{Mn}}^{2+}$ ion,

The atomic number of manganese $\left(\text{Mn}\right)$ is 25. The electronic configuration of manganese is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}{\text{4s}}^{\text{2}}$

It loses four electrons to to form ${\text{Mn}}^{2+}$. The electronic configuration of ${\text{Mn}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{5}}$

The outer most shell is not completely filled. Hence, it will not show noble gas configuration.