Chapter 3, Problem 7P

3-23 Predict which ions are stable:

(a)

(b)

(c)

(d)

(e)

(f)

Interpretation Introduction

Interpretation:

Whether ${\text{i}}^{-}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

${\text{i}}^{-}$ ion is stable.

Explanation of Solution

The atomic number or number of electrons of $\text{i}$ is 53, the electronic configuration of $\text{i}$ is as follows:

$\begin{array}{l}\text{i}-53:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}2p^5\end{array}$

When $\text{i}$ will gain 1 electron; the electronic configuration of its ion become as the nearest noble element Xeon as follows:

$\begin{array}{l}{\text{i}}^{-}-54:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}2p^5\\ [\text{X}\text{e}]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the ${\text{i}}^{-}$ ion is stable.

Interpretation Introduction

Interpretation:

Whether $S{\text{e}}^{2+}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

$S{\text{e}}^{2+}$ ion is not stable.

Explanation of Solution

The atomic number or number of electrons of $S\text{e}$ is 34, the electronic configuration of $S\text{e}$ is as follows:

$\begin{array}{l}S\text{e}-34:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}3d^{10}4s^0\end{array}$

When $S\text{e}$ will lose 2 electrons the electronic configuration of its ion does not become as the nearest noble element Xeon as follows:

$\begin{array}{l}S{\text{e}}^{2+}-32:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}3d^{10}4s^0\\ Kr\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\\ [Kr]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the $S{\text{e}}^{2+}$ ion is not stable.

Interpretation Introduction

Interpretation:

Whether $\text{N}{\text{a}}^{+}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

$\text{N}{\text{a}}^{+}$ ion is stable.

Explanation of Solution

The atomic number or number of electrons of $\text{N}\text{a}$ is 11, the electronic configuration of $\text{N}\text{a}$ is as follows:

$\begin{array}{l}\text{N}\text{a}-11:\\ 1s^{2}2s^{2}2p^{6}3s^1\end{array}$

When $\text{N}\text{a}$ will lose 1 electron; the electronic configuration of its ion become as the nearest noble element Neon as follows:

$\begin{array}{l}\text{N}{\text{a}}^{+}-10:\\ 1s^{2}2s^{2}2p^6\\ [\text{N}\text{e}]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the $\text{N}{\text{a}}^{+}$ ion is stable.

Interpretation Introduction

Interpretation:

Whether $S^{2-}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

$S^{2-}$ ion is stable.

Explanation of Solution

The atomic number or number of electrons of $S$ is 16, the electronic configuration of $S$ is as follows:

$\begin{array}{l}S-16:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^4\end{array}$

When $S$ will gain 2 electrons; the electronic configuration of its ion become as the nearest noble element argon as follows:

$\begin{array}{l}{S}^{2-}-18:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^6\\ [\text{a}r]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the $S^{2-}$ ion is stable.

Interpretation Introduction

Interpretation:

Whether $\text{l}{\text{i}}^{2+}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

$\text{l}{\text{i}}^{2+}$ ion is not stable.

Explanation of Solution

The atomic number or number of electrons of $\text{L}\text{i}$ is 3, the electronic configuration of $\text{L}\text{i}$ is as follows:

$\begin{array}{l}\text{L}\text{i}-3:\\ 1s^{2}2s^1\end{array}$

When $\text{L}\text{i}$ will lose 2 electrons; the electronic configuration of its ion will not become as the nearest noble element helium as follows:

$\begin{array}{l}\text{l}{\text{i}}^{2+}-1:\\ 1s^1\\ \text{H}\text{e}-2:\\ 1s^2\\ [\text{H}\text{e}]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the $\text{l}{\text{i}}^{2+}$ ion is not stable.

Interpretation Introduction

Interpretation:

Whether $B{\text{a}}^{3+}$ ion is stable or not should be determined.

Concept Introduction:

Octet rule: generally all atoms will lose, gain or share electrons to attain 8 valence electrons; the electronic configuration of the nearest noble element.

Duet rule: generally some atoms like hydrogen, lithium, beryllium will lose, gain or share electrons to attain 2 valence electrons; the electronic configuration of the nearest noble element like helium.

Answer to Problem 7P

$B{\text{a}}^{3+}$ ion is not stable.

Explanation of Solution

The atomic number or number of electrons of $B\text{a}$ is 56, the electronic configuration of $B\text{a}$ is as follows:

$\begin{array}{l}B\text{a}-56:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}6s^2\end{array}$

When $B\text{a}$ will lose 3 electrons; the electronic configuration of its ion will not become as the nearest noble element Xeon as follows:

$\begin{array}{l}B{\text{a}}^{3+}-53:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^5\\ [\text{X}\text{e}]-54:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^6\\ [\text{X}\text{e}]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the $B{\text{a}}^{3+}$ ion is not stable.