Chapter 3, Problem 8P

3-24 Predict which ions are stable:

(a) Br²-

(b) C4-

(c) Ca+

(d) Ar+

(e) Na+

(f) Cs+

Interpretation Introduction

Interpretation:

Whether $B{r}^{2-}$ 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 8P

$B{r}^{2-}$ ion is not stable.

Explanation of Solution

The atomic number or number of electrons of $Br$ is 35, the electronic configuration of $Br$ is as follows:

$\begin{array}{l}Br-35:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^5\end{array}$

When $Br$ will gain 2 electrons; the electronic configuration of its ion will not become as the nearest noble element.

$\begin{array}{l}B{r}^{2-}-33:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^3\end{array}$

Hence, the ion $B{r}^{2-}$ is not stable.

Interpretation Introduction

Interpretation:

Whether ${\text{C}}^{4-}$ 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 8P

${\text{C}}^{4-}$ ion is stable.

Explanation of Solution

The atomic number or number of electrons of $\text{C}$ is 6, the electronic configuration of $\text{C}$ is as follows:

$\begin{array}{l}\text{C}-6:\\ 1s^{2}2s^{2}2p^4\end{array}$

When $\text{C}$ will gain 4 electrons; the electronic configuration of its ion will become as the nearest noble element.

$\begin{array}{l}{\text{C}}^{4-}-10:\\ 1s^{2}2s^{2}2p^6\\ [\text{N}\text{e}]\end{array}$

Hence, the ion ${\text{C}}^{4-}$ is stable.

Interpretation Introduction

Interpretation:

Whether $\text{C}{\text{a}}^{+}$ 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 8P

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

Explanation of Solution

The atomic number or number of electrons of $\text{C}\text{a}$ is 20, the electronic configuration of $\text{C}\text{a}$ is as follows:

$\begin{array}{l}\text{C}\text{a}-20:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2\end{array}$

When $\text{C}\text{a}$ will lose 1 electron; the electronic configuration of its ion will not become as the nearest noble element.

$\begin{array}{l}\text{C}{\text{a}}^{+}-19:\\ 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6}4s^1\end{array}$

Hence, the ion $\text{C}{\text{a}}^{+}$ is not stable.

Interpretation Introduction

Interpretation:

Whether $\text{a}{r}^{+}$ 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 8P

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

Explanation of Solution

The atomic number or number of electrons of $\text{a}r$ is 18, the electronic configuration of $\text{a}r$ is as follows:

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

When $\text{a}r$ will gain 1 electron; the electronic configuration of its ion will not become as the nearest noble element.

$\begin{array}{l}\text{a}{r}^{+}-19:\\ 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6}3d^{10}4s^1\end{array}$

Hence, the ion $\text{a}{r}^{+}$ 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 8P

$\text{N}{\text{a}}^{+}$ 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 gain 1 electron; the electronic configuration of its ion will become as the nearest noble element.

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

Hence, the ion $\text{N}{\text{a}}^{+}$ is not stable.

Interpretation Introduction

Interpretation:

Whether $\text{C}{s}^{+}$ 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 8P

$\text{C}{s}^{+}$ ion is stable.

Explanation of Solution

The atomic number or number of electrons of $\text{C}s$ is 55, the electronic configuration of $\text{C}s$ is as follows:

$\begin{array}{l}\text{C}s-55:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}2p^{6}6s^1\end{array}$

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

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

Hence the ion $\text{C}{s}^{+}$ is not stable.