Chapter 3, Problem 3.21P

3-21 Write the formula for the most stable ion formed by each element.

(a) Mg

(b) F

(c) Al

(d)S

(e)K

(f) Br

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $Mg$ 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 3.21P

The formula for the most stable ion formed by $Mg$ is $M{g}^{2+}$.

Explanation of Solution

The atomic number or number of electrons of $Mg$ is 12, The Electronic configuration of $Mg$ is as follows:

$\begin{array}{l}Mg-12:\\ 1s^{2}2s^{2}2p^{6}3s^2\end{array}$

When $Mg$ will lose 2 electrons; the electronic configuration of its ion become as the nearest noble element neon as follows:

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

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the formula for the most stable ion formed by $Mg$ is $M{g}^{2+}$.

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $\text{F}$ 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 3.21P

The formula for the most stable ion formed by $\text{F}$ is ${\text{F}}^{-}$.

Explanation of Solution

The atomic number or number of electrons of $\text{F}$ is 9, the electronic configuration of $\text{F}$ is as follows:

$\begin{array}{l}\text{F}-9:\\ 1s^{2}2s^{2}2p^5\end{array}$

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

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

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the formula for the most stable ion formed by $\text{F}$ is ${\text{F}}^{-}$.

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $\text{a}\text{l}$ 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 3.21P

The formula for the most stable ion formed by $\text{a}\text{l}$ is $\text{a}{\text{l}}^{3+}$.

Explanation of Solution

The atomic number or number of electrons of $\text{a}\text{l}$ is 13, the electronic configuration of $\text{a}\text{l}$ is as follows:

$\begin{array}{l}Mg-12:\\ 1s^{2}2s^{2}2p^{6}3s^2\end{array}$

When $\text{a}\text{l}$ will lose 3 electrons; the electronic configuration of its ion become as the nearest noble element neon as follows:

$\begin{array}{l}\text{a}{\text{l}}^{3+}-10:\\ 1s^{2}2s^{2}2p^6\\ [\text{N}\text{e}]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the formula for the most stable ion formed by $\text{a}\text{l}$ is $\text{a}{\text{l}}^{3+}$.

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $S$ 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 3.21P

The formula for the most stable ion formed by $S$ is $S^{2-}$.

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:

$\begin{array}{l}{S}^{2-}-16:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^6\\ [\text{a}r]\end{array}$

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

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $K$ 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 3.21P

The formula for the most stable ion formed by $K$ is $K^{+}$.

Explanation of Solution

The atomic number or number of electrons of $K$ is 19, the electronic configuration of $K$ is as follows:

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

When $K$ will lose 1 electron; the electronic configuration of its ion become as the nearest noble element argon as follows:

$\begin{array}{l}{K}^{+}-19:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^6\\ [\text{a}r]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the formula for the most stable ion formed by $K$ is $K^{+}$.

Interpretation Introduction

Interpretation:

The formula for the most stable ion formed by $Br$ 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 3.21P

The formula for the most stable ion formed by $Br$ is $B{r}^{-}$.

Explanation of Solution

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

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

When $Br$ will gain 1 electron; the electronic configuration of its ion become as the nearest noble element argon as follows:

$\begin{array}{l}B{r}^{-}-36:\\ 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}3d^{10}\\ [Kr]\end{array}$

The electronic configuration of the nearest noble element like helium, neon argon is the stable configuration hence the formula for the most stable ion formed by $Br$ is $B{r}^{-}$.