Chapter 8, Problem 8.21P

(a)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the outermost electron in $\text{Rb}$ atom is to be determined.

Concept introduction:

The electrons in the outermost occupied shell that determine the chemical properties of the elements are called the outermost electrons.

The quantum numbers provide complete information about the electron. There are four quantum numbers as follows:

1. The principal quantum number and it is represented by n. It tells about the shell to which the electron belongs.

2. The azimuthal quantum number and it is represented by l. It tells about the subshell of the electrons.

3. The magnetic quantum number and it is represented by $m_l$. It tells about the orbitals present in the subshell. The value of $m_l$ ranges from $l$ to $-l$ including 0.

4. The spin quantum number and it is represented by $m_s$. It tells about the spin of the electron and its value can either be $+\frac{\text{1}}{\text{2}}$ or $-\frac{\text{1}}{\text{2}}$.

Answer to Problem 8.21P

The possible quantum numbers for the outermost electron in $\text{Rb}$ atom are $n=5$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

Explanation of Solution

The atomic number of rubidium is 37 and its electronic configuration is $\left[\text{Kr}\right]5s^1$.

Its outermost electron enters in the $5s$ orbital so the value of its principal quantum number $\left(n\right)$ is 5. The value for azimuthal quantum number $\left(l\right)$ for the $s$ orbital is 0. The value of the magnetic quantum number $\left(m_l\right)$ for the $5s$ electron is 0. The value of the spin quantum number $\left(m_s\right)$ for the $5s$ electron can be $+\frac{1}{2}$ or $-\frac{1}{2}$.

Conclusion

The possible quantum numbers for the outermost electron in $\text{Rb}$ atom are $n=5$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

(b)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the electron gained when an ${\text{S}}^{-}$ ion becomes an ${\text{S}}^2{}^{-}$ ion is to be determined.

Concept introduction:

The quantum numbers provide complete information about the electron. There are four quantum numbers as follows:

1. The principal quantum number and it is represented by n. It tells about the shell to which the electron belongs.

2. The azimuthal quantum number and it is represented by l. It tells about the subshell of the electrons.

3. The magnetic quantum number and it is represented by $m_l$. It tells about the orbitals present in the subshell. The value of $m_l$ ranges from $l$ to $-l$ including 0.

4. The spin quantum number and it is represented by $m_s$. It tells about the spin of the electron and its value can either be $+\frac{\text{1}}{\text{2}}$ or $-\frac{\text{1}}{\text{2}}$.

Answer to Problem 8.21P

The possible quantum numbers for the electron gained when an ${\text{S}}^{-}$ ion becomes an ${\text{S}}^2{}^{-}$ ion are $n=3$, $l=1$, $m_l=-1$ , 0, or $+1$, and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

Explanation of Solution

The atomic number of sulfur is 16 so its electronic configuration is $\left[\text{Ne}\right]3s^{2}3p^4$. It gains one electron to become ${\text{S}}^{-}$ whose electronic configuration is $\left[\text{Ne}\right]3s^{2}3p^5$. It further gains one more electron to become ${\text{S}}^{2-}$ whose electronic configuration is $\left[\text{Ne}\right]3s^{2}3p^6$.

The ion formation occurs as:

$\begin{array}{l}\text{S}\left(\left[\text{Ne}\right]3s^{2}3p^4\right)+{\text{e}}^{-}\to {\text{S}}^{-}\left(\left[\text{Ne}\right]3s^{2}3p^5\right)\\ {\text{S}}^{-}\left(\left[\text{Ne}\right]3s^{2}3p^5\right)+{\text{e}}^{-}\to {\text{S}}^{2-}\left(\left[\text{Ne}\right]\text{}3s^{2}3p^6\right)\end{array}$

Its outermost electron enters in the $3p$ orbital so the value of its principal quantum number $\left(n\right)$ is 3. The value of azimuthal quantum number $\left(l\right)$ for the $p$ orbital is 1.

The value of the magnetic quantum number $\left(m_l\right)$ for a given value of $l$ extends from $-l$ to $+l$. Therefore the possible value of magnetic quantum number $\left(m_l\right)$  for $l=1$ are $+1$, 0, or $-1$.

The electron added to the $3p$ orbital is the second electron thus by convention its $m_s$ value should be $-\frac{1}{2}$ but the possible value of the spin quantum number $\left(m_s\right)$ for the $3p$ electron can be $+\frac{1}{2}$ or $-\frac{1}{2}$.

Conclusion

The possible quantum numbers for the electron gained when an ${\text{S}}^{-}$ ion becomes an ${\text{S}}^2{}^{-}$ ion are $n=3$, $l=1$, $m_l=-1$ , 0, or $+1$, and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

(c)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the electron lost when an $\text{Ag}$ atom ionizes is to be determined.

Concept introduction:

The quantum numbers provide complete information about the electron. There are four quantum numbers as follows:

1. The principal quantum number and it is represented by n. It tells about the shell to which the electron belongs.

2. The azimuthal quantum number and it is represented by l. It tells about the subshell of the electrons.

3. The magnetic quantum number and it is represented by $m_l$. It tells about the orbitals present in the subshell. The value of $m_l$ ranges from $l$ to $-l$ including 0.

4. The spin quantum number and it is represented by $m_s$. It tells about the spin of the electron and its value can either be $+\frac{\text{1}}{\text{2}}$ or $-\frac{\text{1}}{\text{2}}$.

Answer to Problem 8.21P

The possible quantum numbers for the electron lost when gained when an $\text{Ag}$ atom ionizes are $n=5$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

Explanation of Solution

The atomic number of silver is 47 so its electronic configuration is $\left[\text{Kr}\right]5s^{1}4d^{10}$. Electron is lost always from the highest energy level. It loses one electron from $5s$ orbital to become ${\text{Ag}}^{+}$ whose electronic configuration is $\left[\text{Kr}\right]4d^{10}$.

The ion formation occurs as:

$\text{Ag}\left(\left[\text{Kr}\right]5s^{1}4d^{10}\right)-{\text{e}}^{-}\to {\text{Ag}}^{+}\left(\left[\text{Kr}\right]4d^{10}\right)$

The electron is lost from the $5s$ orbital so the value of its principal quantum number $\left(n\right)$ is 5. The value of azimuthal quantum number $\left(l\right)$ for the $s$ orbital is 0. The value of the magnetic quantum number $\left(m_l\right)$ for the $5s$ electron is 0. The value of the spin quantum number $\left(m_s\right)$ for the $5s$ electron can be $+\frac{1}{2}$ or $-\frac{1}{2}$.

Conclusion

The possible quantum numbers for the electron lost when gained when an $\text{Ag}$ atom ionizes are $n=5$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

(d)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the electron gained when an ${\text{F}}^{-}$ ion is formed from $\text{F}$ atom is to be determined.

Concept introduction:

The quantum numbers provide complete information about the electron. There are four quantum numbers as follows:

1. The principal quantum number and it is represented by n. It tells about the shell to which the electron belongs.

2. The azimuthal quantum number and it is represented by l. It tells about the subshell of the electrons.

3. The magnetic quantum number and it is represented by $m_l$. It tells about the orbitals present in the subshell. The value of $m_l$ ranges from $l$ to $-l$ including 0.

4. The spin quantum number and it is represented by $m_s$. It tells about the spin of the electron and its value can either be $+\frac{\text{1}}{\text{2}}$ or $-\frac{\text{1}}{\text{2}}$.

Answer to Problem 8.21P

The quantum numbers for the electron gained when an ${\text{F}}^{-}$ ion is formed from $\text{F}$ atom are $n=2$, $l=1$, $m_l=-1$ , 0, or $+1$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.

Explanation of Solution

The atomic number of fluorine is 9 so its electronic configuration is $1s^{2}2s^{2}2p^5$. It gains one electron to form ${\text{F}}^{-}$ ion whose electronic configuration is $1s^{2}2s^{2}2p^6$.

The ion formation occurs as:

$\text{F}\left(1s^{2}2s^{2}2p^5\right)+{\text{e}}^{-}\to {\text{F}}^{-}\left(1s^{2}2s^{2}2p^6\right)$

The electron is added to the $2p$ orbital so the value of its principal quantum number $\left(n\right)$ is 2. The value of azimuthal quantum number $\left(l\right)$ for the $p$ orbital is 1.

The value of the magnetic quantum number $\left(m_l\right)$ for a given value of $l$ extends from $-l$ to $+l$. Therefore the possible value of magnetic quantum number $\left(m_l\right)$  for $l=1$ are $+1$, 0, or $-1$.

The value of the spin quantum number $\left(m_s\right)$ for the $2p$ electron can be $+\frac{1}{2}$ or $-\frac{1}{2}$.

Conclusion

The quantum numbers for the electron gained when an ${\text{F}}^{-}$ ion is formed from $\text{F}$ atom are $n=2$, $l=1$, $m_l=-1$ , 0, or $+1$ and $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$.