Chapter 8, Problem 8.22P

(a)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the outermost electron in $\text{Li}$ 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 8.22P

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

Explanation of Solution

The atomic number of lithium is 3 and its electronic configuration is $1s^{2}2s^1$

Its outermost electron enters in the $2s$ orbital so the value of its principal quantum number $\left(n\right)$ is 2. 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 a given value of $l$ extends from $-l$ to $+l$. Therefore, the value of the magnetic quantum number $\left(m_l\right)$ for $l=0$ is 0.

The possible value of the spin quantum number $\left(m_s\right)$ for the $2s$ electron is $+\frac{1}{2},-\frac{1}{2}$ but by convention its value is taken as positive.

Conclusion

The quantum numbers for the outermost electron in $\text{Li}$ atom are $n=2$, $l=0$, $m_l=0$ 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 $\text{Br}$ ion becomes an ${\text{Br}}^{-}$ ion 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 8.22P

The quantum numbers for the electron gained when an $\text{Br}$ ion becomes an ${\text{Br}}^{-}$ ion are $n=4$, $l=1$, $m_l=-1,0,+1$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.

Explanation of Solution

The atomic number of bromine is 35 so its electronic configuration is $\left[\text{Ar}\right]3d^{10}4s^{2}4p^5$. It gains one electron to become ${\text{Br}}^{-}$ whose electronic configuration is $\left[\text{Ar}\right]3d^{10}4s^{2}4p^6$. The ion formation occurs as:

$\text{Br}\left(\left[\text{Ar}\right]3d^{10}4s^{2}4p^5\right)+{\text{e}}^{-}\to {\text{Br}}^{-}\left(\left[\text{Ar}\right]3d^{10}4s^{2}4p^6\right)$

Its outermost electron enters in the $4p$ orbital so the value of its principal quantum number $\left(n\right)$ is 4. 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 possible value of the spin quantum number $\left(m_s\right)$ for the $4p$ electron is $+\frac{1}{2},-\frac{1}{2}$ but by convention its value is taken as negative.

Conclusion

The quantum numbers for the electron gained when an $\text{Br}$ ion becomes an ${\text{Br}}^{-}$ ion are $n=4$, $l=1$, $m_l=-1,0,+1$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.

(c)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the electron lost when a $\text{Cs}$ atom ionizes 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 8.22P

The quantum numbers for the electron lost when gained when a $\text{Cs}$ atom ionizes are.

$n=6$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.

Explanation of Solution

The atomic number of cesium is 55 so its electronic configuration is $\left[\text{Xe}\right]6s^1$. It loses one electron to become ${\text{Cs}}^{+}$ whose electronic configuration is that of xenon.

The ion formation occurs as:

$\text{Cs}\left(\left[\text{Xe}\right]6s^1\right)\to {\text{Cs}}^{+}\left(\left[\text{Xe}\right]\right)+{\text{e}}^{-}$

The electron is lost from the $6s$ orbital so the value of its principal quantum number $\left(n\right)$ is 6. 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 $6s$ electron is 0. The possible values of the spin quantum number $\left(m_s\right)$ for the $6s$ electron are $+\frac{1}{2},-\frac{1}{2}$ but by convention its value is taken as positive.

Conclusion

The quantum numbers for the electron lost when gained when a $\text{Cs}$ atom ionizes are.

$n=6$, $l=0$, $m_l=0$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.

(d)

Interpretation Introduction

Interpretation:

The full set of possible quantum numbers for the highest energy electron in the ground state of $\text{B}$ 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 8.22P

The quantum numbers for the highest energy electron in the ground state of $\text{B}$ atom are $n=2$, $l=1$, $m_l=-1,0,+1$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.

Explanation of Solution

The atomic number of boron is 5 so its electronic configuration is $\left[\text{He}\right]2s^{2}2p^1$. Its highest energy electron lies in $2p$ orbital.

The electron is present in 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 possible values of the spin quantum number $\left(m_s\right)$ for the $2p$ electron is $+\frac{1}{2},-\frac{1}{2}$ but by convention its value is taken as positive.

Conclusion

The quantum numbers for the highest energy electron in the ground state of $\text{B}$ atom are $n=2$, $l=1$, $m_l=-1,0,+1$ and $m_s=+\frac{1}{2},-\frac{1}{2}$.