Chapter 8, Problem 8C.3ST

(a)

Interpretation Introduction

Interpretation:

The terms that arise from the configuration ${\text{2s}}^{\text{1}}{\text{2p}}^{\text{1}}$ have to be predicted.

Concept introduction:

A term symbol describes the orbital, spin and total angular momentum of an electronic state.  The symbol $L$ represents the vector sum of orbital angular momentum.  The symbol $S$ represents the vector sum of spin angular momentum.  The symbol $J$ represents the total angular momentum.  Therefore, the term symbol is represented as given below.

  ${}^{2S+1}{L}_J$

Answer to Problem 8C.3ST

The possible atomic terms for the electronic configuration ${\text{2s}}^{\text{1}}{\text{2p}}^{\text{1}}$ are ${}^3{\text{P}}_2,{}^3{\text{P}}_1,{}^3{\text{P}}_0$ and ${}^1{\text{P}}_1$.

Explanation of Solution

For the $\text{p}$ electrons the value of orbital angular momentum $l$ is $1$.  For ${\text{p}}^1$ electron configuration, the values of $l$ is shown below.

  $\begin{array}{cccc}l& +1& 0& -1\\ & \boxed{}& \boxed{}& \boxed{}\end{array}$

An electron can occupy any of the orbital.  Therefore, an electron in the orbital $\text{p}$ can have $+1$, $0$, and $-1$ values of $l$.

For the $\text{s}$ electrons the value of orbital angular momentum $l$ is $0$.

The symbol $L$ represents the vector sum of orbital angular momentum.  The maximum value of total orbital angular momentum for the electronic configuration ${\text{2s}}^{\text{1}}{\text{2p}}^{\text{1}}$ is shown below.

    $L=\left|l_1+l_2\right|$                                                                                                      (1)

Where,

• $l_1$ is the maximum value of orbital angular momentum quantum number for $\text{p}$ orbital.
• $l_2$ is the maximum value of orbital angular momentum quantum number for $\text{s}$ orbital.

Substitute the values of $l_1$ and $l_2$ in the equation (1).

  $\begin{array}{c}L=\left|1+0\right|\\ =1\end{array}$

The minimum value of total orbital angular momentum for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ is shown below.

    $L=\left|l_1-l_2\right|$                                                                                                      (2)

Substitute the values of $l_1$ and $l_2$ in the equation (2).

    $\begin{array}{c}L=\left|1-1\right|\\ =0\end{array}$

Therefore, the values of total orbital angular momentum for the electronic configuration ${\text{2s}}^{\text{1}}{\text{2p}}^{\text{1}}$ are $1$ and $0$.

The term corresponds to $L=0$ is $\text{S}$.

The term corresponds to $L=1$ is $\text{P}$.

An electron can have two value of spin angular momentum $\left(s\right)$ that are $\frac{1}{2}$ and $-\frac{1}{2}$.

The possible combinations of spin angular momentum for two electrons are $\left(\frac{1}{2},\frac{1}{2}\right)$, $\left(\frac{1}{2},-\frac{1}{2}\right)$, $\left(\frac{1}{2},-\frac{1}{2}\right)$ and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.

Therefore, the possible values for the vector sum of spin angular momentum $S$ are $1$ and $0$.

The multiplicity is given by the formula as shown below.

  $\text{Multiplicity}=2S+1$                                                                                     (2)

Where,

• $S$ is the total spin angular momentum quantum number.

Substitute the value of $S=1$ in the equation (2).

    $\begin{array}{c}\text{Multiplicity}=2\left(1\right)+1\\ =3\end{array}$

Substitute the value of $S=0$ in the equation (2).

    $\begin{array}{c}\text{Multiplicity}=2\left(0\right)+1\\ =1\end{array}$

The value of total angular momentum $J$ is calculated by the formula given below.

    $J=\left|L+S\right|\mathrm{......}\left|L-S\right|$                                                                                     (3)

Substitute the value of $L=1$ and $S=1$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(1+1\right),\mathrm{......................},\left(1-1\right)\\ =2,\mathrm{................},0\\ =2,1,0\end{array}$

Therefore, the atomic terms will be ${}^3{\text{P}}_2$, ${}^3{\text{P}}_1$ and ${}^3{\text{P}}_0$.

Substitute the value of $L=1$ and $S=0$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(1+0\right),\mathrm{......................},\left(1-0\right)\\ =1\end{array}$

Therefore, the atomic terms will be ${}^1{\text{P}}_1$.

Hence, the possible atomic terms for the electronic configuration ${\text{2s}}^{\text{1}}{\text{2p}}^{\text{1}}$ are ${}^3{\text{P}}_2,{}^3{\text{P}}_1,{}^3{\text{P}}_0$ and ${}^1{\text{P}}_1$.

(b)

Interpretation Introduction

Interpretation:

The terms that arise from the configuration $2{\text{p}}^{1}3{\text{d}}^1$ have to be predicted.

Concept introduction:

A term symbol describes the orbital, spin and total angular momentum of an electronic state.  The symbol $L$ represents the vector sum of orbital angular momentum.  The symbol $S$ represents the vector sum of spin angular momentum.  The symbol $J$ represents the total angular momentum.  Therefore, the term symbol is represented as given below.

  ${}^{2S+1}{L}_J$

Answer to Problem 8C.3ST

The possible atomic terms for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ are ${}^3{\text{F}}_4,{}^3{\text{F}}_3,{}^3{\text{F}}_2,{}^1{\text{F}}_3,{}^3{\text{D}}_3,{}^3{\text{D}}_2,{}^3{\text{D}}_1{,}^1{\text{D}}_2{,}^3{\text{P}}_2,{}^3{\text{P}}_1,{}^3{\text{P}}_0$ and ${}^1{\text{P}}_1$.

Explanation of Solution

For the $\text{p}$ electrons the value of orbital angular momentum $l$ is $1$.  For ${\text{p}}^1$ electron configuration, the values of $l$ is shown below.

  $\begin{array}{cccc}l& +1& 0& -1\\ & \boxed{}& \boxed{}& \boxed{}\end{array}$

An electron can occupy any of the orbital.  Therefore, an electron in the orbital $\text{p}$ can have $+1$, $0$, and $-1$ values of $l$.

For the $\text{d}$ electrons the value of orbital angular momentum $l$ is $2$.  For ${\text{d}}^1$ electron configuration, the values of $l$ is shown below.

  $\begin{array}{cccccc}l& +2& +1& 0& -1& -2\\ & \boxed{}& \boxed{}& \boxed{}& \boxed{}& \boxed{}\end{array}$

An electron can occupy any of the orbital.  Therefore, an electron in the orbital $\text{d}$ can have $+2$, $+1$, $0$, $-1$, and $-2$ values of $l$.

The symbol $L$ represents the vector sum of orbital angular momentum.  The maximum value of total orbital angular momentum for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ is shown below.

    $L=\left|l_1+l_2\right|$                                                                                                      (4)

Where,

• $l_1$ is the maximum value of orbital angular momentum quantum number for $\text{p}$ orbital.
• $l_2$ is the maximum value of orbital angular momentum quantum number for $\text{d}$ orbital.

Substitute the values of $l_1$ and $l_2$ in the equation (4).

  $\begin{array}{c}L=\left|1+2\right|\\ =3\end{array}$

The minimum value of total orbital angular momentum for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ is shown below.

    $L=\left|l_1-l_2\right|$                                                                                                     (5)

Substitute the values of $l_1$ and $l_2$ in the equation (5).

    $\begin{array}{c}L=\left|1-2\right|\\ =\left|-1\right|\\ =1\end{array}$

Therefore, the values of total orbital angular momentum for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ are $3$, $2$ and $1$.

The term corresponds to $L=3$ is $\text{F}$.

The term corresponds to $L=2$ is $\text{D}$.

The term corresponds to $L=1$ is $\text{P}$.

An electron can have two value of spin angular momentum $\left(s\right)$ that are $\frac{1}{2}$ and $-\frac{1}{2}$.

The possible combinations of spin angular momentum for two electrons are $\left(\frac{1}{2},\frac{1}{2}\right)$, $\left(\frac{1}{2},-\frac{1}{2}\right)$, $\left(\frac{1}{2},-\frac{1}{2}\right)$ and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.

Therefore, the possible values for the vector sum of spin angular momentum $S$ are $1$ and $0$.

Substitute the value of $S=1$ in the equation (2).

    $\begin{array}{c}\text{Multiplicity}=2\left(1\right)+1\\ =3\end{array}$

Substitute the value of $S=0$ in the equation (2).

    $\begin{array}{c}\text{Multiplicity}=2\left(0\right)+1\\ =1\end{array}$

Substitute the value of $L=3$ and $S=1$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(3+1\right),\mathrm{......................},\left(3-1\right)\\ =4,\mathrm{................},2\\ =4,3,2\end{array}$

Therefore, the atomic terms will be ${}^3{\text{F}}_4$, ${}^3{\text{F}}_3$ and ${}^3{\text{F}}_2$.

Substitute the value of $L=3$ and $S=0$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(3+0\right),\mathrm{......................},\left(3-0\right)\\ =3\end{array}$

Therefore, the atomic terms will be ${}^1{\text{F}}_3$.

Substitute the value of $L=2$ and $S=1$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(2+1\right),\mathrm{...........}\left(2-1\right)\\ =3,\mathrm{.............},1\\ =3,2,1\end{array}$

Therefore, the atomic terms will be ${}^3{\text{D}}_3$, ${}^3{\text{D}}_2$ and ${}^3{\text{D}}_1$.

Substitute the value of $L=2$ and $S=0$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(2+0\right),\left(2-0\right)\\ =2\end{array}$

Therefore, the atomic terms will be ${}^1{\text{D}}_2$.

Substitute the value of $L=1$ and $S=1$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(1+1\right),\mathrm{...........}\left(1-1\right)\\ =2,\mathrm{.............},0\\ =2,1,0\end{array}$

Therefore, the atomic terms will be ${}^3{\text{P}}_2$, ${}^3{\text{P}}_1$ and ${}^3{\text{P}}_0$.

Substitute the value of $L=1$ and $S=0$ in the equation (3) as shown below.

    $\begin{array}{c}J=\left(1+0\right),\mathrm{...........}\left(1-0\right)\\ =1\end{array}$

Therefore, the atomic terms will be ${}^1{\text{P}}_1$.

Hence, the possible atomic terms for the electronic configuration $2{\text{p}}^{1}3{\text{d}}^1$ are ${}^3{\text{F}}_4,{}^3{\text{F}}_3,{}^3{\text{F}}_2,{}^1{\text{F}}_3,{}^3{\text{D}}_3,{}^3{\text{D}}_2,{}^3{\text{D}}_1{,}^1{\text{D}}_2{,}^3{\text{P}}_2,{}^3{\text{P}}_1,{}^3{\text{P}}_0$ and ${}^1{\text{P}}_1$.