# 8.E: Multielectron Atoms (Exercises)

These are homework exercises to accompany Chapter 8 of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap.

### Q8.4

Prove the speed of electron in first Bohr orbit is e2/4πϵ0ℏ = 2.188e6 m•s-1. The speed is in atomic units.

### S8.4

Use following formula in order to make the calculations

$v=\dfrac{ℏ}{m_e a_0}$

$v=\dfrac{ℏ}{m_e} \left(\dfrac{m_e e^2}{4πϵ_0ℏ^2}\right)$

$v=\dfrac{e^2}{4πϵ_0ℏ}$

Substituting values get the following:

$v=\dfrac{(1.6022 \times 10^{-19}\;C)^2}{(1.1127\times 10^{-10}C^2 \cdot J^{-1} \cdot m^{-1})(1.0546 \times 10^{-34}\;J \cdot s)}$

$v=2.1877 \times 10^{6}\;m \cdot s^{-1}$

This is the speed in atomic units. Are either of the two separate terms in the two term helium Hartree-Fock Orbital acceptable wavefunctions by themselves?

${\psi}\left(r\right)=0.81839e^{-1.44608r}+0.52072e^{-2.86222r}$

### S8.13

Check if they are normalized.

$\int{{\psi}^*\phi dt}=4p\int{{0.81839}^2e^{-2*1.44608r}r^2dr}=4p\left({0.81839}^2\right)\cdot \dfrac{2!}{{\left(2*1.44608\right)}^3}=0.6958\ \ \ No$

$\int{{\psi}^*\phi dt}=4p\int{{0.52072}^2e^{-2*2.86222r}r^2dr}=4p\left({0.52072}^2\right)\cdot \dfrac{2!}{{\left(2*2.86222\right)}^3}=0.0363\ \ \ No$

Since they are not normalized, neither of them are acceptable by themselves.

### Q8.15

Explain why:

$E = \dfrac{\int \Phi^*_2(1,2)\hat{H}\Phi_2(1,2)dr_1 dr_2 d\sigma_1 d\sigma_2 }{\int \Phi^*_2(1,2)\Phi_2(1,2)dr_1 dr_2 d\sigma_1 d\sigma_2}$

can be rewritten as:

$E = \dfrac{\int \Phi^*_2(1,2)\hat{H}\Phi_2(1,2)dr_1 dr_2}{\int \Phi^*_2(1,2)\Phi_2(1,2)dr_1 dr_2}$

### S8.15

The Hamiltonian does not depend on spin, so the spin integral can be factored out.

### Q8.16

Why must you distinguish the two electrons in separated hydrogen atoms?

### S8.16

They must be distinguished from one another because they each belong to a separate nucleus, not to an individual one.

### Q8.17

In both the Hartree-Fock approximation and hydrogen atomic wave function, why is the angular dependence the same?

### S8.17

Since the Hamiltonian used in the approximation only depends on r, the angular dependence is not affected in the Hartree-Fock approximation. So both the Hartree-Fock approximation and hydrogen atom will have the same angular dependence.

### Q8.20

Given the two electron determinate wave function below, determine if the spin component of the system is symmetric, anti-symmetric, or neither.

$\begin{vmatrix} 1s\alpha(1) & 1s\beta(1) \\ 1s\alpha(2) & 1s\alpha(2) \end{vmatrix}$

### S8.20

We have to solve the determinate.

$$1s{1}\alpha(2)1s{2}\beta(2)-1s{1}\beta(2)1s{1}\alpha(2)$$

Now we have to factor out the spatial part from the spin part.

$$1s{1}1s{2}[ \alpha(1)\beta(2)-\beta(1)\alpha(2)]$$

We are focused on the spin part and by observation, we can tell that the spin component is anti-symmetric.

### Q8.23

Given $$\hat{S}_z\alpha = \dfrac{\hbar}{2}\alpha$$ and $$\hat{S}_z\alpha = \dfrac{-\hbar}{2}\beta$$, show that $$\Psi_{200}$$ is an eigenfunction of $$S_{z,total} = \hat{S}_{za} + \hat{S}_{zB}$$.

### S8.23

Start with $$\Psi_{200} = C(\psi_{2sa} + \psi_{2sB})$$.

$\hat{S}_{z.total}\Psi = C(\hat{S}_{za} + \hat{S}_{zB})[\psi_{2sa} + \psi_{2sB}]$

$= C \left(\dfrac{\hbar}{2}-\dfrac{\hbar}{2}\right) [\psi_{2sa} + \psi_{2sB}]$

$=0$

### Q8.24

For the wave function:

$\psi =\begin{vmatrix} \psi _{A}(1)&\psi _{A}(2) \\\psi_{B}(1) &\psi_{B}(2) \end{vmatrix}$

discuss the effect on the wave function of (a) swapping rows of the matrix and (b) swapping columns of the matrix.

### S8.24

Let's expand the determinant to inspect the complete multi-electron wavefunction:

$\psi =\psi_{A}(1)\psi_{B}(2)-\psi_{A}(2)\psi_{B}(1)$

1. If we swap rows: $\psi_{RowSwap} =\psi_{A}(2)\psi_{B}(1)-\psi_{A}(1)\psi_{B}(2)=-\psi$
2. If we swap columns: $\psi_{ColumSwap} =\psi_{A}(2)\psi_{B}(1)-\psi_{A}(1)\psi_{B}(2)=-\psi$

Conclusion: swapping the either two rows or two columns of a Slater determinant changes the sign of the wave function.

### Q8.27

What are the term symbols for carbon and oxygen atoms in the ground state?

### S8.27

Carbon: The electron configuration for carbon at ground state (lowest energy) is $$1s^2 2s^2 2p^2$$

Since the 1s and 2s orbitals are completely filled, they can be neglected when calculating for S. For the lowest energy, you need to use the highest values of S and L, therefore we find that S= 1/2 +1/2=1 and L=1 which corresponds to the letter P. Since the 2p orbital is less than half-way filled, we use J=|L-S|=|1-1|=0.

The term symbol is $$^{2S+1}L_J$$ which is 3P0 for the ground-state of carbon atoms.

Oxygen: The electron configuration for oxygen at ground state (lowest energy) is $$1s^2 2s^2 2p^4$$

Again, the 1s and 2s orbitals may be neglected when calculating for $$S$$.

$S=1/2 +1/2 +1/2 -1/2 =1$

The last electron has a spin down due to the Pauli Exclusion Principle

L=1 which corresponds to the letter P.

Since the 2p orbital for oxygen is more than half-filled,

$J= L+S = 2$

The term symbol for the ground-state oxygen atom is 3P2

### Q8.28

Show that the number of sets of magnetic quantum number ($$m_l$$) and spin quantum number ($$m_s$$) associated with any term symbol is equal to $$(2L+1)(2S+1)$$. Apply this result to the $$np^2$$ case and show that symbol $$^1S$$, $$^3P$$ and $$^1D$$ account for all the possible sets of magnetic quantum numbers and spin quantum numbers

### Q8.29

Calculate all possible numbers of term symbol for an $$np^{1}$$ electron configuration.

### S8.29

$N = \dfrac{G!}{e! (G - e)!}$

where G is the highest number of electrons that an orbital can hold and e is the highest number of electrons that a subshell can hold

$N = \dfrac{6!}{2!(6 - 2)!} = 15$

### Q8.30

Determine the ground state term symbol for the electron configuration of the Halogens.

### S8.30

Halogens have the electron configuration$$np^5$$ we can determine the term symbols for this configuration by mapping out all possible configurations the electrons can fit into the six spin orbitals. To quickly determine how many possible combinations there are we can use a statistical method of $$N_{Comb.}= \binom{spin orbitals}{electrons}$$

$N_{comb} = \binom{6}{5} = \dfrac{6!}{5!(6!-5!)} = 6$

Now we know that there are 6 different configurations that we can map out to determine the term symbols. Since we are working with the $$p$$ orbital we know $$l = 1$$ and $$m_s = -1, \ 0, \ 1$$. I will denote spin using $$\alpha$$ as spin up $$+\dfrac{1}{2}$$ and $$\beta$$ as spin down $$-\dfrac{1}{2}$$. $$M_L$$ is the sum of the $$m_s$$ values corresponding to the number of electrons in that energy level. For example if you have $$\alpha$$$$\beta$$ in $$m_s = +1$$ only, you then have 2 electrons in $$m_s = +1$$ resulting in $$M_L = +1+1=2$$. $$M_s$$ is the sum of the spin up and spin down values.

+1 0 -1 $$M_L$$ $$M_s$$
$$\alpha$$$$\beta$$ $$\alpha$$$$\beta$$ $$\alpha$$ +1 $$+\dfrac{1}{2}$$
$$\alpha$$$$\beta$$ $$\alpha$$ $$\alpha$$$$\beta$$ 0 $$+\dfrac{1}{2}$$
$$\alpha$$ $$\alpha$$$$\beta$$ $$\alpha$$$$\beta$$ -1 $$+\dfrac{1}{2}$$
$$\alpha$$$$\beta$$ $$\alpha$$$$\beta$$ $$\beta$$ +1 $$-\dfrac{1}{2}$$
$$\alpha$$$$\beta$$ $$\beta$$ $$\alpha$$$$\beta$$ 0 $$-\dfrac{1}{2}$$
$$\beta$$ $$\alpha$$$$\beta$$ $$\alpha$$$$\beta$$ -1 $$-\dfrac{1}{2}$$

Now we need to determine the maximum value of $$M_L$$ and$$M_s$$. Looking at the table we see that

$max \ M_L = 1 \\ max \ M_s = \dfrac{1}{2}$

From this we know our maximum value of $$L$$ and $$S$$

$L_{max} = 1 \\ S_{max} = \dfrac{1}{2}$

Our possible values for $$L$$ and $$S$$ are

$L = 1, \ 0\ \\ S = \dfrac{1}{2}$

Since S is only $$\dfrac{1}{2}$$ we know we can only have doublet term symbols since $$2\big(\dfrac{1}{2}\big)+1 = 2$$. L ranges from 1 to 0 so our possible corresponding symbols will be P and S. This leaves us with the possibility of having

$^2P, \ ^2S$

To figure out what is there we start with the term symbol that has the largest L value, with is the  $$P$$. We see that for $$P$$, $$L=1$$ and $$S = \dfrac{1}{2}$$. For a value of $$L=1$$ our $$m_l$$ can be $$+1, \ 0, \ -1$$ and for an $$S= \dfrac{1}{2}$$ our $$m_s = +\dfrac{1}{2}, \ -\dfrac{1}{2}$$. In the table above all rows that contain these possible combination include every row. Therefore all of the configurations are contained in this doublet $$P$$ term symbol. Including values of $$J$$ we know that $$L+S \geq J \geq |L-S|$$. Since $$L=1$$ and $$S = \dfrac{1}{2}$$ our final term symbols are

$^2P_\dfrac{3}{2}, \ ^2P_\dfrac{1}{2}$

Hund's rules say that when L and S are the same with a subshell more than half filled, you look to the largest $$J$$ value to be the most stable. Therefore our final answer and the ground state term symbol for halogens is

$\boxed{^2P_\dfrac{3}{2}}$

### Q8.33

$$^2P$$, $$^2D$$, and $$^4S$$ are the term symbols for an atom with the $$np^{3}$$ electron configuration. Using the term symbols for the $$np^{3}$$ electron configuration, calculate the $$J$$ values associated with each of the term symbol. Then find out which term symbol represent the ground state.

### S8.33

To calculate $$J$$ we use this equation

$J = L + S$

This equation can be further expanded to be

$J = L + S, L + S - 1, L + S - 2, ...\left | L - S \right |$

Term Symbol  J  Full Term Symbol
$$^{2}P$$ 1 $$\dfrac{1}{2}$$ $$\dfrac{3}{2}$$, $$\dfrac{1}{2}$$ $$^2P_{3/2}$$ and $$^2P_{1/2}$$
$$^{2}D$$ 2 $$\dfrac{1}{2}$$ $$\dfrac{5}{2}\, \dfrac{3}{2}$$ $$^4D_{5/2}$$ and $$^4D_{3/2}$$
$$^{4}S$$ 0 $$\dfrac{3}{2}$$ $$\dfrac{3}{2}$$ $$^4S_{3/2}$$

### Q8.34

What are the ground state electron configuration and term symbol for Calcium?

### S9.34

1s22s22p63s23p64s2

or

[Ar]4s2

For the term symbol:

• Spin Multiplicity: S = 0 (all electrons paired)

2S + 1 = 1

• Orbital Angular Multiplicity: L = 0 because we consider an s orbital. This corresponds to S.

J = S + L = 0

There is only one valid value of $$J$$, so the term symbol for this configuration is

2S+1LJ =  1S0

### Q8.34

Find the ground state term symbol for Ca.

### S8.34

$$ns^2$$ electron configurations have the term symbol $$^1S_0$$, so the term symbol for Ca in the ground state is $$^1S_0$$.

### Q8.36

Write the electron configuration for vanadium and use this information to find the ground-state term symbol for V.

### S8.36

The electron configuration for vanadium is $$[Ar]3d^3 4s^3$$

• Spin Multiplicity: The electron configuration predicts three unpaired electrons, so $S= 1/2+1/2+1/2=3/2$ and the multiplicity of $$2S+1$$ predicts that this will be a quartet.
• Orbital Angular Momentum: The electron configuration predicts three electrons with $$l=2$$ and the rest so not contribute, so $$L=1+1+1=3$$, which is an $$F$$ state
• Total Angular Momentum: The shell is half full

The ground-state term symbol for vanadium is $$^4F_{3/2}$$ since vanadium has a half-filled 3d subshell.

### Q8.36

What is the ground-state term symbol for $$Ne$$.

### S8.36

The term symbol is $$^{2S + 1} L_{J}$$

where

• $$S$$ is the total electron spin
• $$L$$ is the total orbital angular momentum

and $J= L+S$

The electron configuration for:$$Ne$$ is $$1s^2 2s^2 2p^6$$ since Ne has spherical symmetry, we get:

$L= 0 + 0= S$

$S = \dfrac{1}{2}+\dfrac{1}{2}+ \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}= 0$

so $$2S+1 = 2\cdot 0+1 = 1$$ and $$J= 0+0=0$$. So the term symbol for $$Ne$$ is $$^{1}S_0$$.

### Q8.37

Looking at the 1$$s$$2$$p$$ electron configuration for He. Solve for the term symbols (states) of the Helium configuration and the degeneracies. If electron spin orbit coupling is included, what effect will this have?

### S8.37

There are two possible sets of $$m$$$$l$$ and $$m$$$$s$$ for the $$ns$$ electron and six possible sets of  $$m$$$$l$$ and $$m$$$$s$$ for the $$np$$ electron, so there are 2 × 6 = 12 possible sets of  $$m$$$$l$$ and $$m$$$$s$$ for the system.  We can denote values for the electron in the $$ns$$ orbital as $$m$$1$$j$$ and those for the electron in the $$np$$ orbital as $$m$$2$$j$$. The allowed values are given below:

Microstate $$m_{l} (1)$$ $$m_{s}(1)$$ $$m_{l}(2)$$ $$m_{s}(2)$$    $$M_{L}$$  $$M_S$$   $$M_J$$
Electron 1 in $$1s$$ orbital Electron 2 in $$2p$$ orbital Combined Angular Momenta of both Electrons
1 0 + 1/2 1 + 1/2 1 1 2
2 0 - 1/2 1 + 1/2 1 0 0
3 0 + 1/2 1 - 1/2 1 0 0
4 0 - 1/2 1 - 1/2 1 -1 1
5 0 + 1/2 0 + 1/2 0 1 0
6 0 - 1/2 0 + 1/2 0 0 1
7 0 + 1/2 0 - 1/2 0 0 0
8 0 - 1/2 0 - 1/2 0 -1 0
9 0 + 1/2 -1 + 1/2 -1 1 0
10 0 - 1/2 -1 + 1/2 -1 0 -1
11 0 + 1/2 -1 - 1/2 -1 0 -2
12 0 - 1/2 -1 - 1/2 -1 -1 -2

Entries 1, 2, 4, 5, 6, 8, 9, 10, and 12 correspond to $$L$$ = 1 and $$S$$ = 1, or 3P term symbol, and entries 3, 7, and 11 correspond to $$L$$ = 1 and $$S$$ = 0, which is a 1P term symbol. The values of $$J$$ can be derived from the table or by using

$$J$$ = ($$L$$ + $$S$$), ($$L$$ + $$S$$ - 1), ($$L$$ + $$S$$ - 2), . . . . . . . , (|$$L$$ + $$S$$|),

The final results given the term symbols below:

3P2                                3P1                            3P0                        1P1

($$L$$ + $$S$$)                ($$L$$ + $$S$$ - 1)               (|$$L$$ - $$S$$|)             (|$$L$$ + $$S$$|)

The states corresponding to this electron configuration and their degeneracies are:

Term symbol: 3P2        3P1            3P0        1P1

Degeneracy:     5           3               1            3

According to Hund's rule, the ground state 3P0. Including the effect of spin-orbit coupling removes the degeneracy of the electronic states, and no spin orbit coupling splits the lines in an atomic spectra.