Solutions 10

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Solution10.1

Find the spectroscopic terms originating from the following configurations (two electrons in two non-equivalent orbitals in the same atom):

1s2s = ³S₁¹S₀
1s3s = ³S₁ ¹S₀
2p3p = ³D₃ ³D₂³D₁ ¹D₂ ³P₂ ³P₁ ³P₀ ¹P₁ ³S₁ ¹S₀
2s3d = ³D₃ ³D₂ ³D₁ ¹D₂

The spectroscopic terms are ^{2S + 1}L_Jwhere S = |M_s1 + M_s2|, |M_s1 + M_s2 - 1|, ...., |M_s1 - M_s2|, L = |l₁ + l₂|, |l₁ + l₂ - 1|, ...., |l₁ - l₂|, and J = |L + S|, |L + S - 1|, ...., |L - S| and since the two electrons in each example are in different orbitals, i.e. 2p3p instead of 2p2p, we do not have to worry about the Pauli exclusion principle rendering certain terms impossible.

Solution 10.2

This is a complex question for several reasons. Previously, we considered excited state as only different electron configurations. However, now we can differentiate levels based on term symbols (microstates). So different microstates will have the same electron configuration. The second aspect is how to identify which state is lower or higher in energy. While technically, Hund's rule applies to only ground-state, the gist of the rules can be be use for higher state (just do not rely exclusively on them).

The first step in identifying microstates and assigning term symbols is to establsih electron configurations

First excited states:

H: This is a one electron system. The first excited state based off of electron configuration will have a 2s configuration. This is a \(^2S_{1/2}\) state.
He: This is a two electron system. The first excited state based off of electron configuration will have a 1s¹2s¹ configuration. There are two microstates for this configuration: \(^3S_{1}\) (spin aligned) and \(^1S_{0}\). Hund's first rule predicts the triplet state is the lowest of this configuration (it is too)
Li: This is a three electron system. The first excited state based off of electron configuration will have a 1s²2s⁰2p¹ configuration. There are two microstates for this configuration: \(^2P_{1/2}\) and \(^2P_{3/2}\). Since the p-orbitals are less than half-full, the term with the lowest J will be lower in energy (Hund's third rule); therefore, the lowest excited state will be \(^2P_{1/2}\).
Be: This is a four electron system. The first excited state based off of electron configuration will have a 1s²2s¹2p¹ configuration. There are two microstates for this configuration: \(^1P_{1}\) and \(^3P_1\). The triplet is preferred via Hund's first rule
B: This is a five electron system. The first excited state based off of electron configuration will have a 1s²2s²3s¹configuration, This would have a \(^2S_{1/2}\) state. However, the ground-state configuration 1s²2s²3p¹ has multiple states in it, so one of those is really the first excited-state.

For C, N and O, the electron configuration is the same as for the ground state, but the occupation of degenerate p-orbitals is not optimal.

C: 2s²2p² which is a ¹D
N: 2s²2p³ which is a ²D
O: 2s²2p⁴ which is a ¹D
F: 2s²2p⁵which is a ²P
Ar: 2s²2p⁵3s¹which is a ³P

Solution 10.3

Find the spectroscopic term originating from the ground states configuration of the sodium atom and the boron atom.

a) Boron: 1s²2s²2p¹so with one 2p electron not in a filled shell our terms are ²P_3/2 and ²P_1/2

b) Sodium: 1s²2s²2p⁶3s¹so with one 3s electron not in a filled shell our term is ²S_1/2

Solution 10.4

a) 2p: ²P_3/2 and ²P_1/2

b) 3d: ²D₅_/2 ²D_3/2

Solution 10.5

a) Be ----> Be⁺ + e^-

b) Be: 1s²2s², so the term symbol is ¹S₀, because it is a closed shell system;

Be+: 1s²2s¹, so L=0, S=1/2, term symbol is: ²S_1/2

c) Electronic energy for Be is: -9143.9 kcal/mol

Electronic energy for Be+ is: -8958.45 kcal/mol

So the ionization energy is: -8958.45 - (-9143.9) = 185.45 kcal/mol.

d) For neutral atom Be, the energy of the highest occupied orbital is -193.8 kcal/mol (the second energy value for "Orbital Energies" on the output page of ChemCompute). So, according to Koopman's theorem, the first ionization energy is -(-193.8)=193.8 kcal/mol.

e) The difference between Koopman theorem value and normal calculation value is: 193.8-185.45=8.35 kcal/mol. It is 8.35/185.45=4.5% of the normal calculation value. Its accurary is due to the assumption of frozen orbitals, i.e., when you remove an electron from the highest occupied orbital, the rest of the orbitals do not change. And this is a pretty good assumption for beryllium. However, if you do not assume frozen orbitals, then, when you remove an electron, the rest of the orbitals are going to change and you will not be able to assume E_Ionization=E_N-1-E_N=-ε_k.

To help you understand better, I will explain the terms in the equation above:

E_Ionization=E_N-1-E_N,this part is the normal way how you calculate ionization energy, without considering Koopman's theorem.

E_Ionization=-ε_kis the energy that you would get if you use Koopman's theorem.

So taken together, according to Koopman's theorem, ideally if the frozen orbitals assumption is true, you should get E_Ionization=E_N-1-E_N=-ε_k.However, if the frozen orbitals assumption is not describing the real situation that is going in the molecule very well, then you will find that E_N-1-E_Nand-ε_khave different values_.But Normally, these two values are very close to each other, becasue the frozen orbitals assumption can usually describe the molecule pretty accurately.

Solution 10.6

For (a) and (b), just expand the original determinant and the determinants with interchanged rows or columns, and compare the results. You will find that interchanging two rows or columns will lead to the change of sign.

(c) Assuming \(\psi_A =\psi_B\), then we get two identical rows in the determinant. According to matrix algebra, the determinant with two identical rows equals to zero. So, we get an invalid wavefunction, because a valid wavefunction cannot be zero all the time.

Solution 10.7

\[\psi_1 = \frac{1}{2}[\phi_1(1)\phi_2(2) + \phi_1(2)\phi_2(1)][\alpha(1)\beta(2)-\beta(1)\alpha(2)]\]

\[=\dfrac{1}{2}[ \phi_1(1)\phi_2(2)\alpha(1)\beta(2)-\phi_1(1)\phi_2(2)\beta(1)\alpha(2)+\phi_1(2)\phi_2(1)\alpha(1)\beta(2) -\phi_1(2)\phi_2(1)\beta(1)\alpha(2) \]

\[=\dfrac{1}{2}[ \phi_1(1)\alpha(1)\phi_2(2)\beta(2)-\phi_1(2)\alpha(2)\phi_2(1)\beta(1) ]-\dfrac{1}{2}[ \phi_1(1)\beta(1)\phi_2(2)\alpha(2)-\phi_1(2)\beta(2)\phi_2(1)\alpha(1) ] \]

\[ = \dfrac{1}{2}\begin{vmatrix} \phi_1(1)\alpha(1) & \phi_2(1)\beta(1) \\ \phi_1(2)\alpha(2) & \phi_2(2)\beta(2) \end{vmatrix}-\dfrac{1}{2} \begin{vmatrix} \phi_1(1)\beta(1) & \phi_2(1)\alpha(1) \\ \phi_1(2)\beta(2) & \phi_2(2)\alpha(2) \end{vmatrix}\]

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\[\psi_2 = \frac{1}{\sqrt{2}}[\phi_1(1)\phi_2(2) - \phi_1(2)\phi_2(1)][\alpha(1)\alpha(2)] \]

\[=\frac{1}{\sqrt{2}}[\phi_1(1)\phi_2(2)\alpha(1)\alpha(2)-\phi_1(2)\phi_2(1)\alpha(1)\alpha(2)] \]

\[=\frac{1}{\sqrt{2}}[\phi_1(1)\alpha(1)\phi_2(2)\alpha(2)-\phi_1(2)\alpha(2)\phi_2(1)\alpha(1)] \]

\[=\frac{1}{\sqrt{2}} \begin{vmatrix} \phi_1(1)\alpha(1) & \phi_2(1)\alpha(1) \\ \phi_1(2)\alpha(2) & \phi_2(2)\alpha(2) \end{vmatrix} \]

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\[\psi_3 = \frac{1}{\sqrt{2}}[\phi_1(1)\phi_2(2) - \phi_1(2)\phi_2(1)][\beta(1)\beta(2)]\]

\[=\frac{1}{\sqrt{2}}[\phi_1(1)\phi_2(2)\beta(1)\beta(2)-\phi_1(2)\phi_2(1)\beta(1)\beta(2)] \]

\[=\frac{1}{\sqrt{2}}[\phi_1(1)\beta(1)\phi_2(2)\beta(2)-\phi_1(2)\beta(2)\phi_2(1)\beta(1)] \]

\[=\frac{1}{\sqrt{2}} \begin{vmatrix} \phi_1(1)\beta(1) & \phi_2(1)\beta(1) \\ \phi_1(2)\beta(2) & \phi_2(2)\beta(2) \end{vmatrix} \]

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\[\psi_4 = \frac{1}{2}[\phi_1(1)\phi_2(2) - \phi_1(2)\phi_2(1)][\alpha(1)\beta(2)+\beta(1)\alpha(2)] \]

\[=\dfrac{1}{2}[ \phi_1(1)\phi_2(2)\alpha(1)\beta(2)+\phi_1(1)\phi_2(2)\beta(1)\alpha(2)-\phi_1(2)\phi_2(1)\alpha(1)\beta(2) -\phi_1(2)\phi_2(1)\beta(1)\alpha(2) \]

\[=\dfrac{1}{2}[ \phi_1(1)\alpha(1)\phi_2(2)\beta(2)-\phi_1(2)\alpha(2)\phi_2(1)\beta(1) ]+\dfrac{1}{2}[ \phi_1(1)\beta(1)\phi_2(2)\alpha(2)-\phi_1(2)\beta(2)\phi_2(1)\alpha(1) ] \]

\[ = \dfrac{1}{2}\begin{vmatrix} \phi_1(1)\alpha(1) & \phi_2(1)\beta(1) \\ \phi_1(2)\alpha(2) & \phi_2(2)\beta(2) \end{vmatrix}+\dfrac{1}{2} \begin{vmatrix} \phi_1(1)\beta(1) & \phi_2(1)\alpha(1) \\ \phi_1(2)\beta(2) & \phi_2(2)\alpha(2) \end{vmatrix}\]