'update all mols'

ch08/note04.md

@@ -1,75 +1,63 @@
 ## Molecular orbital description of hydrogen molecule
 
 
+### Setting up Hamiltonian 
+
+![](images/H2mol-fig1.png)
+
+
 
-Using the Born-Oppenheimer approximation, the electronic Hamiltonian for H$_2$ molecule can be written as:
 
 $${H = -\frac{\hbar^2}{2m_e}\left( \Delta_1 + \Delta_2\right)
 + \frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{R} + \frac{1}{r_{12}} - \frac{1}{r_{A1}} 
 - \frac{1}{r_{A2}} - \frac{1}{r_{B1}} - \frac{1}{r_{B2}}\right)}$$
 
-The distances between the electrons and the nuclei are indicated below.
 
-The main difficulty in the molecular Hamiltonian is the $1/r_{12}$
-term, which connects the two electrons to each other. This means that a simple 
-product wavefunction is not sufficient. No known analytic solutions have been 
-found to the electronic Schr\"odinger equation of $H_2$. For this reason, we will
-attempt to solve the problem approximately by using the LCAO-MO approach that we 
-used previously. For example, the ground state for H$_2$ is obtained by placing 
-two electrons with opposite spins on the $1\sigma_g$ orbital. This assumes that
+- The main difficulty in the molecular Hamiltonian is the $1/r_{12}$
+term, which connects the two electrons to each other. This means that a simple  product wavefunction is not sufficient. No known analytic solutions have been  found to the electronic Schr\"odinger equation of $H_2$. 
+- For this reason, we will attempt to solve the problem approximately by using the LCAO-MO approach that we  used previously. For example, the ground state for H$_2$ is obtained by placing  two electrons with opposite spins on the $1\sigma_g$ orbital. This assumes that
 the wavefunction is expressed as antisymmetrized product (e.g. a Slater determinant).
 
+### Constructing MOs for hydrogen molecule
 
-According to the Pauli principle, two electrons with opposite spins can be assigned
-to a given spatial orbital. As a first approximation, we assume that the molecular
-orbitals in $H_2$ remain the same as in $H_2^+$. Hence we can say that both
-electrons occupy the $1\sigma_g$ orbital (the ground state) and the electronic
-configuration is denoted by ($1\sigma_g$)$^2$. This is a similar notation 
-that we used previously for atoms (for example, He atom is ($1s$)$^2$).
+- According to the Pauli principle, two electrons with opposite spins can be assigned to a given spatial orbital. As a first approximation, we assume that the molecular orbitals in $H_2$ remain the same as in $H_2^+$. Hence we can say that both electrons occupy the $1\sigma_g$ orbital (the ground state) and the electronic configuration is denoted by ($1\sigma_g$)$^2$. This is a similar notation  that we used previously for atoms (for example, He atom is ($1s$)$^2$).
 
 
-The molecular orbital for electron 1 in $1\sigma_g$ molecular orbital is:
+- The molecular orbital for electron 1 in $1\sigma_g$ molecular orbital is:
 
 $${1\sigma_g(1) = \frac{1}{\sqrt{2(1 + S)}}(1s_A(1) + 1s_B(1))}$$
 
 
-Previously we found that the total wavefunction must be antisymmetric with respect to change in electron indices. This can be achieved by using the 
-Slater determinant:
+- Previously we found that the total wavefunction must be antisymmetric with respect to change in electron indices. This can be achieved by using the  Slater determinant:
 
 $${\psi_{MO}^{(1\sigma_g)^2} = \frac{1}{\sqrt{2}}\begin{vmatrix}
 1\sigma_g(1)\alpha (1) & 1\sigma_g(1)\beta (1)\\
 1\sigma_g(2)\alpha (2) & 1\sigma_g(2)\beta (2)\\
 \end{vmatrix}}$$
 
-where $\alpha$ and $\beta$ denote the electron spin. The Slater determinant can be
-expanded as follows:
+- where $\alpha$ and $\beta$ denote the electron spin. The Slater determinant can be expanded as follows:
 
 $${\psi_{MO}^{(1\sigma_g)^2} = \frac{1}{\sqrt{2}} 
  (1\sigma_g(1)1\sigma_g(2)\alpha (1)\beta (2) 
 - 1\sigma_g(1)1\sigma_g(2)\beta (1)\alpha (2))}
 {= \frac{1}{2\sqrt{2}(1 + S_{AB})}(1s_A(1) + 1s_B(1))(1s_A(2) + 1s_B(2))
 (\alpha (1)\beta (2) - \alpha (2)\beta (1))}$$
 
-
-- Note that this wavefunction is only approximate and is definitely not an  eigenfunction of the H$_2$ electronic Hamiltonian. Thus we must calculate the  electronic energy by taking an expectation value of this wavefunction with the
-Hamiltonian given in Eq. (\ref{eq11.37}) (the actual calculation not shown):
+- Note that this wavefunction is only approximate and is definitely not an  eigenfunction of the H$_2$ electronic Hamiltonian. Thus we must calculate the  electronic energy by taking an expectation value of this wavefunction with the Hamiltonian (the actual calculation not shown):
 
 $${E(R) = 2E_{1s} + \frac{e^2}{4\pi\epsilon_0 R} 
 - \textnormal{integrals}}$$
 
 
-where $E_{1s}$ is the electronic energy of one hydrogen atom. The second term 
-represents the Coulomb repulsion between the two positively charged nuclei and
-the last term (``integrals'') contains a series of integrals describing the 
-interactions of various charge distributions with one another (see P. W. 
-Atkins, Molecular Quantum Mechanics, Oxford University Press). With this 
-approach, the minimum energy is reached at $R$ = 84 pm (experimental 74.1 pm)
+- where $E_{1s}$ is the electronic energy of one hydrogen atom. The second term  represents the Coulomb repulsion between the two positively charged nuclei and the last term (``integrals'') contains a series of integrals describing the  interactions of various charge distributions with one another (see P. W.  Atkins, Molecular Quantum Mechanics, Oxford University Press). With this approach, the minimum energy is reached at $R$ = 84 pm (experimental 74.1 pm)
 and dissociation energy $D_e$ = 255 kJ $mol^{-1}$ (experimental 458 kJ 
 $mol^{-1}$).
 
+### Improving upon simple MO approximation
 
-This simple approach is not very accurate but it demonstrates that the method 
-works. To improve the accuracy, ionic and covalent terms should be considered
+![](images/H2mol-fig2.png)
+
+- This simple approach is not very accurate but it demonstrates that the method  works. To improve the accuracy, ionic and covalent terms should be considered
 separately:
 
 $${{1s_A(1)1s_A(2)}_{\textnormal{Ionic (H$^-$ + H$^+$)}}
@@ -98,6 +86,7 @@ requires use of configuration interaction methods, which can yield essentially
 exact results: $D_e$ = 36117.8 $cm^{-1}$ (CI) vs. $36117.3\pm1.0\, cm^{-1}$ 
 (exp) and $R_e$ = 74.140 pm vs. 74.139 pm (exp).
 
+### Term symbols for diatomic molecules
 
 In diatomic (and linear) molecules, the quantization axis is chosen along the 
 molecule. When spin-orbit interaction is negligible, this allows us to define 
@@ -119,36 +108,3 @@ atoms):
 
 $${^{2S+1}\Lambda}$$
 
-### Examples of molecular term symbols 
-
-**Example** What is the term symbol for ground state $H_2$?
-
-:::{dropdwon} **Solution** 
-
-Both electrons are on a $\sigma$ orbital and hence $\lambda_1 = \lambda_2
-= 0$. This gives $\Lambda = 0$, which corresponds to $\Sigma$. The electrons 
-occupy the same molecular orbital with opposite spins and hence $2S + 1 = 1$.
-This gives the term symbol as $^1\Sigma$.
-
-
-For $\Sigma$ terms superscripts $+$ and $-$ are used to express the
-parity of the wavefunction with respect to reflection in the plane 
-containing the internuclear axis. For example, for ground state $H_2$, we would
-have a ``$+$'' symbol. As we have seen before, orbitals in diatomic molecules may be 
-characterized by the $g$/$u$ labels. These labels are often added to term
-symbols as subscripts. If only one unpaired electron is present, the $u$/$g$
-label reflects the symmetry of the unpaired electron orbital. Closed shell 
-molecules have always $g$. With more than one unpaired electron, the overall
-parity should be calculated using the following rules: $g \times g = g$, $g \times 
-u = u$, $u \times g = u$ and $u \times u = g$.
-:::
-
-**Example** What is the term symbol for ground state $O_2$?
-
-::{dropdwon} **Solution** 
-
-:Ground state $O_2$ has two electrons with parallel spins on the $\pi_{+1}$ and $\pi_{-1}$ orbitals. Thus this is a triplet state molecule with the orbital angular momentum from the two $\pi$-electrons being cancelled. This gives a $^3\Sigma$ term.  The two $\pi$'s are anti-bonding and as such they are desginated as $g$ and further $g\times g = g$ (remember that for $\pi$ orbigals the $g/u$ vs. bonding/anti-bonding is reversed from that of $\sigma$ orbitals). To see the $+/-$ symmetry, it is convenient to think about $\pi_x$ and $\pi_y$ Cartesian orbitals (draw a picture!) and see that one of them is $+$ and the other is $-$ (they are perpendicular to each other). Again $+ \times - = -$ and we have the complete term symbol as $^3\Sigma_g^-$.
-
-- When spin-orbit interaction is small, the above term symbols are  adequate (Hund's case (a)).
-- When spin-orbit interaction is large, $S$ and $\Lambda$ can no longer be  specified but their sum $J = |S + \Lambda|$ is a good quantum number.
-:::

ch08/note05.md

@@ -1,9 +1,8 @@
-## Electron configurations of homonuclear diatomic molecules
+## Orbitals of homonuclear diatomic molecules
 
 
 Which atomic orbitals mix to form molecular orbitals and what are their
-relative energies? The graph on the left can be used to obtain the energy order
-of molecular orbitals and indicates the atomic orbital limits.
+relative energies? The graph on the left can be used to obtain the energy order of molecular orbitals and indicates the atomic orbital limits.
 
 
 ### The non-crossing rule: States with the same symmetry never cross.
@@ -12,10 +11,12 @@ of molecular orbitals and indicates the atomic orbital limits.
 |-----------------------|------------------------------------------------|
 | Antibonding orbitals: | $1\sigma_u^*$, $2\sigma_u^*$, $1\pi_g^*$, etc. |
 
+![](images/non-crossing-rule.png)
 
 ### Table of MOs
 
 
+
 The orbitals should be filled with electrons in the order of increasing energy. 
 Note that $\pi$, $\delta$, etc. orbitals can hold a total of 4 electrons. If only 
 one bond is formed, we say that the bond order (BO) is 1. If two bonds form (for 

ch08/note06.md

@@ -1,45 +1,44 @@
 ## Electronic structure of polyatomic molecules: the valece bond method
 
 
-The valence bond method is an approximate approach, which can be used in  understanding formation of chemical bonding. In particular, concepts like 
-hybrid orbitals follow directly from it.
+### Creating hybrid orbitals centered on atoms
 
+- The valence bond method is an approximate approach, which can be used in  understanding formation of chemical bonding. In particular, concepts like  hybrid orbitals follow directly from it.
 
-The valence bond method is based on the idea that a chemical bond is formed 
-when there is non-zero overlap between the atomic orbitals of the participating 
-atoms. Note that the the atomic orbitals must therefore have the same symmetry in 
-order to gain overlap.
+- The valence bond method is based on the idea that a chemical bond is formed  when there is non-zero overlap between the atomic orbitals of the participating  atoms. Note that the the atomic orbitals must therefore have the same symmetry in  order to gain overlap.
 
+- Hybrid orbitals are essentially linear combinations of atomic orbitals that  belong to a single atom. Note that hybrid orbitals are not meaningful for free  atoms as they only start to form when other atoms approach. The idea is best  illustrated through the following examples.
 
-Hybrid orbitals are essentially linear combinations of atomic orbitals that 
-belong to a single atom. Note that hybrid orbitals are not meaningful for free 
-atoms as they only start to form when other atoms approach. The idea is best 
-illustrated through the following examples.
+### 1. $BeH_2$ molecule.
 
+- $Be$ atoms have atomic electron configuration of He$2s^2$. The two approaching hydrogen perturb the atomic orbitals and the two outer shell electrons reside on the two hybrid orbitals formed ($z$-axis is along the molecular axis):
 
-1. $BeH_2$ molecule. $Be$ atoms have atomic electron configuration of He$2s^2$. The two approaching hydrogen perturb the atomic orbitals and the two outer shell electrons reside on the two hybrid orbitals formed ($z$-axis is along the molecular axis):
-
+![](images/BeH1.png)
 
 $${\psi_{sp}^1 = \frac{1}{\sqrt{2}}(2s + 2p_z)}$$
 
 $${\psi_{sp}^2 = \frac{1}{\sqrt{2}}(2s - 2p_z)}$$
 
 The hybrid orbitals further form two molecular $\sigma$ orbitals:
 
+![](images/BeH2.png)
+
 $${\psi = c_11s_A + c_2\psi_{sp}^1}$$
 
 $${\psi' = c_1'1s_B + c_2\psi_{sp}^2}$$
 
-**This form of hybridization is called $sp$.** This states that one 
-$s$ and one $p$ orbital participate in forming the hybrid orbitals. For $sp$
+![](images/BeH2-orbs.png)
+
+- **This form of hybridization is called $sp$.** This states that one  $s$ and one $p$ orbital participate in forming the hybrid orbitals. For $sp$
 hybrids, linear geometries are favored and here H--Be--H is indeed linear. 
 Here each MO between Be and H contain two shared electrons. Note that the 
 number of initial atomic orbitals and the number of hybrid orbitals formed must
 be identical. Here $s$ and $p$ atomic orbitals give two $sp$ hybrid orbitals. 
 Note that hybrid orbitals should be orthonormalized.
 
+### 2. $BH_3$ molecule.
 
-2. $BH_3$ molecule. All the atoms lie in a plane (i.e. planar 
+- All the atoms lie in a plane (i.e. planar 
 structure) and the angles between the H atoms is 120\degree . The 
 boron atom has electron configuration $1s^22s^22p$. Now three
 atomic orbitals ($2s$, $2p_z$, $2p_x$) participate in forming three hybrid orbitals:
@@ -52,15 +51,17 @@ $${\psi^2_{sp^2} = \frac{1}{\sqrt{3}}2s - \frac{1}{\sqrt{6}}2p_z
 $${\psi^3_{sp^2} = \frac{1}{\sqrt{3}}2s - \frac{1}{\sqrt{6}}2p_z 
 - \frac{1}{\sqrt{2}}2p_x}$$
 
-
 The three orbitals can have the following spatial orientations:
 
+![](images/BeH3.png)
 
 - Each of these hybrid orbitals bind form $\sigma$ bonds with H atoms. This is 
 called $sp^2$ hybridization because two $p$ orbitals and one $s$ orbital 
 participate in the hybrid.
 
-3. $CH_4$ molecule. The electron configuration of carbon atom is
+### 3. $CH_4$ molecule. 
+
+- The electron configuration of carbon atom is
 $1s^22s^22p^2$. The outer four valence electrons should be placed on four
 $sp^3$ hybrid orbitals:
 
@@ -75,13 +76,16 @@ $${\psi^4_{sp^3} = \frac{1}{2}(2s - 2p_x + 2p_y - 2p_z)}$$
 
 These four hybrid orbitals form $\sigma$ bonds with the four hydrogen atoms.
 
+![](images/CH4.png)
 
 The $sp^3$ hybridization is directly responsible for the geometry of CH$_4$ 
 molecule. Note that for other elements with $d$-orbitals, one can also get 
 bipyramidal (coordination 5) and octahedral (coordination 6) structures.
 
 
-4. NH$_3$ molecule. In this molecule, nitrogen is also $sp^3$ 
+### 4. NH$_3$ molecule. 
+
+- In this molecule, nitrogen is also $sp^3$ 
 hybridized. The N atom electron configuration is $1s^22s^22p_x^12p_y^12p_z^1$. Thus
 a total of 5 electrons should be placed on the four hybrid orbitals. One of the
 hybrid orbitals becomes doubly occupied (``lone-pair electrons'') and the three
@@ -90,13 +94,16 @@ Because of the lone-pair electrons, the geometry of $NH_3$ is tetrahedral with
 a bond angle of 109\degree (experimental value 107\degree).
 
 
-5. $H_2 O$ molecule. The oxygen is $sp^3$ hybridized with O atom 
+### 5. $H_2 O$ molecule. 
+
+- The oxygen is $sp^3$ hybridized with O atom 
 electron configuration: $1s^22s^22p^4$. Now two of
 the four hybrid orbitals are doubly occupied with the electrons from 
 oxygen atom and the remaining two hybrid orbitals can participate in $\sigma$ bonding 
 with two H atoms. This predicts the bond angle H--O--H as $109\degree $(experimental 
 value $104\degree$). Thus $H_2O$ has two lone-pair electrons.
 
+### Numerical Calculations 
 
 In numerical quantum chemical calculations, basis sets that resemble 
 linear combinations of atomic orbitals are typically used (LCAO-MO-SCF). The