22.2.6: vi. Problem Solutions

The above diagram indicates how the SALC-AOs are formed from the 1s, 2s, and 2p N atomic orbitals. It can be seen that there are $$3\sigma_g$$, $$3\sigma_u$$, $$1\pi_{ux}$$, $$1\pi_{uy}$$, $$1\pi_{gx}$$, and $$1\pi_{gy}$$ SALC - AOs. The Hamiltonian matrices (Fock matrices) are given. Each of these can be diagonalized to give the following MO energies:

3$$\sigma_g$$; -15.52, -1.45, and -0.54 (hartrees)
3$$\sigma_u$$; -15.52, -0.72, and 1.13
$$1\pi_{ux};$$ -0.58
$$1\pi_{uy};$$ -0.58
$$1\pi_{gx};$$ 0.28
$$1\pi_{gy};$$ 0.28

It can be seen that the 3$$\sigma_g$$ orbitals are bonding, the 3$$\sigma_u$$ orbitals are antibonding, the $$1\pi_{ux}$$ and $$1\pi_{uy}$$ orbitals are bonding, and the $$1\pi_{gx}$$ and $$1\pi_{gy}$$ orbitals are antibonding. The eigenvectors one obtains are in the orthogonal basis and therefore pretty meaningless. Back transformation into the original basis will generate the expected results for the 1e$$^-$$ MOs (expected combinations of SALC-AOs).

2. Using these approximate energies we can draw the following MO diagram:

This MO diagram is not an orbital correlation diagram but can be used to help generate one. The energy levels on each side (C and $$H_2$$) can be "superimposed" to generate the left side of the orbital correlation diagram and the center $$CH_2$$ levels can be used to form the right side. Ignoring the core levels this generates the following orbital correlation diagram.

3.

Using $$D_{2h}$$ symmetry and labeling the orbitals $$(f_1-f_{12})$$ as shown above proceed by using the orbitals to define a reducible representation which may be subsequently reduced to its irreducible components. Use projectors to find the SALC-AOs for these irreps.

3. a. The $$2P_x$$ orbitals on each carbon form the following reducible representation:

\begin{align} & D_{2h} & \text{ } & E & \text{ } & C_2(z) & \text{ } & C_2(y) & \text{ } & C_2(x) & \text{ } & i & \text{ } & \sigma (xy) & \text{ } & \sigma (xz) & \text{ } & \sigma (yz) \\ & \Gamma_{2p_x} & \text{ } & 2 & \text{ } & -2 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 0 & \text{ } & 2 & \text{ } & -2 \end{align}
The number of irreducible representations may be found by using the following formula:
$n_{irrep} = \dfrac{1}{g} \sum\limits_{R} \chi_{red}(R)\chi_{irrep}(R) ,$
where g = the order of the point group (8 for $$D_{2h}$$).

\begin{align} n_{A_g} &=& & \dfrac{1}{8}\sum\limits_R \Gamma_{2p_x} (R)A_g(R) \\ &=& & \dfrac{1}{8}\left[ (2)(1)+(-2)(1)+(0)(1)+(0)(1)+(0)(1)+(0)(1)+(2)(1)+(-2)(1) \right] = 0 \end{align}

Similarly,
\begin{align} &n_{B_{1g}} &=& &0& \\ &n_{B_{2g}} &=& &1& \\ &n_{B_{3g}} &=& &0& \\ &n_{A_u} &=& &0& \\ &n_{B_{1u}} &=& &0& \\ &n_{B_{2u}} &=& &0& \\ &n_{B_{3u}} &=& &1& \end{align}

Projectors using the formula:

$P_{irrep} = \sum\limits_R \chi_{irrep}(R)R ,$

may be used to find the SALC-AOs for these irreducible representations.

$P_{B_{2g}} = \sum\limits_R \chi_{B_{2g}}(R)R ,$

\begin{align} P_{B_{2g}} &=& &(1)Ef_1 + (-1)C_2(z)f_1 + (1)C_2(y)f_1 + (-1)C_2(x)f_1 + (1)if_1 + (-1)\sigma (xy)f_1 + (1)\sigma (xz)f_1 + (-1)\sigma (yz)f_1 \\ &=& &(1)f_1 + (-1)-f_1 + (1)-f_2 + (-1)f_2 + (1)-f_2 + (-1)f_2 + (1)f_1 + (-1)-f_1 \\ &=& &f_1 + f_1 - f_2 - f_2 - f_2 - f_2 + f_1 + f_1 \\ &=& &4f_1 - 4f_2 \end{align}

Normalization of this SALC-AO (and representing the SALC-AOs with $$\phi$$) yields:

$\int N(f_1 - f_2)N(f_1 - f_2)d\tau = 1$ $N^2 \left( \int f_1f_1d\tau - \int f_1f_2d\tau - \int f_2f_1d\tau + \int f_2f_2d\tau \right) = 1$ $N^2(1 + 1) = 1$ $2N^2 = 1$ $N = \dfrac{1}{\sqrt{2}}$ $\phi_{1b_{2g}} = \dfrac{1}{\sqrt{2}}(f_1 - f_2).$

The $$B_{3u}$$ SALC-AO may be found in a similar fashion:

\begin{align} P_{B_{3u}}f_1 &=& &(1)f_1 + (-1)-f_1 + (-1)-f_2 + (1)f_2 + (-1)-f_2 + (1)f_2 + (1)f_1 + (-1)-f_1 \\ &=& &f_1 + f_1 + f_2 + f_2 + f_2 + f_2 + f_1 + f_1 \\ &=& &4f_1 + 4f_2 \end{align}

Normalization of this SALC-AO yields:

$\phi_{1b_{3u}} = \dfrac{1}{\sqrt{2}} (f_1 + f_2).$

Since there are only two SALC-AOs and both are of different symmetry types these SALC-AOs are MOs and the 2x2 Hamiltonian matrix reduces to 2 1x1 matricies.

\begin{align} H_{1b_{2g},1b_{2g}} &=& &\int \dfrac{1}{\sqrt{2}}(f_1 - f_2)H\dfrac{1}{\sqrt{2}}(f_1 - f_2)d\tau \\ &=& &\dfrac{1}{2}\left( \int f_1Hf_1d\tau - 2\int f_1Hf_2d\tau + \int f_2Hf_2d\tau \right) \\ &=& &\dfrac{1}{2}\left( \alpha_{2p\pi} - 2\beta_{2p\pi - 2p\pi} + \alpha_{2p\pi} \right) \\ &=& &\alpha_{2p\pi} - \beta_{2p\pi - 2p\pi} \\ &=& &-11.4 - (-1.2) = -10.2 \end{align}

\begin{align} H_{1b_{3u},1b_{3u}} &=& &\int \dfrac{1}{\sqrt{2}} (f_1 + f_2 )H\dfrac{1}{\sqrt{2}}(f_1 + f_2)d\tau \\ &=& &\dfrac{1}{2}\left( \int f_1 H f_1d\tau + 2\int f_1Hf_2d\tau + \int f_2Hf_2 d\tau \right) \\ &=& &\dfrac{1}{2}\left( \alpha_{2p\pi} + 2\beta_{2p\pi -2p\pi} + \alpha_{2p\pi} \right) \\ &=& &\alpha_{2p\pi} + \beta_{2p\pi-2p\pi} \\ &=& &-11.4 + (-1.2) = -12.6 \end{align}

This results in a $$\pi \rightarrow \pi^{\text{*}}$$ splitting of 2.4 eV.

3. b. The $$sp^2$$ orbitals forming the C-C bond generate the following reducible representation:

\begin{align} &D_{2h}& &E& &C_2(z)& &C_2(y)& &C_2(x)& &i& &\sigma (xy)& &\sigma (xz)& &\sigma (yz)& \\ &\Gamma_{C_{sp}}^2 &2& &2& &0& &0& &0& &0& &2& &2& \end{align}

This reducible representation reduces to $$1A_g \text{ and } 1B_{1u}$$ irreducible representations.

Projectors are used to find the SALC-AOs for these irreducible representations.

\begin{align} P_{A_g}f_3 &=& &(1)Ef_3 + (1)C_2(z)f_3 + (1)C_2(y)f_3 + (1)C_2(x)f_3 + (1)if_3 + (1)\sigma (xy)f_3 + (1)\sigma (xz)f_3 + (1)\sigma (yz)f_3 \\ (1)f_3 + (1)f_3 + (1)f_4 + (1)f_4 + (1)f_4 + (1)f_4 + (1)f_3 + (1)f_3 \\ &=& &4f_3 + 4f_4 \end{align}

Normalization of this SALC-AO yields:

$\phi_{1a_{g}} = \dfrac{1}{\sqrt{2}}(f_3 + f_4).$

The $$B_{1u}$$ SALC-AO may be found in a similar fashion:

\begin{align} P_{B_{1u}}f_3 &=& &(1)f_3 + (1)f_3 + (-1)f_4 + (-1)f_4 + (-1)f_4 + (-1)f_4 + (1)f_3 + (1)f_3 \\ &=& &4f_3 - 4f_4\end{align}

Normalization of this SALC-AOs yields:

$\phi_{1b_{3u}} = \dfrac{1}{\sqrt{2}} (f_3 - f_4).$

Again sine there are only two SALC-AOs and both are of different symmetry types these SALC-AOs are MOs and the 2x2 Hamiltonian matrix reduces to 2 1x1 matrices.

\begin{align} H_{1a_g,1a_g} &=& &\int \dfrac{1}{\sqrt{2}}(f_3 + f_4)H\dfrac{1}{\sqrt{2}}(f_3 + f_4)d\tau \\ &=& &\dfrac{1}{2}\left( \int f_3Hf_3d\tau + 2\int f_3Hf_4d\tau + \int f_4Hf_4d\tau \right) \\ &=& &\dfrac{1}{2} \left( \alpha_{sp^2} + 2\beta_{sp^2-sp^2} + \alpha_{sp^2} \right) \\ &=& &\alpha_{sp^2} + \beta_{sp^2-sp^2} \\ &=& &-14.7 + (-5.0) = -19.7 \\ H_{1b_{1u},1b_{1u}} &=& &\int \dfrac{1}{\sqrt{2}}(f_3 - f_4)H\dfrac{1}{\sqrt{2}}(f_3 - f_4)d\tau \\ &=& &\dfrac{1}{2}\left( \int f_3Hf_3d\tau - 2\int f_3Hf_4d\tau + \int f_4Hf_4d\tau \right) \\ &=& &\dfrac{1}{2}\left( \alpha_{sp^2} - 2\beta_{2p^2-2p^2} + \alpha_{sp^2} \right) \\ &=& &\alpha_{sp^2} - \beta_{sp^2-sp^2} \\ &=& &-14.7 - (-5.0) = -9.7 \end{align}

3. c. The C $$sp^2$$ orbitals and the H s orbitals forming the C-H bonds generate the following reducible representation:

\begin{align} &D_{2h}& &E& &C_2(z)& &C_2(y)& &C_2(x)& &i& &\sigma (xy)& &\sigma (xz)& &\sigma (yz)& \\ &\Gamma_{sp^2-s}& &8& &0& &0& &0& &0& &0& &0& &8& \end{align}

This reducible representation reduces to $$2A_g, 2B_{3g}, 2B_{1u}, \text{ and } 2B_{2u}$$ irreducible representation.

Projectors are used to find the SALC-AOs for these irreducible representations.

\begin{align} P_{A_g} f_6 &=& &(1)Ef_6 + (1)C_2(z)f_6 + (1)C_2(y)f_6 + (1)C_2(x)f_6 + (1)if_6 (1)\sigma (xy)f_6 + (1)\sigma (xz)f_6 + (1)\sigma (yz)f_6 \\ &=& &(1)f_6 + (1)f_5 + (1)f_7 + (1)f_8 + (1)f_8 + (1)f_7 + (1)f_5 + (1)f_6 \\ &=& &2f_5 + 2f_6 + 2f_7 + 2f_8 \end{align}

Normalization yields: $$\phi 2a_g = \dfrac{1}{2}( f_5 + f_6 + f_7 + f_8 ).$$

\begin{align} P_{A_g}f_{10} &=& &(1)Ef_{10} + (1)C_2(z)f_{10} + (1)C_2(y)f_{10} + (1)C_2(x)f_{10} + (1)if_{10} + (1)\sigma (xy)f_{10} + (1)\sigma (xz)f_{10} + (1)\sigma (yz)f_{10} \\ &=& &(1)f_{10} + (1)f_9 + (1)f_{11} + (1)f_{12} + (1)f_{12} + (1)f_{11} + (1)f_9 + (1)f_{10} \\ &=& &2f_9 + 2f_{10} + 2f_{11} + 2f_{12} \end{align}

Normalization yields: $$\phi_{3a_g} = \dfrac{1}{2}( f_9 + f_{10} + f_{11} + f_{12} ).$$

\begin{align} P_{B_{3g}}f_6 &=& &(1)f_6 + (-1)f_5 + (-1)f_7 + (1)f_8 + (1)f_8 +(-1)f_7 + (-1)f_5 + (1)f_6 \\ &=& &-2f_5 + 2f_6 - 2f_7 + 2f_8 \end{align}

Normalization yields: $$\phi_{1b_{3g}} = \dfrac{1}{2}(-f_5 + f_6 - f_7 + f_8).$$

\begin{align} P_{B_{3g}}f_{10} &=& &(1)f_{10} + (-1)f_9 + (-1)f_{11} + (1)f_{12} + (1)f_{12} + (-1)f_{11} + (-1)f_9 + (1)f_{10} \\ &=& &-2f_9 + 2f_{10} - 2f_{11} + 2f_{12} \end{align}

Normalization yields: $$\phi_{2b_{3g}} = \dfrac{1}{2}(-f_9 + f_{10} - f_{11} + f_{12}).$$

\begin{align} P_{B_{1u}}f_6 &=& &(1)f_6 + (1)f_5 + (-1)f_7 + (-1)f_8 + (-1)f_8 + (-1)f_7 + (1)f_5 + (1)f_6 \\ &=& &2f_5 + 2f_6 - 2f_7 - 2f_8 \end{align}

Normalization yields: $$\phi_{2b_{1u}} = \dfrac{1}{2}(f_5 + f_6 - f_7 - f_8).$$

\begin{align} P_{B_{1u}}f_10 &=& &(1)f_{10} + (1)f_9 + (-1)f_{11} + (-1)f_{12} + (-1)f_{12} + (-1)f_{11} + (1)f_9 + (1)f_{10} \\ &=& &2f_9 + 2f_{10} - 2f_{11} - 2f_{12} \end{align}

Normalization yields: $$\phi_{3b_{1u}} = \dfrac{1}{2}(f_9 + f_{10} - f_{11} - f_{12}).$$

\begin{align} P_{B_{2u}}f_6 = (1)f_6 + (-1)f_5 + (1)f_7 + (-1)f_8 + (-1)f_8 + (1)f_7 + (-1)f_5 + (1)f_6 &=& &-2f_5 + 2f_6 + 2f_7 - 2f_8 \end{align}

Normalization yields: $$\phi_{1b_{2u}} = \dfrac{1}{2}(-f_5 + f_{6} - f_{7} - f_{8}).$$

\begin{align} P_{B_{2u}}f_{10} = (1)f_{10} + (-1)f_9 + (1)f_{11} + (-1)f_{12} + (-1)f_{12} + (1)f_{11} + (-1)f_9 + (1)f_{10} &=& &-2f_9 + 2f_{10} + 2f_{11} - 2f_{12} \end{align}

Normalization yields: $$\phi_{2b_{2u}} = \dfrac{1}{2}(-f_9 + f_{10} + f_{11} - f_{12}).$$

Each of these four 2x2 symmetry blocks generate identical Hamitonian matrices. This will bve demonstrated for the $$B_{3g}$$ symmetry, the others proceed analogously:

\begin{align} H_{1b_{3g},1b_{3g}} &=& &\int \dfrac{1}{2}(-f_5 + f_6 - f_7 + f_8)H\dfrac{1}{2}(-f_5 + f_6 - f_7 + f_8)d\tau \\ &=& &\dfrac{1}{4}\bigg[ \int f_5Hf_5d\tau - \int f_5Hf_6d\tau + \int f_5Hf_7d\tau - \int f_5Hf_8d\tau \\ & &-&\int f_6Hf_5d\tau + \int f_6Hf_6d\tau - \int f_6Hf_7d\tau + \int f_5Hf_8d\tau \\ & & +&\int f_7Hf_9d\tau - \int f_7Hf_{10}d\tau + \int f_7Hf_{11}d\tau - \int f_7Hf_{12}d\tau \\ & & -& \int f_8Hf_9d\tau + \int f_8Hf_{10}d\tau - \int f_8Hf_{11}d\tau + \int f_8Hf_{12}d\tau \bigg] \\ &=& &\dfrac{1}{4}\left[ \beta_{sp^2-s} - 0 + 0 - 0 - 0 + \beta_{sp^2-s} - 0 + 0 + 0 - 0 + \beta_{sp^2-s} - 0 - 0 + 0 - 0 + \beta_{sp^2-s} \right] = \beta_{sp^2-s} \end{align}

\begin{align} H_{1b_{3g},2b_{3g}} &=& &\int \dfrac{1}{2}(-f_5 + f_6 - f_7 + f_8)H\dfrac{1}{2}(-f_9 + f_{10} - f_{11} + f_{12})d\tau \\ &=& &\dfrac{1}{4}\bigg[ \int f_5Hf_9d\tau - \int f_5Hf_{10}d\tau + \int f_5Hf_{11}d\tau - \int f_5Hf_{12}d\tau \\ & &-&\int f_6Hf_{9}d\tau + \int f_6Hf_{10}d\tau - \int f_6Hf_{11}d\tau + \int f_5Hf_{12}d\tau \\ & & +&\int f_7Hf_9d\tau - \int f_7Hf_{10}d\tau + \int f_7Hf_{11}d\tau - \int f_7Hf_{12}d\tau \\ & & -& \int f_8Hf_9d\tau + \int f_8Hf_{10}d\tau - \int f_8Hf_{11}d\tau + \int f_8Hf_{12}d\tau \bigg] \\ &=& &\dfrac{1}{4}\left[ \beta_{sp^2-s} - 0 + 0 - 0 - 0 + \beta_{sp^2-s} - 0 + 0 + 0 - 0 + \beta_{sp^2-s} - 0 - 0 + 0 - 0 + \beta_{sp^2-s} \right] = \beta_{sp^2-s} \end{align}

\begin{align} H_{2b_{3g},2b_{3g}} &=& &\int \dfrac{1}{2}(-f_9 + f_{10} - f_{11} + f_{12})H\dfrac{1}{2}(-f_9 + f_{10} - f_{11} + f_{12})d\tau \\ &=& &\dfrac{1}{4}\bigg[ \int f_9Hf_9d\tau - \int f_9Hf_{10}d\tau + \int f_9Hf_{11}d\tau - \int f_9Hf_{12}d\tau \\ & &-&\int f_{10}Hf_{9}d\tau + \int f_{10}Hf_{10}d\tau - \int f_{10}Hf_{11}d\tau + \int f_{10}Hf_{12}d\tau \\ & & +&\int f_{11}Hf_9d\tau - \int f_{11}Hf_{10}d\tau + \int f_{11}Hf_{11}d\tau - \int f_{11}Hf_{12}d\tau \\ & & -& \int f_{12}Hf_9d\tau + \int f_{12}Hf_{10}d\tau - \int f_{12}Hf_{11}d\tau + \int f_{12}Hf_{12}d\tau \bigg] \\ &=& &\dfrac{1}{4}\left[ \alpha_s - 0 + 0 - 0 - 0 + \alpha_s - 0 + 0 + 0 - 0 + \alpha_s - 0 - 0 + 0 - 0 + \alpha_s \right] = \alpha_s \end{align}

This matrix eigenvalue problem then becomes:

$\begin{vmatrix} & \alpha_{sp^2} -\epsilon & \beta_{sp^2-s} & \\ & \beta_{sp^2-s} & \alpha_s-\epsilon & \end{vmatrix} = 0$

$\begin{vmatrix} & -14.7 -\epsilon & -4.0 & \\ & -4.0 & -13.6 & \end{vmatrix} = 0$

Solving this yields eigenvalues of:

\begin{vmatrix} & -18.19 & -10.11 & \end{vmatrix}

and corresponding eigenvectors:

\begin{vmatrix} & -0.7537 & -0.6572 & \\ & -0.6572 & 0.7537 & \end{vmatrix}

This results in an orbital energy diagram:

For the ground state of ethylene you would fill the bottom 3 levels (the C-C, C-H, and $$\pi$$ bonding orbitals), with 12 electrons.

4.

Using the hybrid atomic orbitals as labeled above (functions $$f_1-f_7$$) and the $$D_{3h}$$ point group

symmetry it is easiest to construct three sets of reducible representations:

i. the B $$2p_z$$ orbital (labeled function 1)
ii. the 3 B $$sp^2$$ hybrids (labeled functions 2 - 4)
iii. the 3 H 1s orbitals (labeled functions 5 - 7).

i. The B $$2p_z$$ orbital generates the following irreducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v &\\ & \Gamma_{2p_{z}} & & 1 & & 1 & & -1& & -1 & & -1 & & 1 & \end{align}

This irreducible representation is $$A_2''$$ and is its own SALC-AO.

ii. The B $$sp^2$$ orbitals generate the following reducible representation:

$\begin{vmatrix} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v &\\ & \Gamma_{2p} & & 3 & & 0 & & 1& & 3 & & 0 & & 1 & \end{vmatrix}$

This reducible representation reduces to $$1A_1'$$ and 1E'
irreducible representations.
Projectors are used to find the SALC-AOs for these irreducible representations.
Define: $$C_3$$ = 120 degree rotations, $$C_3' = 240$$ degree rotation,
$$C_2$$ = rotation around $$f_4 , C_2'$$ = rotation around $$f_2$$, and
$$C_2$$ = rotation adound $$f_3$$. $$S_3$$ and $$S_3'$$ are defined analogous
to $$C_3$$ and $$C_3'$$ with accompanying horizontal reflection.
$$\sigma_v$$ = a reflection plane through $$f_4, \sigma_v'$$ = a reflection plane
through $$f_2,$$ and $$\sigma_v''$$ = a reflection plan through $$f_3$$

$$P_{A_1'} f_2 = (1)E f_2 + (1)C_3f_2 + (1)C_3'f_2 + (1)C_2f_2 + (1)C_2'f_2 + (1)C_2''f_2 + (1)\sigma_hf_2 + (1)S_3f_2 +(1)S_3'f_2 + (1)\sigma_vf_2 + (1)\sigma_v'f_2 + (1)\sigma_v''f_2$$

$$= (1)f_2 + (1)f_3 + (1)f_4 + (1)f_3 + (1)f_2 + (1)f_4 + (1)f_2 + (1) f_3 + (1)f_4 + (1)f_3 + (1)f_2 + (1)\sigma f_4$$

$$= 4f_2 + 4f_3 + 4f_4$$

Normalization yields: $$\phi_{1e'} = \dfrac{1}{\sqrt{6}}\left( 2f_2 - f_3 -f_4 \right) .$$

To find the second e' (orthogonal to the first), projection of $$f_3$$ yields ($$2f_3 - f_2 -f_4$$) and projection on $$f_4$$ yields $$(2f_4 - f_2 - f_3$$). Neither of these functions are orthogonal to the first, but a combination of the two ($$2f_4 - f_2 - f_4) - (2f_4 - f_2 - f_3)$$ yields a function which is orthogonal to the first.

Normalization yields: $$\phi_{2e'} = \dfrac{1}{\sqrt{2}}\left( f_3 - f_4\right).$$\

iii. The H 1s orbitals generate the following reducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v & \\ & \Gamma_{sp^2} & & 3 & & 0 & & 1 & & 3 & & 0 & & 1 & \end{align}

This reducible representation reduces to $$1A_1'$$ and 1E' irreducible representations exactly like part ii. and in addition the projectors used to find the SALC-AOs for these irreducible representations is exactly analogous to part ii.

$\phi_{2a_1'} = \dfrac{1}{\sqrt{3}} (f_5 + f_6 + f_7)$

$\phi_{3e'} = \dfrac{1}{\sqrt{6}}(2f_5 - f_6 - f_7).$

$\phi_{4e'} = \dfrac{1}{\sqrt{2}}(f_6 - f_7).$

So, there are $$1A_2'', 2A_1'$$ and 2E' orbitals. Solving the Hamiltonian matrix for each symmetry block yields:

$$A_2''$$ Block:

\begin{align} H_{1a_{2'},1a_{2'}} & = & \int f_1Hf_1d\tau \\ & = & \alpha_{2p\pi} \\ & = & -8.5\end{align}

$$A_1'$$ Block:

\begin{align} H_{1_{a1'},1_{a2'}} & = & \int \dfrac{1}{\sqrt{3}}(f_2 +f_3+f_4)H\dfrac{1}{\sqrt{3}}(f_2 + f_3 + f_4)d\tau & = & \dfrac{1}{3} [ \int f_2Hf_2d\tau + \int f_2Hf_3d\tau + \int f_2Hf_4d\tau + \\ & + & \int f_3 Hf_2d\tau +\int f_3 Hf_3d\tau + \int f_3Hf_4d\tau + \\ & + & \int f_4Hf_2d\tau + \int f_4Hf_3d\tau + \int f_4Hf_4d\tau ] \\ & = & \dfrac{1}{3}\left[ \alpha_{sp^2} + 0 + 0 + 0 + \alpha_{sp^2} + 0 + 0 + 0 + \alpha_{sp^2} \right] = \alpha_{sp^2} \end{align}

\begin{align} H_{1_{a1'},2_{a1'}} & = & \int \dfrac{1}{\sqrt{3}}(f_2 +f_3+f_4)H\dfrac{1}{\sqrt{3}}(f_5 + f_6 + f_7)d\tau & = & \dfrac{1}{3} [ \int f_2Hf_5d\tau + \int f_2Hf_6d\tau + \int f_2Hf_7d\tau + \\ & + & \int f_3 Hf_5d\tau +\int f_3 Hf_6d\tau + \int f_3Hf_7d\tau + \\ & + & \int f_4Hf_5d\tau + \int f_4Hf_6d\tau + \int f_4Hf_7d\tau ] \\ & = & \dfrac{1}{3}\left[ \beta_{sp^2-s} + 0 + 0 + 0 + \beta_{sp^2-s} + 0 + 0 + 0 + \beta_{sp^2-s} \right] = \beta_{sp^2-s} \end{align}

\begin{align} H_{2_{a1'},2_{a1'}} & = & \int \dfrac{1}{\sqrt{3}}(f_5 +f_6+f_7)H\dfrac{1}{\sqrt{3}}(f_5 + f_6 + f_7)d\tau & = & \dfrac{1}{3} [ \int f_5Hf_5d\tau + \int f_5Hf_6d\tau + \int f_5Hf_7d\tau + \\ & + & \int f_6 Hf_5d\tau +\int f_6 Hf_6d\tau + \int f_6Hf_7d\tau + \\ & + & \int f_7Hf_5d\tau + \int f_7Hf_6d\tau + \int f_7Hf_7d\tau ] \\ & = & \dfrac{1}{3}\left[ \alpha_{s} + 0 + 0 + 0 + \alpha_{s} + 0 + 0 + 0 + \alpha_{s} \right] = \alpha_{s} \end{align}

This matrix eigenvalue problem then becomes:

\befin{align} \begin{vmatrix} & \alpha_{sp^2 -\epsilon} & \beta_{sp^2-s} & \\ & \beta_{sp^2-s} & \alpha_s-\epsilon & \end{vmatrix} & = & 0 \\ \begin{vmatrix} & -10.7 - \epsilon & -3.5 & \\ & -3.5 & -13.6 - \epsilon & \end{vmatrix} & = & 0 \end{align}

Solving this yields eigenvalues of:

\begin{vmatrix} & -15.94 & -8.36 & \end{vmatrix}

and corresponding eigenvectors:

\begin{vmatrix} & -0.5555 & -0.8315 & \\ & -0.8315 & 0.5555 & \end{vmatrix}

E' Block:

This 4x4 symmetry block factors to two 2x2 blocks: where one 2x2 block includes the SALC-AOs

\begin{align} \phi_{e'} & = & \dfrac{1}{\sqrt{6}}(2f_2 - f_3 - f_4) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{6}}(2f_5 - f_6 - f_7),\end{align}

and the other includes the SALC-AOs

\begin{align} \phi_{e'} & = & \dfrac{1}{\sqrt{2}}(f_3 - f_4) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{2}}(f_6 - f_7). \end{align}

Both of these 2x2 matrices are identical to the $$A_1$$' 2x2 array and therefore yield identical energies and MO coefficients.
This results in an orbital energy diagram:

For the ground state of $$BH_3$$ you would fill the bottom level (B-H bonding), $$a_1'$$ and e' orbitals, with 6 electrons.

5.

5. a. The two F p orbitals (top and bottom) generate the following reducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v & \\ & \Gamma_p & & 2 & & 2 & & 0 & & 0 & & 0 & & 2 & \end{align}

This reducible representation reduces to $$1A_1'$$ and $$1A_2''$$ irreducible representations.
Projectors may be used to find the SALC-AOsfor these irreducible representations.

\begin{align} \phi_{a_1'} & = & \dfrac{1}{\sqrt{2}}(f_1 - f_2) \\ \phi_{a_2}'' & = & \dfrac{1}{\sqrt{2}}(f_1 + f_2) \end{align}

5. b. The three trigonal F p orbitals generate the following reducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v & \\ & \Gamma_p & & 3 & & 0 & & 1 & & 3 & & 0 & & 1 & \end{align}

This reducible representation reduces to $$1A_1'$$ and 1E' irreducible representations.
Projectors may be used to find the SALC-AOs for these irreducible representations (but they are exactly analogous to the previous few problems):

\begin{align} \phi_{a_1'} & = & \dfrac{1}{\sqrt{3}}(f_3 + f_4 + f_5) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{6}}(2f_3 - f_4 - f_5) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{2}}(f_4 - f_5). \end{align}

5. c. The 3 P $$sp^2$$ orbitals generate the following reducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v & \\ & \Gamma_p & & 3 & & 0 & & 1 & & 3 & & 0 & & 1 & \end{align}

This reducible representation reduces to $$1A_1'$$ and 1E' irreducible representations. Again, projectors may be used to find the SALC-AOs for these irreducible representations.(but again they are exactly analogous to the previous few problems):

\begin{align} \phi_{a_1'} & = & \dfrac{1}{\sqrt{3}}(f_6 + f_7 + f_8) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{6}}(2f_6 - f_7 - f_8) \\ \phi_{e'} & = & \dfrac{1}{\sqrt{2}}(f_7 - f_8). \end{align}

The leftover P $$p_z$$ orbital generate the following irreducible representation:

\begin{align} & D_{3h} & & E & & 2C_3 & & 3C_2 & & \sigma_h & & 2S_3 & & 3\sigma_v & \\ & \Gamma_p & & 1 & & 1 & & -1 & & -1 & & -1 & & 1 & \end{align}

This irreducible representation is an $$A_2''$$

$\phi_{a_2''} = f_9.$

Drawing an energy level diagram using these SALC-AOs would result in the following: