Determine the angles of \(SF_{6}\) using the vector approach (dot product formula).
S10.9
The \(s\) orbitals are spherical and therefor do not contribute to the directional vectors for this problem. The \(d_{z^2}\) orbital only has \(z\) directionality and the \(d_{x^{2}-y^{2}}\) orbital has equal parts \(x\) and \(y\) directionality. The equations can be rewritten:
The square root constants are the necessary normalization constants.
Q10.11
Given that \[ \psi_{1} = 0.71j + 0.55k\] and \[ \psi_{2} = -0.71j + 0.55k\] Find the bond angle between \(\psi_1\) and \(\psi_2\) using the vector approach.
Additionally, find \(\gamma\) assuming the bond angles to be \(104.5°\).
S10.12
Because the molecules are in the yz plane, the \(p_{x}\) orbital can be neglected because it is completely orthogonal. The hybrid orbitals are a linear combination of the \(2s,\:2p_{y},\text{ and }2p_{z}\) orbitals.
\(c_{1}\) and \(c_2\) can be found by thinking about how the bonds are oriented. The two bonds both have \(cos[\theta]\) character in the z-direction and have \(\pm sin[\theta]\) in the y-direction. The two functions can therefore be written as
The atomic orbitals within the linear combination of the lone pair wave functions are normalized and orthogonal, zeroing out cross products and leaving only the squares of the coefficients.
10.14
Molecular orbitals for a linear $$XY_{2}$$ molecule can be represented as
Draw a schematic representation for the $$3\sigma_{g}, 4\sigma_{g}, 1\pi_{g}, 2\sigma_{g}$$ orbitals
Which has the highest energy?
S10.14
\[ 3\sigma_{g}\]
NOTE:
\[3\sigma_g\]
$$4\sigma_{g}$$
$$1\pi_{g}$$
$$2\sigma_{g}$$
$$4\sigma_{g}$$ has the highest energy
The \[4\sigma_g\] moleclar orbital has the highest energy; as expected since it more nodes.
Q10.18
Use the given Walsh diagram to predict the geometry of the following molecules:
\(H_{2}O\)
\(H_{2}S\)
\(H_{2}Be\)
Walsh Diagram of an HAH molecule.Public Domain
S10.18
The Walsh Diagram predicts the geometry of a molecule by assigning its valence electrons to the appropriate energy levels. In general, the lowest energy configuration is preferred.
\(H_{2}O\) has 8 valence electrons, which corresponds to the 4th highest orbital on the diagram. The bent configuration (90º) is lower in energy in this case.
\(H_{2}S\) has 8 valence electrons as well, because sulfur and oxygen are in the same periodic group. Therefore, the bent configuration will be favored.
\(H_{2}Be\) has 4 valence electrons, which corresponds to the 2nd highest orbital on the diagram. The linear configuration (180º) is lower in energy in this case.
Q10.19
Use the Walsh diagram for the valence electrons of a XY2 molecule to predict whether the following molecules are linear or bent:
a. (CO2) b. (CO2+) c. (CO2-) d. (SO2-) b. (CF2+)
S10.19
Valence Electrons
Geometry
2-16
linear
17-20
bent
21-24
linear
CO2 16 valence electrons Linear
CO2+ 15 valence electronsLinear
CO2- 17 valence electrons Bent
SO2- 19 valence electrons Bent
CF2+ 18 valence electrons Bent
Q10.20
Walsh correlation diagrams can be used to predict the shapes of polyatomic molecules that contain more than three atoms. In this and the following three problems we consider molecules that have the general formula XH3. We will restrict our discussion to XH3 molecules, where all the H—X—H bond angles are the same. If the molecule is planar, then the H—X—H bond angle is 120°. A nonplanar XH3 molecule, then, has an H—X—H bond angle that is less than 120°. Figure 10.26 shows the Walsh correlation diagram that describes how the energies of the molecular orbitals for an XH3 molecule change as a function of the H—X—H bond angle. Note that because XH3 is not linear, the labels used to describe the orbitals on the two sides of the correlation diagram do not have designations such as and . We see that the lowest-energy molecular orbital is insensitive to the H—X—H bond angle. Which atomic orbital(s) contribute to the lowest-energy molecular orbital? Explain why the energy of this molecular orbital is insensitive to changes in the H—X—H bond angle.
S10.20
The lowest energy molecular orbital is the 1s orbital, which is a core atomic orbital instead of a bonding atomic orbital.
Q10.21
Consider the BH2 where the general Walsh diagram for a XH2 is shown below. What is the geometric preference of the molecule in ground and excited state?
S10.21
The BH2 molecule has the same geometric shape as water, it is bent where the HOMO is \({\pi_u}\). The first excited state relies on the degree of bending, and the 2a1 is unoccupied, and the next 1b2 is the highest occupied where the preferred geometry is linear, so at the first excited state will be linear.
\(BH_{2}\) is a linear molecule. It has 4 valance electrons which fill 2 of the lines in the Walsh diagram. This second line has lower energy towards linear conformation and \(1 \sigma_{u}\).
Q10.24
Solve for \(\psi_\pi \) corresponding to the energy \(E = \alpha + \beta \) for ethene.
S10.24
The bonding Huckel molecular orbitals is
\[ \psi_\pi = c_1 2p_{z1} + c_2 2p_{z2}\]
the relationship of the coefficients can be defined as the following from the secular determinate:
\[c_1 (\alpha - E) + c_2 \beta = 0 \]
\[c_1 \beta+ c_2 (\alpha - E) = 0 \]
Substituting \(E = \alpha + \beta \) into these expressions and solving gives
\[-c_1\beta+c_2\beta = 0 \]
\[c_1 = c_2 \]
Then plugging back into the original equation gives
\[ \psi_\pi = c_1 (2p_{zA} + 2p_{zB} ) \]
Now we can solve for \(c_1\) by normalizing the wavefunction
Generalize the molecular orbital treatment of propene allyl cation. Find the energies and wave function of this molecule.
S10.25
The Huckel secular determinant for propene is \[\left. \begin{vmatrix} \alpha - E & \beta & 0 \\\ \beta & \alpha - E & \beta \\ 0 & \beta & \alpha - E \end{vmatrix} \right. = 0 \]
making a substitution for \( x = \frac{\alpha - E }{\beta} \) the secular determinant becomes \[\left. \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} \right. = 0 \]
Solving the determinant yields a cubic polynomial \(x^3 - 2x = 0\) the roots of this polynomial are \( x = 0, \pm \sqrt{2} \)
Replacing x with the previous substitution made it is found that \[E = \alpha \pm \sqrt{2} \beta \] and \[E = 0\]
To find the wave function of propene you must find the constants \[\left. \begin{vmatrix} c_{1}(x) & c_{2}& 0 \\ c_{1} & c_{2}(x) & c_{3} \\ 0 & c_{2}& c_{3}(x) \end{vmatrix} \right. = 0 \]