(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the 1 s orbital of hydrogen with the 2 s orbital of fluorine. The 1 s orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a 1 sorbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a 1 s orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HE. These are called "nonbonding orbitals." Sketch the energy- level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HE. Where are the nonbonding electrons?

Step by step solution

01

Determine the number of MOs in the HF molecule

We will first determine the number of molecular orbitals in the HF molecule using valence atomic orbitals of hydrogen and fluorine atoms. For hydrogen (H), the valence atomic orbitals is \(1s\). For fluorine (F), valence atomic orbitals are \(2s\) and three \(2p\) orbitals (\(2p_x\), \(2p_y\), and \(2p_z\)). When atomic orbitals combine in the formation of a molecule, they combine one on one, so since we have four atomic orbitals in total (1 for H and 3 for F), there will be four molecular orbitals overall. Thus, the number of MOs for the HF molecule is 4.

02

Determine the number of occupied MOs

Now let's determine how many of the molecular orbitals in part (a) would be occupied by electrons. For hydrogen (H), there is 1 electron in the \(1s\) orbital. For fluorine (F), there are 2 electrons in the \(2s\) orbital and 6 electrons in the \(2p\) orbitals (2 electrons in each of the three \(2p\) orbitals). Together, H and F have a total of 9 electrons, which means 9 electrons will occupy the molecular orbitals formed.

03

Draw orientation pictures and identify bonding orbitals

To draw pictures showing the proper orientation of all three \(2p\) orbitals on F interacting with a \(1s\) orbital on H, we must identify the coordinates axis. Assuming that the hydrogen (H) and fluorine (F) atoms lie on the \(z\)-axis, we have: 1. \(2p_x\) orbital of F oriented along the \(x\)-axis, which is perpendicular to the z-axis and not overlapping with the \(1s\) orbital of H. 2. \(2p_y\) orbital of F oriented along the \(y\)-axis, which is also perpendicular to the z-axis and not overlapping with the \(1s\) orbital of H. 3. \(2p_z\) orbital of F oriented along the \(z\)-axis, which directly overlaps with the \(1s\) orbital of H. Based on this, only the \(2p_z\) orbital can make a bond with the \(1s\) orbital of H, since their overlap is maximized.

04

Sketch energy-level diagram and calculate bond order

To sketch the energy-level diagram for the HF molecule and calculate the bond order, we need to account for the molecular orbitals and their electron occupancy: 1. The \(1s\) orbital from H completely overlaps with the \(2p_z\) orbital from F, forming a sigma bonding MO (\(\sigma\)) and a sigma antibonding MO (\(\sigma^*\)) along the \(z\)-axis. 2. The interaction between \(1s\) of H with \(2s\) of F is neglected due to the significant energy difference. 3. The remaining \(2p_x\) and \(2p_y\) orbitals on F are nonbonding orbitals (as per exercise information). Now we can calculate the bond order: Bond Order = (number of electrons in bonding MOs - number of electrons in antibonding MOs) / 2 Bond Order = (2 - 1) / 2 = 1 / 2 Therefore, the bond order for the HF molecule is 0.5.

05

Identify the location of nonbonding electrons

To identify the location of nonbonding electrons in the Lewis structure for HF, examine the atomic orbitals contributing to the HF molecule and the electrons in the MOs. The \(2p_x\) and \(2p_y\) orbitals on fluorine (F) do nothing, resulting in no bonding. The electrons in these orbitals contribute to the nonbonding electrons. Thus, the nonbonding electrons in the HF Lewis structure can be found in the \(2p_x\) and \(2p_y\) orbitals on fluorine (F).

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