Show how two \(2 p\) atomic orbitals can combine to form a \(\sigma\) or a \(\pi\) molecular orbital.

Short Answer

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Two 2p atomic orbitals can combine to form a σ molecular orbital through head-to-head overlap along the internuclear axis, resulting in electron density symmetric around this axis. They can form a π molecular orbital through side-to-side overlap, maintaining their orientation perpendicular to the internuclear axis, with electron density accumulating above and below the axis in case of 2px orbitals or in front and behind the axis in case of 2py orbitals.

Step by step solution

01

Understand 2p orbitals characteristics

2p orbitals have two lobes on opposite sides of the nucleus, with a nodal plane passing through the nucleus where the electron probability is zero. There are three 2p orbitals, one along each of the x, y, and z axes: 2px, 2py, and 2pz. In forming molecular orbitals, the orbitals must have the same energy and proper orientation.
02

Formation of a sigma (σ) molecular orbital

Sigma (σ) molecular orbitals are formed by the head-to-head (or end-to-end) overlap of two atomic orbitals along their internuclear axis. The electron density is symmetric along this axis. For forming σ molecular orbitals using 2p orbitals, we will consider the overlap of two 2pz orbitals: 1. Consider two atoms A and B, each with a 2pz orbital. 2. Align the internuclear axis (the axis connecting the nuclei of both atoms) along the z-axis. 3. Overlap the 2pz orbital of atom A with the 2pz orbital of atom B along the z-axis in a head-to-head fashion. 4. The resulting electron density will be symmetric around the internuclear axis, forming a sigma (σ) molecular orbital.
03

Formation of a pi (π) molecular orbital

Pi (π) molecular orbitals are formed by the side-to-side overlap of two atomic orbitals, leading to electron density above and below (or in front and behind) the internuclear axis. For forming π molecular orbitals using 2p orbitals, we will consider the overlap of two 2px and two 2py orbitals: 1. Consider two atoms A and B, each with a set of 2px and 2py orbitals. 2. Align the internuclear axis along the z-axis (we will demonstrate with 2px orbitals, but the process is the same for 2py orbitals). 3. Overlap the 2px orbital of atom A with the 2px orbital of atom B in a side-to-side fashion, maintaining their orientation perpendicular to the internuclear axis. 4. With this side-to-side overlap, electron density will be above and below the internuclear axis, forming a pi (π) molecular orbital. The process for 2py orbitals is the same, with electron density accumulating in front and behind the internuclear axis in the resulting π molecular orbital. In summary, two 2p atomic orbitals can combine to form a σ molecular orbital by head-to-head overlap along the internuclear axis, or a π molecular orbital by side-to-side overlap perpendicular to the internuclear axis.

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Most popular questions from this chapter

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{array}{cc} \mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta H=375 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & \Delta H=158 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

Draw the Lewis structures for \(\mathrm{TeCl}_{4}, \mathrm{ICl}_{5}, \mathrm{PCl}_{5}, \mathrm{KrCl}_{4}\), and \(\mathrm{XeCl}_{2}\). Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit \(d^{2} s p^{3}\) hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) than the hybrid orbital model?

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}{ }^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) EX- plain your answer.

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi\).

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