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Show how a \(d_{x z}\) atomic orbital and a \(p_{z}\) atomic orbital combine to form a bonding molecular orbital. Assume the \(x\) -axis is the internuclear axis. Is a \(\sigma\) or a \(\pi\) molecular orbital formed? Explain.

Short Answer

Expert verified
A \(d_{xz}\) atomic orbital and a \(p_z\) atomic orbital combine to form a bonding molecular orbital through the overlap of their lobes with the same sign, which occurs in the same plane. There is no direct interaction along the internuclear axis (x-axis). Therefore, a π (pi) molecular orbital is formed in this process.

Step by step solution

01

Visualize the orbitals involved

Firstly, let's visualize the \(d_{xz}\) and \(p_z\) atomic orbitals. The \(d_{xz}\) orbital lies in the xz-plane and has four lobes. In contrast, the \(p_z\) orbital has two lobes along the z-axis. Keep in mind that the internuclear axis is the x-axis.
02

Determine the possible overlap

To form a molecular orbital, orbitals must overlap in space. In this case, since the \(d_{xz}\) orbital lies in the xz-plane and the \(p_z\) orbital is aligned along the z-axis, there is no direct overlap along the internuclear (x) axis. The overlap occurs between \(d_{xz}\) and \(p_z\) lobes lying in the same plane.
03

Analyze the orbital overlap

The bonding molecular orbital formed results from the constructive interference of wave functions. This constructive interference happens when the wave functions of the orbitals have the same sign (either positive or negative). In this case, an overlap between one lobe of the \(d_{xz}\) and one lobe of the \(p_z\) orbital with the same sign forms the bonding molecular orbital.
04

Determine the type of molecular orbital formed

Now we need to decide if the molecular orbital formed is a σ (sigma) or a π (pi) orbital. A σ molecular orbital occurs when there is symmetry around the internuclear axis, while a π molecular orbital occurs when there is no symmetry around the internuclear axis. Since the overlap occurs only between lobes lying in the same plane and there is no direct interaction along the internuclear axis, the molecular orbital formed is a π (pi) molecular orbital.
05

Conclusion

In summary, a \(d_{xz}\) atomic orbital and a \(p_z\) atomic orbital can combine to form a bonding molecular orbital through the overlap of their lobes with the same sign. Since there is no direct interaction along the internuclear axis (x-axis), a π (pi) molecular orbital is formed in this process.

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Most popular questions from this chapter

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with \(25-\mathrm{nm}\) light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{aligned} \mathrm{N}_{2}(g) & \longrightarrow 2 \mathrm{~N}(g) & \Delta H &=941 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) & \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1501 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}(g) & \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1402 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{~kJ} / \mathrm{mol})\) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{~kJ} / \mathrm{mol})\).

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d\), or \(6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

In Exercise 89 in Chapter 8, the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is \(717 \mathrm{~kJ} / \mathrm{mol}\). The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is \(83 \mathrm{~kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Use the localized electron model to describe the bonding in \(\mathrm{C}_{2} \mathrm{H}_{2}\) (exists as \(\mathrm{HCCH}\) ).

Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

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