Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

Short Answer

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Bonding molecular orbitals (BMOs) form due to in-phase overlapping of atomic orbitals, resulting in lower energy levels and electron density between atomic nuclei. This stabilizes the molecule. Antibonding molecular orbitals (ABMOs) form due to out-of-phase overlapping of atomic orbitals, resulting in higher energy levels and decreased electron density between atomic nuclei. This destabilizes the molecule. Electrons fill bonding orbitals before antibonding orbitals, according to the Aufbau principle. The bond order is determined by \(\dfrac{(n_{BMO} - n_{ABMO})}{2}\) and indicates a stable molecule when positive.

Step by step solution

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1. Formation of Molecular Orbitals

Bonding and antibonding molecular orbitals are formed due to the overlapping of atomic orbitals in a molecule. When two atomic orbitals overlap in-phase (constructive interference), a bonding molecular orbital is formed. When two atomic orbitals overlap out-of-phase (destructive interference), an antibonding molecular orbital is formed.
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2. Energy Levels

Bonding molecular orbitals (BMOs) are formed due to constructive interference and are lower in energy than the individual atomic orbitals (AOs) contributing to the BMO. Antibonding molecular orbitals (ABMOs) are formed due to destructive interference and are higher in energy than the individual atomic orbitals contributing to the ABMO.
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3. Electron Occupancy

Electrons in a molecule occupy molecular orbitals according to the Aufbau principle, which states that electrons fill orbitals in order of increasing energy. Since bonding molecular orbitals have lower energy than antibonding molecular orbitals, they are filled with electrons before the antibonding orbitals are filled.
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4. Effects on Bond Order

The bond order of a molecule is determined by the difference between the number of electrons in bonding orbitals and the number of electrons in antibonding orbitals, divided by 2: \(Bond\: order = \dfrac{(n_{BMO} - n_{ABMO})}{2}\). A positive bond order indicates a stable molecule, whereas a negative or zero bond order indicates an unstable or non-existent molecule. Bonding molecular orbitals stabilize the molecule by increasing the electron density between the atomic nuclei, which causes an attractive force between the positively charged nuclei of both atoms. Antibonding molecular orbitals destabilize the molecule by decreasing the electron density between the atomic nuclei, causing a repulsive force between the positively charged nuclei of both atoms.

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Most popular questions from this chapter

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}^{+}, \mathrm{F}_{2},\) and \(\mathrm{F}_{2}^{-},\) predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

Predict the molecular structure for each of the following. (See Exercises 25 and \(26 .\) ) a. \(\mathrm{BrFI}_{2}\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) c. \(\operatorname{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Use the localized electron model to describe the bonding in \(\mathrm{H}_{2} \mathrm{CO}\) (carbon is the central atom).

Which of the following statements is/are true? Correct the false statements. a. The molecules \(\operatorname{SeS}_{3}, \operatorname{SeS}_{2}, \operatorname{PCl}_{5}, \operatorname{TeCl}_{4},\) ICl \(_{3}\), and \(\mathrm{XeCl}_{2}\) all exhibit at least one bond angle which is approximately \(120^{\circ} .\) b. The bond angle in \(\mathrm{SO}_{2}\) should be similar to the bond angle in \(\mathrm{CS}_{2}\) or \(\mathrm{SCl}_{2}\) c. Of the compounds \(\mathrm{CF}_{4}, \mathrm{KrF}_{4},\) and \(\mathrm{SeF}_{4},\) only \(\mathrm{SeF}_{4}\) exhibits an overall dipole moment (is polar). d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to maximize electron repulsions.

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{kJ} / \mathrm{mol},\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen 1s character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

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