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Chapter 9: Problem 60

(a) If you combine two atomic orbitals on two different atoms to make a new orbital, is this a hybrid orbital or a molecular orbital? (b) If you combine two atomic orbitals on one atom to make a new orbital, is this a hybrid orbital or a molecular orbital? (c) Does the Pauli exclusion principle (Section 6.7) apply to MOs? Explain.

Short Answer

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(a) Combining two atomic orbitals on two different atoms forms a Molecular Orbital (MO), as electrons in these orbitals are shared between the atoms involved. (b) Combining two atomic orbitals on one atom forms a Hybrid Orbital, which results from the mixing of atomic orbitals on the same atom. (c) Yes, the Pauli Exclusion Principle applies to MOs, meaning no two electrons can have the same set of quantum numbers in a molecular orbital, and only two electrons with opposite spins can occupy the same MO.

Step by step solution

01

a. Combination of atomic orbitals from different atoms

When two atomic orbitals from two different atoms combine, they form what is called a Molecular Orbital (MO). Molecular orbitals are orbitals that result from the overlap of atomic orbitals in molecules, and electrons in these orbitals are shared between the atoms involved.
02

b. Combination of atomic orbitals from one atom

When two atomic orbitals combine on the same atom, they form a Hybrid Orbital. Hybridization involves the mixing of two or more atomic orbitals on the same atom to form new orbitals with different geometries and energies. These hybrid orbitals can then form bonds with other atoms.
03

c. Pauli exclusion principle and Molecular Orbitals

Yes, the Pauli Exclusion Principle applies to Molecular Orbitals (MOs). The Pauli Exclusion Principle states that no two electrons in an atom or molecule can have the same set of quantum numbers. This means that in a molecular orbital, at most, two electrons with opposite spins can occupy the same MO. Electrons in MOs will fill the lowest energy levels first before moving on to higher ones, which ensures that the overall energy of the molecule remains as low as possible.

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Most popular questions from this chapter

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6 r}\) you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in \(\mathrm{XeF}_{6}\). (e) The molecule \(\mathrm{IF}_{7}\) has a pentagonal- bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\)

Why are there no \(s p^{4}\) or \(s p^{5}\) hybrid orbitals?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\mathrm{a}} ; \mathrm{PCl}_{3}\) \(100.3^{\circ} ; \mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{Pl}_{3}, 102.0^{\circ} .\) The trend is generally at- tributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(X-P-X\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCI}_{4}\)

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

Predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{CCl}_{4}\), (b) \(\mathrm{NH}_{3}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{XeF}_{4}\) (e) \(\mathrm{CH}_{3} \mathrm{Br}\), (f) \(\mathrm{GaH}_{3}\).
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