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Chapter 9: Problem 74

Indicate whether each statement is true or false. (a) \(s\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is 100\(\%\) for finding an electron at the nucleus in a \(\pi^{*}\) orbital. (c) Antibonding orbitals are higher in energy than bonding orbitals (if all orbitals are created from the same atomic orbitals). (d) Electrons cannot occupy an antibonding orbital.

Short Answer

Expert verified
(a) True: s orbitals can only form σ and σ* molecular orbitals due to their spherical and symmetric nature around the nucleus. (b) False: The probability of finding an electron at the nucleus in a π* orbital is zero due to the nodal plane passing through the nucleus. (c) True: Antibonding orbitals have higher energy than bonding orbitals when both are created from the same atomic orbitals. (d) False: Electrons can occupy antibonding orbitals, but their presence weakens the overall bond.

Step by step solution

01

Statement (a) Analysis

The statement says that s orbitals can only make σ or σ* molecular orbitals. To analyze this statement, let's first recall what σ and σ* orbitals are. σ orbitals are bonding molecular orbitals that are symmetric around the bond axis, while σ* orbitals are antibonding orbitals that are also symmetric around the bond axis. Now we consider s orbitals which are spherical and symmetric around the nucleus. When these s orbitals combine, they can form molecular orbitals that are symmetric around the bond axis, which are σ and σ* orbitals. Therefore, the statement seems to be true.
02

Statement (a) Conclusion

Statement (a) is true, as s orbitals can only form σ and σ* molecular orbitals due to their spherical and symmetric nature around the nucleus.
03

Statement (b) Analysis

The statement says that the probability is 100% for finding an electron at the nucleus in a π* orbital. π* orbitals are antibonding molecular orbitals that arise from the out-of-phase combination of atomic p orbitals. The nature of π* orbitals is such that they have a nodal plane going through the nuclei of the atoms involved in the bond. This means that the probability of finding an electron in the region along the nodal plane, including the nucleus, is zero.
04

Statement (b) Conclusion

Statement (b) is false, as the probability of finding an electron at the nucleus in a π* orbital is zero due to the nodal plane passing through the nucleus.
05

Statement (c) Analysis

The statement says that antibonding orbitals are higher in energy than bonding orbitals if all orbitals are created from the same atomic orbitals. By definition, bonding orbitals result from the constructive interference of atomic orbitals, leading to increased electron density between the nuclei and hence a stabilization of the energy. Antibonding orbitals, on the other hand, result from destructive interference, causing the electron density between the nuclei to decrease and the energy to increase. So, antibonding orbitals should have higher energy than bonding orbitals when created from the same atomic orbitals.
06

Statement (c) Conclusion

Statement (c) is true, as antibonding orbitals have higher energy than bonding orbitals when both are created from the same atomic orbitals.
07

Statement (d) Analysis

The statement says that electrons cannot occupy an antibonding orbital. This statement is not accurate. Electrons can occupy antibonding orbitals, but the more electrons that occupy antibonding orbitals, the weaker the overall bond between the atoms. In molecular orbital theory, we fill up orbitals according to the Aufbau principle, starting with the lowest energy orbitals first. Electrons will fill the bonding orbitals before filling anti-bonding orbitals. However, when filling higher energy orbitals, electrons can and do occupy antibonding orbitals.
08

Statement (d) Conclusion

Statement (d) is false, as electrons can occupy antibonding orbitals, but their presence weakens the overall bond.

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Most popular questions from this chapter

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

(a) An AB \(_{6}\) molecule has no lone pairs of electrons on the A atom. What is its molecular geometry? (b) An AB \(_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four B atoms). What is the electron-domain geometry around the A atom? (c) For the AB \(_{4}\) molecule in part (b), predict the molecular geometry.

What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) Three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than the corresponding one in \(\mathrm{PH}_{3}\) ?

How does a trigonal pyramid differ from a tetrahedron so far as molecular geometry is concerned?
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