Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

Short Answer

Expert verified
a) False. \(p\) orbitals can also make \(\pi\) or \(\pi^{*}\) molecular orbitals. b) False. Electrons can occupy antibonding orbitals, with a non-zero probability of being found there. c) Partially true, molecules with electrons in antibonding orbitals are not necessarily unstable, but they tend to be less stable compared to molecules without electrons in antibonding orbitals. d) False, electrons can occupy nonbonding orbitals.

Step by step solution

01

a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals.

The statement is false. Although \(p\) orbitals can form \(\sigma\) and \(\sigma^{*}\) molecular orbitals, they can also form other types of molecular orbitals, such as \(\pi\) and \(\pi^{*}\) molecular orbitals. In the formation of a molecular orbital, \(p\) orbitals may align along the axis and overlap end-to-end, which results in the formation of a \(\sigma\) or \(\sigma^{*}\) molecular orbital. Besides this, they can also interact side-by-side and form \(\pi\) or \(\pi^{*}\) molecular orbitals.
02

b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital.

The statement is false. Antibonding orbitals are characterized by their higher energy compared to bonding orbitals and are generally less stable. However, electrons can and do occupy antibonding orbitals. The probability of finding an electron in an antibonding orbital is not 0. The key is the number of electrons in a given molecule – if there are more electrons than can be accommodated by the bonding orbitals, some electrons will be forced into antibonding orbitals.
03

c) Molecules containing electrons that occupy antibonding orbitals must be unstable.

The statement is partially true. While it is true that having electrons in antibonding orbitals can make a molecule less stable compared to a molecule with no electrons in antibonding orbitals, the presence of electrons in antibonding orbitals doesn't necessarily make the molecule unstable. A molecule's stability is affected by many factors, not only the occupancy of antibonding orbitals. In some cases, the overall bonding situation of a molecule might be stable despite the presence of electrons in antibonding orbitals.
04

d) Electrons cannot occupy a nonbonding orbital.

The statement is false. Nonbonding orbitals are orbitals that do not participate in bonding within a molecule, but they can still be occupied by electrons. These orbitals are generally at an energy level between bonding and antibonding orbitals, so they do not contribute to the overall bonding or antibonding properties of the molecule. However, their occupancy is still essential in determining the overall electron configuration and molecular properties. After analyzing all the statements, we find that: a) False. \(p\) orbitals can also make \(\pi\) or \(\pi^{*}\) molecular orbitals. b) False. Electrons can occupy antibonding orbitals, with a non-zero probability of being found there. c) Partially true, molecules with electrons in antibonding orbitals are not necessarily unstable, but they tend to be less stable compared to molecules without electrons in antibonding orbitals. d) False, electrons can occupy nonbonding orbitals.

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Most popular questions from this chapter

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (b) hydrogen cyanide, HCN; (c) sulphur trioxide, \(\mathrm{SO}_{3} ;\) (d) ozone, \(\mathrm{O}_{3}\).

Indicate the hybridization of the central atom in (a) $\mathrm{H}_{2} \mathrm{~S}$, (d) \(\mathrm{AlI}_{3}\). (b) \(\mathrm{SeF}_{6},(\mathbf{c}) \mathrm{P}(\mathrm{OH})_{3}\)

(a) The nitric oxide molecule, NO, readily loses one electron to form the NO \(^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in NO, NO^, and NO', and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the NO \(^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene $(148 \mathrm{pm})\( is a little shorter than the average \)\mathrm{C}-\mathrm{C}$ single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2}\) ? Explain your answer.

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