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Chapter 9: Problem 14

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?

Short Answer

Expert verified
In summary, single bonds can rotate about the internuclear axis without breaking the bond due to their symmetric electron distribution and the presence of only sigma bonds. On the other hand, double and triple bonds have more complex electron distributions involving pi bonds, which do not allow for free rotation about the internuclear axis without breaking the bond.

Step by step solution

01

Understanding single, double, and triple bonds

In a covalent bond, atoms share electrons to gain stability. A single bond forms when two atoms share one pair of electrons, a double bond forms when two atoms share two pairs of electrons, and a triple bond forms when two atoms share three pairs of electrons. The electron distribution in single, double, and triple bonds plays a crucial role in their flexibility.
02

Electron distribution in single bond

In a single bond, electrons are shared in the form of a sigma bond, which is formed by the direct overlap of atomic orbitals. The electron density is symmetric along the internuclear axis, and this allows the atoms to rotate around the bond without much resistance or change in energy. This is why a single bond can rotate freely about the internuclear axis.
03

Electron distribution in double and triple bonds

In double and triple bonds, the electron distribution is more complex. In addition to the sigma bond present in single bonds, double and triple bonds also involve one or more pi bonds. A pi bond is formed by the parallel overlap of atomic orbitals. The electron cloud is distributed above and below the internuclear axis, causing more electron density and a higher energy state.
04

Double and triple bonds rotation restriction

Unlike single bonds, the electron distribution in double and triple bonds does not allow for rotation about the internuclear axis. The pi bonds rely on the atomic orbitals maintaining fixed orientations relative to each other. If one atom in the bond starts to rotate, the overlap of orbitals forming the pi bond would be disrupted, leading to a rise in energy and the possible breaking of the bond. Hence, atoms in double and triple bonds cannot rotate about the internuclear axis without breaking the bond.
05

Conclusion

In summary, single bonds can rotate about the internuclear axis without breaking the bond due to their symmetric electron distribution and the presence of only sigma bonds. On the other hand, double and triple bonds have more complex electron distributions involving pi bonds, which do not allow for free rotation about the internuclear axis without breaking the bond.

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Most popular questions from this chapter

The molecules \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) are isoelectronic but their properties are quite different. Although as a first approximation we often use the same MO diagram for both, suggest how the \(\mathrm{MOs}\) in \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) might be different.

Use the localized electron model to describe the bonding in \(\mathrm{CCl}_{4}\) .

Sodium can react with oxygen to form sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) which is composed of \(\mathrm{Na}^{+}\) and \(\mathrm{O}_{2} 2-\) ions. Potassium can react with oxygen to form potassium superoxide \(\left(\mathrm{KO}_{2}\right)\) which is composed of \(\mathrm{K}^{+}\) and \(\mathrm{O}_{2}-\) ions. Does the peroxide ion or the superoxide ion have the shorter bond length? Explain.

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CO} \quad \text { b. } \mathrm{CO}^{+} \quad \text { c. } \mathrm{CO}^{2+} $$
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