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Chapter 10: Problem 15

Explain why an atom cannot have a permanent dipole moment.

Short Answer

Expert verified
An atom cannot have a permanent dipole moment due to the symmetric distribution of its electrons around the nucleus and their continuous motion which prevents permanent separation of charges.

Step by step solution

01

Understand the structure of an atom

An atom is composed of a central nucleus, which contains protons and neutrons, and electrons that revolve around the nucleus in different energy levels or shells. The protons in the nucleus have a positive charge, while the electrons have a negative charge. Since protons and electrons are equal in quantity, the charges cancel each other out, resulting in a neutral atom.
02

Analyze electron distribution

Electrons within the atom are distributed evenly around the nucleus and are constantly in motion. As such, any temporary imbalance in the charges due to the position of an electron will be neutralized as the electron moves.
03

Understand the lack of permanent dipole moment

For an atom to have a permanent dipole moment, there would need to be a permanent separation of charge within the atom. However, as discussed, the electrons are not stationary and their continuous motion around the nucleus means that any momentary separation of charge cannot persist, maintaining the overall neutrality of the atom. Thus, it is impossible for an atom to have a permanent dipole moment.

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Most popular questions from this chapter

List these molecules in order of increasing dipole moment: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CBr}_{4}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{HF}, \mathrm{NH}_{3}, \mathrm{CO}_{2}\)

Use molecular orbital theory to explain why the \(\mathrm{Be}_{2}\) molecule does not exist

Consider the reaction $$ \mathrm{BF}_{3}+\mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B}-\mathrm{NH}_{3} $$ Describe the changes in hybridization (if any) of the \(B\) and \(N\) atoms as a result of this reaction

Explain why the bond order of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+},\) but the bond order of \(\mathrm{O}_{2}\) is less than that of \(\mathrm{O}_{2}^{+}\)

Discuss the basic features of the VSEPR model. Explain why the magnitude of repulsion decreases in this order: lone pair-lone pair \(>\) lone pair-bonding pair \(>\) bonding pair-bonding pair.
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