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Chapter 2: Problem 126

Which has the more negative electron affinity, the oxygen atom or the \(\mathrm{O}^{-}\) ion? Explain your answer.

Short Answer

Expert verified
The neutral oxygen atom (O) has a more negative electron affinity compared to the O⁻ ion. This is because the electrostatic repulsion experienced by an incoming electron in the O⁻ ion is greater due to the presence of an additional electron, leading to a less negative electron affinity in the O⁻ ion than in the neutral oxygen atom.

Step by step solution

01

Recall the electron configurations

The first step is to recall the electron configurations of an oxygen atom and an O⁻ ion. An oxygen atom has 8 electrons, with the electron configuration of 1s²2s²2p⁴. When an electron is added to an oxygen atom, it forms an O⁻ ion with 9 electrons and the electron configuration becomes 1s²2s²2p⁵.
02

Compare the electron affinities for an oxygen atom and an O⁻ ion

When adding an electron to an oxygen atom, the electron experiences electrostatic attractions from the positive nucleus and electrostatic repulsions from the other electrons in the atom. In the case of an O⁻ ion, there is an additional repulsion force between the incoming electron and the excess electron already in the ion. Since the O⁻ ion has one more electron than the neutral oxygen atom, the electrostatic repulsion experienced by the incoming electron in the O⁻ ion will be greater compared to that in the neutral oxygen atom. As a result, adding an electron to the O⁻ ion will require more energy (less negative electron affinity) than adding an electron to the neutral oxygen atom.
03

Identify which has the more negative electron affinity

Based on the comparison made in Step 2, we can conclude that the neutral oxygen atom (O) has a more negative electron affinity compared to the O⁻ ion. The reason is that the neutral oxygen atom experiences less electrostatic repulsion between the incoming electron and its existing electrons than the O⁻ ion.

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Most popular questions from this chapter

For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$ \psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta} $$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{m}\right)\) Calculate the position of the nodes for this wave function.

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