What is the relationship between the ionization energy of an anion with a 1 - charge such as \(\mathrm{F}^{-}\) and the electron affinity of the neutral atom, \(\mathrm{F}\) ?

Short Answer

Expert verified
The relationship between the ionization energy of a F- anion and the electron affinity of a neutral F atom is given by: \[ IE_1 (\mathrm{F}^{-}) = -EA(\mathrm{F}) \] This means that the ionization energy of the fluoride anion is equal to the negative value of the electron affinity of the neutral fluorine atom.

Step by step solution

01

Understand Ionization Energy

Ionization energy is the minimum amount of energy required to remove an electron from a gaseous atom or ion. The greater the ionization energy, the more difficult it is to remove an electron from the atom or ion. Generally, ionization energies increase across a period and decrease down a group in the periodic table. For the fluoride anion (F-), we can talk about the first ionization energy (IE₁), which corresponds to the energy required to remove the first electron: \[ \mathrm{F}^{-} \rightarrow \mathrm{F} + e^{-} \] Here, IE₁ will be positive, as energy is required to remove the electron.
02

Understand Electron Affinity

Electron affinity is the energy change associated with the addition of an electron to a gaseous atom. It represents the attraction between the incoming electron and the nucleus of the atom. Generally, electron affinities become more negative across a period and less negative down a group in the periodic table. For the neutral fluorine atom (F), we can talk about the electron affinity (EA) corresponding to the energy change when the first electron is added: \[ \mathrm{F} + e^{-} \rightarrow \mathrm{F}^{-} \] Here, EA will be negative, as energy is released when the electron is added to the atom.
03

Compare Ionization Energy and Electron Affinity

Now that we understand both ionization energy and electron affinity, we can compare the two processes for the given examples of fluoride anion (F-) and neutral fluorine atom (F). When an electron is removed from the fluoride anion (F-), it results in a neutral fluorine atom (F) and a free electron: \[ \mathrm{F}^{-} \rightarrow \mathrm{F} + e^{-} \quad (IE_1, \, energy \, required) \] When an electron is added to the neutral fluorine atom (F), it results in a fluoride anion (F-): \[ \mathrm{F} + e^{-} \rightarrow \mathrm{F}^{-} \quad (EA, \, energy \, released) \]
04

Determine the Relationship

From the comparison we made above, we can observe that the ionization energy (IE₁) and the electron affinity (EA) represent opposite processes involving the fluoride anion (F-) and the neutral fluorine atom (F). Moreover, the ionization energy corresponds to the energy required to remove an electron, whereas electron affinity corresponds to the energy released upon the addition of an electron. Therefore, the relationship between the ionization energy of a F- anion and the electron affinity of a neutral F atom can be expressed as: \[ IE_1 (\mathrm{F}^{-}) = -EA(\mathrm{F}) \] In other words, the ionization energy of the fluoride anion is equal to the negative value of the electron affinity of the neutral fluorine atom.

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