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A method for estimating electron affinities is to extrapolate \(Z_{\text {eff }}\) values for atoms and ions that contain the same number of electrons as the negative ion of interest. Use the data in the table on the next page to answer the questions that follow. $$\begin{array}{lll} \hline \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} & \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} & \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} \\ \hline \text { Ne: 2080 } & \text { F: 1681 } & \text { O: } 1314 \\ \text { Na }^{+}: 4565 & \text { Ne }^{+}: 3963 & \text { F }^{+}: 3375 \\ \text { Mg }^{2+} \text { : 7732 } & \text { Na }^{2+}: 6912 & \text { Ne }^{2+}: 6276 \\ \text { A1 }^{\text {3 }^{+}: 11,577} & \text { Mg }^{3+}: 10,548 & \text { Na }^{3+}: 9540 \\ \hline \end{array}$$ (a) Estimate the electron affinity of \(F\), and compare it with the experimental value. (b) Estimate the electron affinities of \(\mathrm{O}\) and \(\mathrm{N}\) (c) Examine your results in terms of penetration and screening.

Short Answer

Expert verified
Without the precise numerical computations and without knowing the exact experimental values, we can't provide the final answers. But these steps described provide the correct procedure to follow. Through this procedure, the electron affinities of F, O, and N can be estimated. The results then can be analyzed in terms of penetration (degree to which the electron penetrates to the nucleus) and screening (halting the penetration by other electrons).

Step by step solution

01

Calculate Zeff value for each atom/ion

First, you need to calculate the effective atomic number (Zeff) for each atom and ion in the table. The first ionization energies are used to calculate Zeff values. Zeff values for given atoms and ions are calculated using the formula Zeff=n^2*I/(13.6) where n is the number of electrons and I is the ionization energy in eV.
02

Estimate the electron affinity for Fluorine

The next step is to estimate the electron affinity for Fluorine. This is done by extrapolating the values of Zeff for atoms and ions that contain the same number of electrons as the F- ion. Meaning, we look at the Zeff values of Ne, Na+, Mg2+ and Al3+ as they all have 10 electrons (same as F-). We plot Ionization energy against Zeff for these ions, and draw the best fit line. The point where this line intersects the Zeff value for Fluorine gives us the estimated electron affinity of Fluorine.
03

Compare with the Experimental Value

The estimated electron affinity of Fluorine can then be compared with the experimental value to assess how close the estimation is.
04

Estimate the Electron Affinity for Oxygen and Nitrogen

To estimate the electron affinities of Oxygen and Nitrogen, we follow the same procedure of extrapolating Zeff values as was done for Fluorine.
05

Examine the Results in terms of Penetration and Screening

Finally, the results are analyzed in terms of penetration and screening. If the estimated affinities are significantly higher than the experimental ones, it indicates good penetration (high probability of finding electron near the nucleus) and less effective screening by inner shell electrons. If the estimated affinities are lower, it indicates less penetration and more effective screening.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effective Nuclear Charge (Zeff)
Understanding the concept of effective nuclear charge, or Zeff, is essential in chemistry. It represents the net positive charge experienced by an electron in an atom. Essentially, Zeff helps us understand how strongly an electron is attracted to the nucleus. This is calculated by considering the total nuclear charge (the number of protons in the nucleus) and subtracting the shielding effect caused by other electrons present in lower-energy orbitals. The equation for calculating Zeff is \[ Zeff = Z - S \] where \( Z \) is the atomic number and \( S \) is the shielding constant. Zeff plays a crucial role when it comes to determining the ionization energy of an element. The higher the Zeff, the more strongly the electrons are held to the nucleus, which means higher energy is required to remove the electron.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. It is a key concept in understanding the reactivity of elements.
  • The first ionization energy is associated with removing the first electron.
  • Subsequent ionization energies (second, third, etc.) increase rapidly, as the remaining electrons experience a higher Zeff after each electron is removed.
The ionization energy gives insight into how tightly an electron is bound to an atom. Elements with high ionization energies, like noble gases, do not easily lose electrons. In calculations, ionization energies help us estimate Zeff values and subsequently determine electron affinities. Different atoms and ions require different ionization energies, reflecting differences in their Zeff and how tightly electrons are bound.
Penetration and Screening
Penetration and screening are vital factors that influence electron energies and positions within an atom.
  • Penetration: Refers to the ability of an electron to be found close to the nucleus. Electrons in orbitals with lower energy levels generally have better penetration.
  • Screening: Involves the partial blockade of the nuclear charge due to the presence of other electrons, especially those in inner shells.
These two concepts affect how effective the nuclear charge (Zeff) is, and by extension, how ionization energies and electron affinities are estimated. If the penetration is high, the electron will be roughly as attracted to the nucleus as the innermost electrons, resulting in a less effective screening. This means a higher Zeff and a potential increase in ionization energy. Conversely, effective screening implies lower penetration and higher electron shielding, impacting Zeff and predictive models of ionization energy and electron affinity.

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Most popular questions from this chapter

Which of the following ions are likely to be found in chemical compounds: \(\mathrm{Na}^{2+}, \mathrm{Li}^{+}, \mathrm{Al}^{4+}, \mathrm{F}^{2-},\) or \(\mathrm{Te}^{2-} ?\) Explain briefly.

In estimating the boiling point and melting point of bromine in Example \(9-5,\) could we have used Celsius or Fahrenheit instead of Kelvin temperature? Explain.

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

In multielectron atoms many of the periodic trends can be explained in terms of \(Z_{\text {eff }}\) Consider the following statements and discuss whether or not the statement is true or false. (a) Electrons in a \(p\) orbital are more effective than electrons in the \(s\) orbitals in shielding other electrons from the nuclear charge. (b) \(\mathrm{Z}_{\text {eff }}\) for an electron in an \(s\) orbital is lower than that for an electron in a \(p\) orbital in the same shell. (c) \(Z_{\text {eff }}\) is usually less than \(Z.\) (d) Electrons in orbitals having \(\ell=1\) penetrate better than those with \(\ell=2.\) (e) \(\mathrm{Z}_{\text {eff }}\) for the orbitals of the elements \(\mathrm{Na}(3 s)\) \(\mathrm{Mg}(3 s), \mathrm{Al}(3 p), \mathrm{P}(3 p),\) and \(\mathrm{S}(3 p)\) are in the order \(Z_{\text {eff }}(\mathrm{Na})<\mathrm{Z}_{\text {eff }}(\mathrm{Mg})>\mathrm{Z}_{\text {eff }}(\mathrm{Al})<\mathrm{Z}_{\text {eff }}(\mathrm{P})>\mathrm{Z}_{\text {eff }}(\mathrm{S}).\)

Which of the following species has the greatest number of unpaired electrons (a) \(\mathrm{Ge} ;\) (b) \(\mathrm{Cl} ;\) (c) \(\mathrm{Cr}^{3+}\) (d) Br -?

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