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Chapter 7: Problem 53

Although the electron affinity of bromine is a negative quantity, it is positive for \(\mathrm{Kr}\). Use the electron configurations of the two elements to explain the difference.

Short Answer

Expert verified
In conclusion, the difference in electron affinity between bromine and krypton is due to their electron configurations. Bromine has a partially filled 4p subshell, making it favorable to accept an additional electron and release energy, resulting in a negative electron affinity. Krypton has a completely filled 4p subshell, so adding an electron requires placing it in a higher energy level (5s subshell), necessitating energy input and leading to a positive electron affinity.

Step by step solution

01

Understand Electron Affinity

Electron affinity is the amount of energy released or absorbed when an electron is added to a neutral atom, forming a negative ion. In general, nonmetals have higher (more negative) electron affinity than metals. A negative value for electron affinity means that energy is released when an electron is added to an atom, making it more favorable. A positive value means that energy needs to be added to make the process possible, which is less favorable.
02

Write the Electron Configurations of Bromine and Krypton

Bromine (Br) has an atomic number of 35, and Krypton (Kr) has an atomic number of 36. We can write their electron configurations as follows: Bromine (Br): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\) Krypton (Kr): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\)
03

Analyze Energy Released or Absorbed in the Process

For bromine, adding an electron means placing it in the 4p subshell. Since the 4p subshell in Br is not completely filled, the electron can be added to the existing 4p orbital, and there is a release in energy, thus making its electron affinity negative. Krypton, on the other hand, has a completely filled 4p subshell. Adding an electron to Kr would mean adding it to a new 5s subshell instead of the existing 4p subshell. This process requires energy, as the electron-electron repulsions in the 5s orbital are greater than the attractive forces between the nucleus and the added electron. Therefore, the electron affinity of Kr is positive.
04

Conclusion

In conclusion, the difference in electron affinity between bromine and krypton can be explained by their electron configurations. Bromine has a partially filled 4p subshell, and it can accept an additional electron, thereby releasing energy. Krypton, however, has a completely filled 4p subshell, and adding an electron would require placing the electron in a higher energy level, which requires energy input, giving it a positive electron affinity.

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Most popular questions from this chapter

Arrange the following oxides in order of increasing acidity: \(\mathrm{CO}_{2}, \mathrm{CaO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}, \mathrm{SiO}_{2},\) and \(\mathrm{P}_{2} \mathrm{O}_{5}\)

We can draw an analogy between the attraction of an electron to a nucleus and seeing a lightbulb -in essence, the more \(n u=\) clear charge the electron "sees," the greater the attraction. (a) Within this analogy, discuss how the screening by core electrons is analogous to putting a frosted-glass lampshade between the lightbulb and your eyes, as shown in the illustration. (b) Explain how we could mimic moving to the right in a row of the periodic table by changing the wattage of the lightbulb. (c) How would you change the wattage of the bulb and/or the frosted glass to mimic the effect of moving down a column of the periodic table? [Section 7.2]

Find three atoms in the periodic table whose ions have an electron configuration of \(n d^{6}(n=3,4,5 \ldots) .\)

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

For each of the following pairs, indicate which element has the smaller first ionization energy: (a) \(\mathrm{Ti}, \mathrm{Ba} ;(\mathrm{b}) \mathrm{Ag}, \mathrm{Cu} ;(\mathrm{c}) \mathrm{Ge}\) \(\mathrm{Cl} ;\) (d) \(\mathrm{Pb},\) Sb. (In each case use electron configuration and effective nuclear charge to explain your answer.)
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