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Chapter 7: Problem 17

Which will experience the greater effect nuclear charge, the electrons in the \(n=2\) shell in \(\mathrm{F}\) or the \(n=2\) shell in \(\mathrm{B}\) ? Which will be closer to the nucleus?

Short Answer

Expert verified
The electrons in the n=2 shell of Boron (B) will experience a greater effective nuclear charge (3.45) than those in the n=2 shell of Fluorine (F) (3.2), and they will be closer to the nucleus.

Step by step solution

01

Determine the Atomic Number

First, let's find the atomic number (often denoted as Z) of Fluorine (F) and Boron (B). You can find atomic numbers in the periodic table. For Fluorine (F), the atomic number is 9. For Boron (B), the atomic number is 5.
02

Calculate the Effective Nuclear Charge using Slater's Rules

The effective nuclear charge (Z_eff) can be calculated using Slater's Rules. The general formula for Z_eff is: Z_eff = Z - S where Z is the atomic number of the element and S is the shielding constant. For the n=2 shell, we can calculate S with the following formula: S = (0.35 * number of n=1 electrons) + (0.85 * number of n=2 electrons, not counting the electron of interest) For Fluorine: S_F = (0.35 * 2) + (0.85 * (7-1)) = 0.7 + 5.1 = 5.8 Z_eff_F = Z_F - S_F = 9 - 5.8 = 3.2 For Boron: S_B = (0.35 * 2) + (0.85 * (2-1)) = 0.7 + 0.85 = 1.55 Z_eff_B= Z_B - S_B = 5 - 1.55 = 3.45
03

Compare the Effective Nuclear Charge

Now we can see that the effective nuclear charge of Fluorine's n=2 shell electrons is 3.2, and Boron's n=2 shell electrons is 3.45. So, the electrons in Boron's n=2 shell will experience a greater effective nuclear charge.
04

Determine Which Electrons are Closer to the Nucleus

As effective nuclear charge increases, the electrons in the shell experience a stronger attraction to the nucleus, which results in the electrons being pulled closer to the nucleus. Since Boron's n=2 shell electrons have a higher effective nuclear charge than Fluorine's n=2 shell electrons (3.45 > 3.2), the n=2 shell electrons in Boron (B) are closer to the nucleus than those in Fluorine (F).

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Most popular questions from this chapter

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, \(\mathrm{F}\) ?

Mercury in the environment can exist in oxidation states \(0,\) \(+1,\) and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy $\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{~J}\right)$ The oxygen on the mineral surface gave emitted electron energies at $531 \mathrm{eV}\(, corresponding to the \)1 s$ orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for O. (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case?

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