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Chapter 7: Problem 115

The electron affinity for sulfur is more exothermic than that for oxygen. How do you account for this?

Short Answer

Expert verified
Sulfur's electron affinity is more exothermic than that of oxygen because its larger atomic size and greater electron shielding effect result in a weaker attraction between its valence electrons and nucleus. This makes it relatively easier for sulfur to accommodate an additional electron, releasing more energy compared to oxygen.

Step by step solution

01

Understanding Electron Affinity

Electron affinity is the amount of energy released or spent when an electron is added to a neutral atom in the gaseous state to form a negative ion. So, a more exothermic reaction means a larger release of energy when an electron is added to the atom. Therefore, sulfur's more exothermic electron affinity means that it releases more energy when an electron is added to it than oxygen does.
02

Comparing Oxygen and Sulfur in the Periodic Table

Oxygen and sulfur belong to the same group (Group 16) in the periodic table, which means they share some properties, such as the number of valence electrons (6 in this case). Oxygen is in period 2 and has an atomic number of 8, while sulfur is in period 3 and has an atomic number of 16.
03

Atomic Size and its Importance

When moving down a group in the periodic table, the atomic size increases due to an increase in the number of energy levels. Sulfur, being in period 3, is larger in size than oxygen, which is in period 2. The larger atomic size of sulfur results in its valence electrons experiencing a weaker attraction to the nucleus compared to the valence electrons of smaller oxygen atom.
04

Electron Shielding Effect

Electron shielding is the phenomenon where inner electrons shield outer electrons from the attractive force of the nucleus. In the case of oxygen and sulfur, sulfur has more inner electrons (10) than oxygen (2), thus exhibiting a greater shielding effect for its valence electrons.
05

Explaining Sulfur's More Exothermic Electron Affinity

By considering electron shielding and atomic size, we can account for sulfur's more exothermic electron affinity. The larger atomic size and more significant shielding effect of sulfur translate into a weaker attraction between its valence electrons and nucleus. This makes it relatively easier for sulfur to accommodate an additional electron in its outer energy level, releasing more energy in the process compared to oxygen. Therefore, sulfur's electron affinity is more exothermic than that of oxygen.

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Most popular questions from this chapter

Octyl methoxycinnamate and oxybenzone are common ingredients in sunscreen applications. These compounds work by absorbing ultraviolet (UV) B light (wavelength \(280-320 \mathrm{~nm}\) ), the UV light most associated with sunburn symptoms. What frequency range of light do these compounds absorb?

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).

In each of the following sets, which atom or ion has the smallest ionization energy? a. \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) b. \(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) c. \(\mathrm{N}, \mathrm{O}, \mathrm{F}\) d. \(\mathrm{S}^{2-}, \mathrm{S}, \mathrm{S}^{2+}\) e. \(\mathrm{Cs}\), Ge, Ar

In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?

For each of the following pairs of elements \((\mathrm{C}\) and \(\mathrm{N}) \quad(\mathrm{Ar}\) and \(\mathrm{Br})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.
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