Chapter 9: Problem 92
Use the electron configuration of oxygen to explain why it tends to form a 2 - ion.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 9: Problem 92
Use the electron configuration of oxygen to explain why it tends to form a 2 - ion.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
You learned in this chapter that ionization generally increases as you move from left to right across the periodic table. However, consider the following data, which shows the ionization energies of the period 2 and 3 elements: $$ \begin{array}{ccccc} \text { Group } & \begin{array}{c} \text { Period 2 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { lonization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} & \begin{array}{c} \text { Period 3 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { Ionization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} \\ \text { 1A } & \text { Li } & 520 & \text { Na } & 496 \\ \text { 2A } & \text { Be } & 899 & \text { Mg } & 738 \\ \text { 3A } & \text { B } & 801 & \text { Al } & 578 \\ \text { 4A } & \text { C } & 1086 & \text { Si } & 786 \\ \text { 5A } & \text { N } & 1402 & \text { P } & 1012 \\ \text { 6A } & \text { 0 } & 1314 & \text { S } & 1000 \\ \text { 7A } & \text { F } & 1681 & \text { Cl } & 1251 \\ \text { 8A } & \text { Ne } & 2081 & \text { Ar } & 1521 \\ \hline \end{array} $$ Notice that the increase is not uniform. In fact, ionization energy actually decreases a bit in going from elements in group \(2 \mathrm{~A}\) to \(3 \mathrm{~A}\) and then again from \(5 \mathrm{~A}\) to \(6 \mathrm{~A}\). Use what you know about electron configurations to explain why these dips in ionization energy exist.
The particle nature of light was first proposed by Albert Einstein, who suggested that light could be described as a stream of particles called photons. A photon of wavelength \(\lambda\) has an energy (E) given by the equation: \(E=h c / \lambda\), where \(E\) is the energy of the photon in \(\mathrm{J}, h\) is Planck's constant \(\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\), and \(c\) is the speed of light \(\left(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\). Calculate the energy of \(1 \mathrm{~mol}\) of photons with a wavelength of \(632 \mathrm{~nm}\).
Write orbital diagrams for the valence electrons and indicate the number of unpaired electrons for each element. (a) \(\mathrm{Br}\) (b) \(\mathrm{Kr}\) (c) \(\mathrm{Na}\) (d) In
Explain why Group 1 elements tend to form \(1+\) ions and Group 7 elements tend to form 1- ions.
Which type of electromagnetic radiation has the shortest wavelength? (a) radio waves (b) microwaves (c) infrared (d) ultraviolet
What do you think about this solution?
We value your feedback to improve our textbook solutions.