Chapter 7: Problem 124
Predict the atomic number of the next alkali metal after francium and give its ground-state electron configuration.
/*! 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 7: Problem 124
Predict the atomic number of the next alkali metal after francium and give its ground-state electron configuration.
All the tools & learning materials you need for study success - in one app.
Get started for free
Draw atomic orbital diagrams representing the ground-state electron configuration for each of the following elements. a. \(\mathrm{Na}\) b. \(\mathrm{Co}\) c. \(\mathrm{Kr}\) How many unpaired electrons are present in each element?
Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ $$ \begin{array}{ll} \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\). c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)
For each of the following pairs of elements \((\mathrm{Mg}\) and \(\mathrm{K}) \quad(\mathrm{F}\) and \(\mathrm{Cl})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.
An ionic compound of potassium and oxygen has the empirical formula KO. Would you expect this compound to be potassium(II) oxide or potassium peroxide? Explain.
One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?
What do you think about this solution?
We value your feedback to improve our textbook solutions.