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Chapter 8: Problem 65

Use the alkali metals and alkaline earth metals as examples to show how we can predict the chemical properties of elements simply from their electron configurations.

Short Answer

Expert verified
Chemical properties of elements can be predicted by their electron configurations. Alkali metals, having 1 valence electron, are highly reactive and tend to form +1 cations. Alkaline earth metals, with 2 valence electrons, are less reactive but form +2 cations. The same concept could be used to predict the chemical properties of other elements.

Step by step solution

01

Understand Electron Configuration

Electron configuration refers to the arrangement of electrons in the atomic orbitals. Alkali metals, found in Group 1 of the periodic table, have an electron configuration of [noble gas]ns1, where 'noble gas' represents the electron configuration of the closest preceding noble gas and 'ns1' shows that alkali metals have 1 electron in their outermost (valence) shell. On the other hand, Alkaline earth metals, in Group 2, possess the electron configuration [noble gas]ns2, meaning they have 2 electrons in their valence shell.
02

Relate Electron Configuration to Chemical Properties

The chemical properties of an element are predominantly determined by the electrons in the valence shell. Elements with similar electron configurations exhibit similar chemical properties. Alkali metals are highly reactive due to the single electron in their outer shell, which is readily given up to form a positive ion (cation). Similarly, alkaline earth metals tend to lose both electrons in their valence shell, forming a +2 cation. They are less reactive than alkali metals though, as the removal of two electrons requires more energy.
03

Predict Chemical Properties

Using alkaline metals, we can predict that elements with a single valence electron (ns1 configuration) will be highly reactive and form +1 cations upon losing that electron. From the alkaline earth metals, we infer that elements with two valence electrons (ns2 configuration) will be less reactive than those in the first group and form +2 cations upon ionization. This principle can be applied to predict the chemical properties of other elements based on their electron configurations.

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Most popular questions from this chapter

In each of the following pairs, indicate which one of the two species is smaller: (a) \(\mathrm{Cl}\) or \(\mathrm{Cl}^{-},\) (b) Na or \(\mathrm{Na}^{+}\), (c) \(\mathrm{O}^{2-}\) or \(\mathrm{S}^{2-},\) (d) \(\mathrm{Mg}^{2+}\) or \(\mathrm{Al}^{3+},\) (e) \(\mathrm{Au}^{+}\) or \(\mathrm{Au}^{3+}\).

Draw a rough sketch of a periodic table (no details are required). Indicate where metals, nonmetals, and metalloids are located.

For each pair of elements listed here, give three properties that show their chemical similarity: (a) sodium and potassium and (b) chlorine and bromine.

Which elements are more likely to form acidic oxides? basic oxides? amphoteric oxides?

The ionization energies of sodium (in \(\mathrm{kJ} / \mathrm{mol}\) ), starting with the first and ending with the eleventh, are 495.9,4560,6900,9540,13,400,16,600,20,120 \(25,490,28,930,141,360,170,000 .\) Plot the log of ionization energy \((y\) axis \()\) versus the number of ionization \((x\) axis \() ;\) for example, \(\log 495.9\) is plotted versus 1 (labeled \(I_{1}\), the first ionization energy), log 4560 is plotted versus 2 (labeled \(I_{2}\), the second ionization energy), and so on. (a) Label \(I_{1}\) through \(I_{11}\) with the electrons in orbitals such as \(1 s, 2 s, 2 p,\) and \(3 s .\) (b) What can you deduce about electron shells from the breaks in the curve?
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