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Chapter 8: Problem 32

Group the species that are isoelectronic: \(\mathrm{Be}^{2}, \mathrm{~F}\) \(\mathrm{Fe}^{2+}, \mathrm{N}^{3-}, \mathrm{He}, \mathrm{S}^{2-}, \mathrm{Co}^{3+}, \mathrm{Ar}\).

Short Answer

Expert verified
The isoelectronic groups are: \[\{\mathrm{Be}^{2+}, \mathrm{He}\}, \{\mathrm{F}, \mathrm{N}^{3-}\}, \{\mathrm{Fe}^{2+}, \mathrm{Co}^{3+}\}, \{\mathrm{S}^{2-}, \mathrm{Ar}\}\]

Step by step solution

01

Determine the total number of electrons for each species

First up, identify the atomic numbers for each atom: Be is 4, F is 9, Fe is 26, N is 7, He is 2, S is 16, Co is 27, and Ar is 18. By considering the charges, calculate the total number of electrons for each ion or atom. \(\mathrm{Be}^{2+}\) has 2 electrons, \(\mathrm{F}\) has 9 electrons, \(\mathrm{Fe}^{2+}\) has 24 electrons, \(\mathrm{N}^{3-}\) has 10 electrons, \(\mathrm{He}\) has 2 electrons, \(\mathrm{S}^{2-}\) has 18 electrons, \(\mathrm{Co}^{3+}\) has 24 electrons, and \(\mathrm{Ar}\) also has 18 electrons.
02

Group the isoelectronic species

After the computation, we can group the species with the same total number of electrons. This way, \(\mathrm{Be}^{2+}\) and \(\mathrm{He}\) form one group (2 electrons), \(\mathrm{F}\) and \(\mathrm{N}^{3-}\) form another group (9 and 10 electrons), and \(\mathrm{Fe}^{2+}\) and \(\mathrm{Co}^{3+}\) form the third group (24 electrons). Lastly, \(\mathrm{S}^{2-}\) and \(\mathrm{Ar}\) form the final group (18 electrons). The isoelectronic groups are therefore: \[\{\mathrm{Be}^{2+}, \mathrm{He}\}, \{\mathrm{F}, \mathrm{N}^{3-}\}, \{\mathrm{Fe}^{2+}, \mathrm{Co}^{3+}\}, \{\mathrm{S}^{2-}, \mathrm{Ar}\}\]

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Most popular questions from this chapter

A metal ion with a net +3 charge has five electrons in the \(3 \mathrm{~d}\) subshell. Identify the metal.

A technique called photo electron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with ultraviolet (UV) light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write \( h v=I E+\frac{1}{2} m u^{2} \) in which \(\nu\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?

The \(\mathrm{H}^{-}\) ion and the He atom have two \(1 \mathrm{~s}\) electrons each. Which of the two species is larger? Explain.

Write the outer electron contigurations for (a) the alkali metals, (b) the alkaline earth metals, (c) the halogens, (d) the noble gases.

Without referring to a periodic table, write the electron configurations of elements with these atomic numbers: (a) \(9,\) (b) \(20,\) (c) \(26,\) (d) 33. Classify the elements.
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