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Chapter 9: Problem 19

All the isoelectronic species illustrated in the text had the electron configurations of noble gases. Can two ions be isoelectronic without having noble-gas electron configurations? Explain.

Short Answer

Expert verified
Yes, two ions can be isoelectronic without having noble-gas electron configurations. This is possible when two ions have the same total number of electrons but these electrons do not occupy a completely filled electron shell configuration, as is the case with noble gasses.

Step by step solution

01

Understanding Isoelectronic Terms

Isoelectronic refers to different atoms, ions, or molecules that have the same total number of electrons. The challenge here is to think about whether these atoms or ions must necessarily have a noble-gas electron configuration.
02

Analyzing the Electron Configurations

The electron configurations of noble gases are characterized by completely filled electron shells. An atom becomes an ion by losing or gaining electrons to achieve a 'stable' configuration, usually similar to the nearest noble gas. However, this process is not limited to imitating noble gas configurations.
03

Conclusion

So, two ions can indeed be isoelectronic without having noble gas electron configurations. This becomes possible when two ions have the same total number of electrons, but those electrons do not fully fill their respective shells (which would be the case for a noble gas configuration). In other words, they may both lack electrons to become truly stable (resembling a noble-gas configuration), but they could still have the same total number of electrons.

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Most popular questions from this chapter

Give the symbol of the element (a) in group 14 that has the smallest atoms; (b) in period 5 that has the largest atoms; (c) in group 17 that has the lowest first ionization energy.

Which of the following species has the greatest number of unpaired electrons (a) \(\mathrm{Ge} ;\) (b) \(\mathrm{Cl} ;\) (c) \(\mathrm{Cr}^{3+}\) (d) Br -?

Write electron configurations to show the first two ionizations for Cs. Explain why the second ionization energy is much greater than the first.

Plot a graph of the square roots of the ionization energies versus the nuclear charge for the two series \(\mathrm{Li}, \mathrm{Be}^{+}, \mathrm{B}^{2+}, \mathrm{C}^{3+},\) and \(\mathrm{Na}, \mathrm{Mg}^{+}, \mathrm{Al}^{2+}, \mathrm{Si}^{3+} .\) Explain the observed relationship with the aid of Bohr's expression for the binding energy of an electron in a one electron atom.

Mendeleev's periodic table did not preclude the possibility of a new group of elements that would fit within the existing table, as was the case with the noble gases. Moseley's work did preclude this possibility. Explain this difference.
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