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Chapter 8: Problem 30

Name the ions with +3 charges that have the following electron configurations: (a) \([\mathrm{Ar}] 3 d^{3}\), (b) [Ar], (c) \([\mathrm{Kr}] 4 d^{6}\), (d) \([\mathrm{Xe}] 4 f^{14} 5 d^{6}\).

Short Answer

Expert verified
The ions with +3 charges that correspond to the given electron configurations are \(Cr^{+3}\), \(Ar^{+3}\), \(Ru^{+3}\), and \(Fm^{+3}\).

Step by step solution

01

Identify Element for [Ar] 3d^3

From the electron configuration, \([Ar] 3d^3\), we can tell that it corresponds to the element with atomic number 24, which is Chromium (Cr) adding 3 electrons. But in this case, we are dealing with a \(+3\) ion so we would need to take three electrons away from Chromium. This means the ion is \(Cr^{+3}\).
02

Identify Element for [Ar]

In this case, the electron configuration corresponds to the noble gas Argon, which has an atomic number of 18. If we remove three electrons, we get the \(Ar^{+3}\) ion.
03

Identify Element for [Kr] 4d^6

The electron configuration \([Kr] 4d^6\) corresponds to the 44th element, Ruthenium (Ru). For a \(+3\) ion charge, we have to subtract three electrons from Ruthenium. This means the ion is \(Ru^{+3}\).
04

Identify Element for [Xe] 4f^14 5d^6

The electron configuration \([Xe] 4f^{14} 5d^6\) corresponds to the 100th element, Fermium (Fm). If you remove three electrons to obtain a \(+3\) ion, the ion is \(Fm^{+3}\).

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Most popular questions from this chapter

The air in a manned spacecraft or submarine needs to be purified of exhaled carbon dioxide. Write equations for the reactions between carbon dioxide and (a) lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right),\) (b) sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) and \((\mathrm{c})\) potassium superoxide \(\left(\mathrm{KO}_{2}\right)\)

What factors account for the unique nature of hydrogen?

Some chemists think that helium should properly be called "helon." Why? What does the ending in helium (-ium) suggest?

In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group 7A. Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F}\) ions by electrolytic means. Then, in \(1986,\) a chemist reported that by reacting potassium hexafluoromanganate(IV) \(\left(\mathrm{K}_{2} \mathrm{MnF}_{6}\right)\) with antimony pentafluoride \(\left(\mathrm{SbF}_{5}\right)\) at \(150^{\circ} \mathrm{C},\) he had generated elemental fluorine. Balance the following equation representing the reaction: $$ \mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2} $$

How do the chemical properties of oxides change from left to right across a period? From top to bottom within a particular group?
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