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Chapter 23: Problem 1

By means of orbital diagrams, write electron configurations for the following transition element atom and ions: \((a) \mathrm{Ti} ;(\mathbf{b}) \mathrm{V}^{3+} ;(\mathrm{c}) \mathrm{Cr}^{2+} ;(\mathrm{d}) \mathrm{Mn}^{4+} ;(\mathrm{e}) \mathrm{Mn}^{2+} ;(\mathrm{f}) \mathrm{Fe}^{3+}\).

Short Answer

Expert verified
The electron configurations using orbital diagrams for the given atoms and ions are: (a) Ti: [Ar] 4s^2 3d^2 (b) V3+: [Ar] 3d^2 (c) Cr2+: [Ar] 3d^4 (d) Mn4+: [Ar] 3d^3 (e) Mn2+: [Ar] 3d^5 (f) Fe3+: [Ar] 3d^3.

Step by step solution

01

Titanium (Ti)

The atomic number of Titanium is 22, so it has 22 electrons in the ground state. The electron configuration is: \( [Ar] 4s^2 3d^2 \)
02

Vanadium Ion (V3+)

The atomic number of Vanadium is 23. As it's V3+, this means 3 electrons are removed. First, the 2 electrons from the 4s orbital will be lost, and then 1 from the 3d orbital. Thus, the electron configuration becomes: \( [Ar] 3d^2 \)
03

Chromium Ion (Cr2+)

The atomic number of Chromium is 24, but Chromium is a special case where one electron from 4s orbital moves to 3d orbital to attain half-full stability. But, here we have Cr2+ so 2 electrons are removed from the 4s orbital (since ionization starts from the outermost shell). Thus, the configuration changes to: \( [Ar] 3d^4 \)
04

Manganese Ion (Mn4+)

The atomic number of Manganese is 25. As we have Mn4+, it means 4 electrons are removed. First, 2 electrons are removed from the 4s orbital. After that, the next 2 are removed from the 3d orbital. Hence, the configuration becomes: \( [Ar] 3d^3 \)
05

Manganese Ion (Mn2+)

As previously discussed for Mn4+, here we have Mn2+ so, it means 2 electrons are removed. Firstly, the 2 electrons removed will be from 4s orbital. Thus, the configuration converts to: \( [Ar] 3d^5 \)
06

Iron Ion (Fe3+)

The atomic number of Iron is 26. Now, for Fe3+, 3 electrons are removed. First the electrons from the 4s orbital are lost, followed by one from the 3d. So, the electron configuration is: \( [Ar] 3d^3 \)

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Most popular questions from this chapter

Equation \((23.18),\) which represents the chromatedichromate equilibrium, is actually the sum of two equilibrium expressions. The first is an acid-base reaction, \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\). The second reaction involves elimination of a water molecule between two \(\mathrm{HCrO}_{4}^{-}\) ions (a dehydration reaction), \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} .\) If the ionization constant, \(K_{\mathrm{a}},\) for \(\mathrm{HCrO}_{4}^{-}\) is \(3.2 \times 10^{-7},\) what is the value of \(K\) for the dehydration reaction?

Suggest a series of reactions, using common chemicals, by which each of the following syntheses can be performed. (a) \(\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s})\) from \(\mathrm{CuO}(\mathrm{s})\) (b) \(\operatorname{Cr} \mathrm{Cl}_{3}\left(\text { aq) } \text { from }\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s})\right.\)

Briefly describe each of the following ideas, phenomena, or methods: (a) lanthanide contraction; (b) zone refining; (c) basic oxygen process; (d) slag formation.

One of the simplest metals to extract from its ores is mercury. Mercury vapor is produced by roasting cinnabar ore (HgS) in air. Alternatives to this simple roasting, designed to reduce or eliminate \(\mathrm{SO}_{2}\) emissions, is to roast the ore in the presence of a second substance. For example, when cinnabar is roasted with quicklime, the products are mercury vapor and calcium sulfide and calcium sulfate. Write equations for the two reactions described here.

Which of the first transition series elements exhibits the greatest number of different oxidation states in its compounds? Explain.
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