How many electrons are in the valence \(d\) orbitals in these transition-metal ions? (a) \(\mathrm{Co}^{3+},(\mathbf{b}) \mathrm{Cu}^{+}\), (d) \(\mathrm{Os}^{3+}\). (c) \(\mathrm{Cd}^{2+}\)

Short Answer

Expert verified
The number of electrons in the valence \(d\) orbitals for each transition-metal ion is as follows: (a) \(\mathrm{Co}^{3+}\): 6 electrons (b) \(\mathrm{Cu}^{+}\): 10 electrons (c) \(\mathrm{Cd}^{2+}\): 10 electrons (d) \(\mathrm{Os}^{3+}\): 5 electrons

Step by step solution

01

Determine the ground-state electron configurations of the given metal ions

First, we need to determine the ground-state electron configurations of each metal ion before losing any electrons. For this, we can use the periodic table to identify the atomic number \(Z\) which represents the number of protons and, in a neutral atom, the number of electrons.
02

Determine the electronic configuration of each ion

Subtract the number of lost electrons from each ion's electron total to find the electronic configuration of each ion. This is important to define the number of electrons in the valence shell of each ion.
03

Identify the number of electrons in the valence d orbitals for each ion

Finally, find the number of electrons in the valence \(d\) orbitals for each ion. Now let's find the number of electrons in the valence \(d\) orbitals for each ion:
04

(a) \(\mathrm{Co}^{3+}\)

1. Ground-state electron configuration of \(\mathrm{Co}\) (cobalt): \(\mathrm{Co}\) has an atomic number of 27, which means it has 27 electrons in a neutral state. Thus, the electron configuration of neutral \(\mathrm{Co}\) is \([\mathrm{Ar}]4s^23d^7\). 2. Electronic configuration of \(\mathrm{Co}^{3+}\): Since \(\mathrm{Co}^{3+}\) has lost 3 electrons, its electron configuration is \([\mathrm{Ar}]4s^03d^6\). 3. Number of valence \(d\) electrons in \(\mathrm{Co}^{3+}\): There are 6 electrons in the valence \(d\) orbitals.
05

(b) \(\mathrm{Cu}^{+}\)

1. Ground-state electron configuration of \(\mathrm{Cu}\) (copper): \(\mathrm{Cu}\) has an atomic number of 29, which means it has 29 electrons in a neutral state. Thus, the electron configuration of neutral \(\mathrm{Cu}\) is \([\mathrm{Ar}]4s^23d^9\). 2. Electronic configuration of \(\mathrm{Cu}^{+}\): Since \(\mathrm{Cu}^{+}\) has lost 1 electron, its electron configuration is \([\mathrm{Ar}]4s^03d^{10}\). 3. Number of valence \(d\) electrons in \(\mathrm{Cu}^{+}\): There are 10 electrons in the valence \(d\) orbitals.
06

(c) \(\mathrm{Cd}^{2+}\)

1. Ground-state electron configuration of \(\mathrm{Cd}\) (cadmium): \(\mathrm{Cd}\) has an atomic number of 48, which means it has 48 electrons in a neutral state. Thus, the electron configuration of neutral \(\mathrm{Cd}\) is \([\mathrm{Kr}]5s^24d^10\). 2. Electronic configuration of \(\mathrm{Cd}^{2+}\): Since \(\mathrm{Cd}^{2+}\) has lost 2 electrons, its electron configuration is \([\mathrm{Kr}]5s^04d^{10}\). 3. Number of valence \(d\) electrons in \(\mathrm{Cd}^{2+}\): There are 10 electrons in the valence \(d\) orbitals.
07

(d) \(\mathrm{Os}^{3+}\)

1. Ground-state electron configuration of \(\mathrm{Os}\) (osmium): \(\mathrm{Os}\) has an atomic number of 76, which means it has 76 electrons in a neutral state. Thus, the electron configuration of neutral \(\mathrm{Os}\) is \([\mathrm{Xe}]6s^24f^{14}5d^6\). 2. Electronic configuration of \(\mathrm{Os}^{3+}\): Since \(\mathrm{Os}^{3+}\) has lost 3 electrons, its electron configuration is \([\mathrm{Xe}]6s^04f^{14}5d^5\). 3. Number of valence \(d\) electrons in \(\mathrm{Os}^{3+}\): There are 5 electrons in the valence \(d\) orbitals. In summary, the number of electrons in the valence \(d\) orbitals for each ion is: (a) \(\mathrm{Co}^{3+}\): 6 electrons (b) \(\mathrm{Cu}^{+}\): 10 electrons (c) \(\mathrm{Cd}^{2+}\): 10 electrons (d) \(\mathrm{Os}^{3+}\): 5 electrons

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transition Metal Ions
Transition metal ions play a vital role in chemistry, in part due to the unique properties of their valence d orbitals. These ions are formed when atoms of transition metals lose electrons, often from their outermost shells, resulting in a positively charged ion. The term 'transition' denotes their position on the periodic table, sandwiched between the s-block and p-block elements. They are known for their color, catalytic properties, and the ability to form a variety of complex compounds.

The valence d orbitals are particularly important because they can house up to ten electrons and are largely responsible for the magnetic and chemical behaviors of these metals and their ions. During ion formation, the energy required to remove electrons usually means that the s orbital electrons are lost before the d orbital electrons, despite the counterintuitive order on the periodic table. This behavior of transition metal ions is crucial in fields such as materials science, biochemistry, and industrial chemistry, where they are employed in everything from electronic structures to enzyme functions.
Electron Configuration
The electron configuration of an atom or ion tells us how electrons are distributed among various orbitals. For transition metals, this configuration becomes more complex due to the involvement of d orbitals, which are of higher energy than s and p orbitals in the same shell. The unique arrangement of electrons in these d orbitals gives transition metals their distinctive properties, including variable oxidation states.

When transition metals form ions, they tend to lose the s orbital electrons first, followed by the d orbital electrons, if necessary. Understanding and predicting the electron configuration of transition metal ions is central to explaining their reactivity, the types of bonds they can form, and their overall chemistry. For instance, the loss of electrons to form \( \(\mathrm{Co}^{3+}\), \(\mathrm{Cu}^{+}\), \(\mathrm{Cd}^{2+}\), and \(\mathrm{Os}^{3+}\) \) ions alters the electron configuration and directly affects the number of electrons in their valence d orbitals, which can be determined using the specifically defined steps in the step-by-step solution.
Periodic Table
The periodic table is an essential tool for chemists and a systematic way to organize the elements according to their atomic structure and properties. Its layout is such that it reflects the periodicity of element properties, which is instrumental in predicting the behavior of atoms during chemical reactions. For transition metals, the periodic table provides information on their position between Groups 3 to 12.

Through the periodic table, we can determine an element's atomic number, which is equal to its number of protons and, in an electrically neutral atom, its number of electrons. This assists us in figuring out the ground-state electron configurations of elements and their ions, as well as provides clues about their possible interactions with other elements. For example, knowing the position of Cobalt, Copper, Cadmium, and Osmium on the periodic table, and understanding their propensity to form specific ions, we can deduce their valence electron configurations and make predictions about their physical and chemical behavior.

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Most popular questions from this chapter

Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}(o\) -phen \()\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \mathrm{Br}\) (c) \(\left[\mathrm{Cr}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\) (d) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right]\left(\mathrm{ClO}_{4}\right)_{2}\)

(a) A complex absorbs light in the range of \(200-300 \mathrm{nm} . \mathrm{Do}\) you expect it to have visible color? (b) A solution of a compound appears green. Does this observation necessarily mean that all colors of visible light other than green are absorbed by the solution? Explain. (c) What information is usually presented in a visible absorption spectrum of a compound? (d) What energy is associated with an absorption at \(440 \mathrm{nm}\) in \(\mathrm{kJ} / \mathrm{mol} ?\)

The red color of ruby is due to the presence of Cr(III) ions at octahedral sites in the close-packed oxide lattice of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Draw the crystal-field splitting diagram for Cr(III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.

The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a \(0.100-\mathrm{L}\) sample of the water with a solution of EDTA \(^{4-}\). The EDTA \(^{4-}\) chelates the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow &[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow &[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second \(0.100-\mathrm{L}\) sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) \([\mathrm{EDTA}]^{4-},\) Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in \(\mathrm{mg} / \mathrm{L}\).

Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl},\) on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of AgCl per mole of Cr present. (Write two chemical equations.) (c) When an NaOH solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) a precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

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