Which \(\mathrm{M}^{2}\) ions are predicted to have the following ground-state electron configurations: (a) \([\mathrm{Ar}] 3 d^{7} ;\) (b) \([\mathrm{Kr}] 4 d^{7} ;\) (c) \([\mathrm{Kr}] 4 d^{10} 5 s^{2} ;\) (d) [Xe] \(5 d^{10}\) ?

Short Answer

Expert verified
The \( \mathrm{M}^{2+} \) ions with the ground-state electron configurations are: (a) Cu, (b) Ag, (c) Sn, (d) Hg.

Step by step solution

01

Identifying the Element for Configuration (a)

The electron configuration given in part (a) is \( [\mathrm{Ar}] 3d^{7} \) for a divalent metal ion \( \mathrm{M}^{2+} \). This configuration corresponds to an atom that has two more electrons in its neutral state, or \( [\mathrm{Ar}] 3d^{9} \). This is the electron configuration of the neutral copper (Cu) atom. Therefore, when Cu loses two electrons to become Cu^{2+}, its electron configuration is \( [\mathrm{Ar}] 3d^{7} \).
02

Identifying the Element for Configuration (b)

The electron configuration for part (b) is \( [\mathrm{Kr}] 4d^{7} \) for the ion \( \mathrm{M}^{2+} \). Similar to step 1, this configuration indicates that the neutral atom originally had \( [\mathrm{Kr}] 4d^{9} \) configuration. The element with this configuration is silver (Ag). Upon losing two electrons to form \( \mathrm{Ag}^{2+} \) the remaining electron configuration becomes \( [\mathrm{Kr}] 4d^{7} \).
03

Identifying the Element for Configuration (c)

For part (c), the electron configuration is \( [\mathrm{Kr}] 4d^{10} 5s^{2} \) for a divalent metal ion \( \mathrm{M}^{2+} \). In order to have this configuration as a \( \mathrm{M}^{2+} \) ion, the neutral atom would need to have two additional electrons, making the electron configuration \( [\mathrm{Kr}] 4d^{10} 5s^{2} 5p^{2} \) which would be the element tin (Sn). Upon losing two electrons, typically from the outermost p-orbital, the ion \( \mathrm{Sn}^{2+} \) retains the configuration \( [\mathrm{Kr}] 4d^{10} 5s^{2} \).
04

Identifying the Element for Configuration (d)

Finally, in part (d), the given configuration is \( [\mathrm{Xe}] 5d^{10} \) for an ion \( \mathrm{M}^{2+} \). For an ion with this configuration, the neutral atom would have two more electrons, being \( [\mathrm{Xe}] 5d^{10} 6s^{2} \). This configuration corresponds to the element mercury (Hg). Once mercury ionizes to become \( \mathrm{Hg}^{2+} \) it would have the electron configuration \( [\mathrm{Xe}] 5d^{10} \) as two electrons are removed from the s-orbital.

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Key Concepts

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

Electronic Configuration
Understanding the electron configuration of atoms is fundamental in chemistry and physics. It describes how electrons are distributed in an atom's orbitals. For an atom in its ground state, or most stable state, electrons fill the lowest energy orbitals first, following a set of rules known as the Aufbau principle, Pauli exclusion principle, and Hund’s rule.

As seen in the exercise solutions, determining the electron configuration for ions, such as divalent metal ions, requires knowing the configuration of the neutral atom first. Electrons are lost from the outermost orbitals when atoms form cations, generally maintaining the configuration of the nearest noble gas with some additional d-electrons, depending on the element. This method of losing electrons to achieve a noble gas configuration is common among transition elements, which we will explore further.
Divalent Metal Ions
Divalent metal ions, denoted as M^{2+}, are ions that have lost two electrons, resulting in a +2 charge. This occurs because metals typically exhibit electropositive character and tend to lose electrons to attain a more stable electron configuration. For transition metals, this is particularly interesting as the electron loss comes from a combination of their s and d orbitals.

In the provided examples from the textbook exercise, divalent ions like Cu^{2+} and Ag^{2+} had complex configurations involving the loss of electrons from d orbitals. The stability of these ions within the context of the solution is due to the arrangement that minimizes the energy according to the established rules of electron distribution. In the case of tin (Sn^{2+}) and mercury (Hg^{2+}), the loss of p or s electrons respectively shows the diversity within divalent metal ions and the influence of the underlying electronic structure.
Transition Elements
Transition elements, often found in the d-block of the periodic table, are characterized by their partially filled d orbitals. This unique feature allows transition metals to exhibit a wide range of oxidation states, and they frequently form colored compounds and complex ions.

The ground-state electron configurations provided in the exercise demonstrate the variations among transition elements as they form divalent metal ions. The presence of a d^7 configuration in both copper (Cu) and silver (Ag) divalent ions highlights the flexibility of these elements' valence electrons. Additionally, the d^10 configuration seen in mercury (Hg^{2+}) is indicative of a filled d orbital, which confers stability. Transition elements often engage in complex formation, and the differing stability of these ions is fundamentally connected to their unique electronic configurations, which plays a critical role in their chemical behavior.

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