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Some ions do not have a corresponding neutral atom that has the same electron configuration. For each of the following ions, identify the neutral atom that has the same number of electrons and determine if this atom has the same electron configuration. \((\mathbf{a}) \mathrm{Cl}^{-},(\mathbf{b}) \mathrm{Sc}^{3+},(\mathbf{c}) \mathrm{Fe}^{2+},(\mathbf{d}) \mathrm{Zn}^{2+},(\mathbf{e}) \mathrm{Sn}^{4+}\)

Short Answer

Expert verified
(a) The neutral atom corresponding to Cl⁻ is argon (Ar) and they have the same electron configuration: [\(\ce{Ne}\)] 3s² 3p⁶. (b) The neutral atom corresponding to Sc³⁺ is argon (Ar) and they have the same electron configuration: [\(\ce{Ne}\)] 3s² 3p⁶. (c) The neutral atom corresponding to Fe²⁺ is chromium (Cr), but their electron configurations are not the same, as Fe²⁺ has [\(\ce{Ar}\)] 3d⁶, and Cr has [\(\ce{Ar}\)] 3d⁵ 4s¹. (d) The neutral atom corresponding to Zn²⁺ is nickel (Ni), but their electron configurations are not the same, as Zn²⁺ has [\(\ce{Ar}\)] 3d¹⁰, and Ni has [\(\ce{Ar}\)] 3d⁸ 4s². (e) The neutral atom corresponding to Sn⁴⁺ is palladium (Pd) and they have the same electron configuration: [\(\ce{Kr}\)] 4d¹⁰.

Step by step solution

01

(a) Identify the neutral atom corresponding to Cl⁻ and compare their electron configurations

1. Find the number of electrons in Cl⁻ ion: A Cl⁻ ion has one extra electron compared to a neutral chlorine atom. Since a neutral chlorine atom has 17 electrons, the Cl⁻ ion has 18 electrons. 2. Identify the neutral atom with 18 electrons: The neutral atom with 18 electrons is argon (Ar), which has an atomic number of 18. 3. Compare electron configurations: Cl⁻ ion has the electron configuration of [\(\ce{Ne}\)] 3s² 3p⁶, whereas Ar has the electron configuration of [\(\ce{Ne}\)] 3s² 3p⁶. Both Cl⁻ and Ar have the same electron configuration.
02

(b) Identify the neutral atom corresponding to Sc³⁺ and compare their electron configurations

1. Find the number of electrons in Sc³⁺ ion: A Sc³⁺ ion has lost 3 electrons compared to a neutral scandium atom. Since a neutral scandium atom has 21 electrons, the Sc³⁺ ion has 18 electrons. 2. Identify the neutral atom with 18 electrons: The neutral atom with 18 electrons is argon (Ar), which has an atomic number of 18. 3. Compare electron configurations: Sc³⁺ ion has the electron configuration of [\(\ce{Ne}\)] 3s² 3p⁶, whereas Ar has the electron configuration of [\(\ce{Ne}\)] 3s² 3p⁶. Both Sc³⁺ and Ar have the same electron configuration.
03

(c) Identify the neutral atom corresponding to Fe²⁺ and compare their electron configurations

1. Find the number of electrons in Fe²⁺ ion: An Fe²⁺ ion has lost 2 electrons compared to a neutral iron atom. Since a neutral iron atom has 26 electrons, the Fe²⁺ ion has 24 electrons. 2. Identify the neutral atom with 24 electrons: The neutral atom with 24 electrons is chromium (Cr), which has an atomic number of 24. 3. Compare electron configurations: Fe²⁺ ion has the electron configuration of [\(\ce{Ar}\)] 3d⁶, whereas Cr has the electron configuration of [\(\ce{Ar}\)] 3d⁵ 4s¹. The electron configurations of Fe²⁺ and Cr are not the same.
04

(d) Identify the neutral atom corresponding to Zn²⁺ and compare their electron configurations

1. Find the number of electrons in Zn²⁺ ion: A Zn²⁺ ion has lost 2 electrons compared to a neutral zinc atom. Since a neutral zinc atom has 30 electrons, the Zn²⁺ ion has 28 electrons. 2. Identify the neutral atom with 28 electrons: The neutral atom with 28 electrons is nickel (Ni), which has an atomic number of 28. 3. Compare electron configurations: Zn²⁺ ion has the electron configuration of [\(\ce{Ar}\)] 3d¹⁰, whereas Ni has the electron configuration of [\(\ce{Ar}\)] 3d⁸ 4s². The electron configurations of Zn²⁺ and Ni are not the same.
05

(e) Identify the neutral atom corresponding to Sn⁴⁺ and compare their electron configurations

1. Find the number of electrons in Sn⁴⁺ ion: A Sn⁴⁺ ion has lost 4 electrons compared to a neutral tin atom. Since a neutral tin atom has 50 electrons, the Sn⁴⁺ ion has 46 electrons. 2. Identify the neutral atom with 46 electrons: The neutral atom with 46 electrons is palladium (Pd), which has an atomic number of 46. 3. Compare electron configurations: Sn⁴⁺ ion has the electron configuration of [\(\ce{Kr}\)] 4d¹⁰, whereas Pd has the electron configuration of [\(\ce{Kr}\)] 4d¹⁰. Both Sn⁴⁺ and Pd have the same electron configuration.

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Key Concepts

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

Electron Configuration Comparison
Understanding the intricacies of electron configurations is fundamental in chemistry, especially when comparing the electronic arrangements of ions to their parent atoms. At its core, an electron configuration describes how electrons are distributed among the atomic orbitals of an atom or ion. For instance, let's take the case of the Cl- ion. With 18 electrons in its configuration, one more than its neutral form, the Cl- ion has an electron configuration identical to the noble gas argon (Ar).

Similarly, Sc3+ loses three electrons when it becomes an ion, dropping down to the electron count of argon as well. However, the situation is different with Fe2+ and Zn2+. Both these ions have fewer electrons than their neutral counterparts, leading to different electron configurations when compared to any neutral atom with the same number of electrons. Fe2+ aligns with chromium (Cr) in number but not in configuration, highlighting how electron placement in orbitals can vary depending on multiple factors, such as the energy of the orbitals and Hund's rules. In contrast, Sn4+ sheds four electrons to achieve the same configuration as neutral palladium (Pd), a rare match between an ion and a different element in its neutral state. Understanding these comparisons not only solidifies one's grasp on how electron configurations are formulated but also illustrates the fascinating diversity within the periodic table.
Ions and Neutral Atoms
The transition from neutral atoms to ions involves the gain or loss of electrons, which impacts their electron configuration. In chemistry, this is a transformative event that dictates an element's chemical properties and reactivity. For example, when atoms turn into ions, they don't necessarily resemble the nearest noble gas in terms of electron configuration, as demonstrated by Fe2+ and Zn2+. The removal or addition of electrons to achieve stability carries significant implications in the formation of ionic compounds and their resulting behavior.

It is imperative to highlight that ions seek the most stable electronic arrangement, which can lead to configurations that are sometimes unexpected when only considering the 'octet rule'. This foundational concept is evident in the exercise solutions above, where ions such as Cl- and Sc3+ achieve the electron configurations of noble gases to stabilize, while others like Fe2+ do not have an exact match in terms of electron distribution with any neutral atom.
Chemistry Education
Teaching concepts such as electron configurations to students requires a clear and methodical approach. Effective chemistry education hinges on making these abstract concepts tangible for learners. Visual aids, like orbital diagrams, and analogies can play a key role in enhancing understanding. It's crucial for educators to explain that while the 'octet rule' is a helpful guideline, it is not absolute. There are exceptions, and this complexity reflects the true nature of atomic behavior in chemistry.

The exercise from the textbook effectively demonstrates the nuanced relationship between ions and neutral atoms, providing a practical application of electron configurations. It offers students a deeper dive, beyond memorization, into critical thinking as they discern patterns and make comparisons. Such exercises build foundational knowledge that students can apply in advanced topics like molecular orbital theory and chemical bonding. The pedagogy of chemistry, therefore, must evolve to include a mix of theoretical knowledge, practical application, and engagement with real-world applications to foster a comprehensive understanding of the subject matter.

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Most popular questions from this chapter

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measurehow well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\) . (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S\) , calculate \(Z_{\text { eff}}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S.\) (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Which of the following is the expected product of the reaction of \(\mathrm{K}(s)\) and \(\mathrm{H}_{2}(g) ?(\mathbf{i}) \mathrm{KH}(s),(\mathbf{i} \mathbf{i}) \mathrm{K}_{2} \mathrm{H}(s),\) (iii) \(\mathrm{KH}_{2}(s),\) \((\mathbf{i} \mathbf{v}) \mathrm{K}_{2} \mathrm{H}_{2}(s), \mathrm{or}(\mathbf{v}) \mathrm{K}(s)\) and \(\mathrm{H}_{2}(g)\) will not react with one another.

Write the electron configurations for the following ions, and determine which have noble-gas configurations: \((\mathbf{a})\mathrm{Co}^{2+}\) \((\mathbf{b})\mathrm{Sn}^{2+},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ag}^{+},(\mathbf{e}) \mathrm{S}^{2-}.\)

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol},\) for the group 1 \(\mathrm{B}\) and group 2 \(\mathrm{B}\) metals are as follows: (a) Why are the electron affinities of the group 2 \(\mathrm{B}\) elements greater than zero? (b) Why do the electron affinities of the group 1 \(\mathrm{B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

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