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Chapter 7: Problem 58

Write the ground-state electron configurations of the following ions: (a) \(\mathrm{Li}^{+},(\mathrm{b}) \mathrm{H}^{-},(\mathrm{c}) \mathrm{N}^{3-},(\mathrm{d}) \mathrm{F}^{-},\) (e) \(\mathrm{S}^{2-}\) (f) \(\mathrm{Al}^{3+},(\mathrm{g}) \mathrm{Se}^{2-}\) (h) \(\mathrm{Br}^{-}\) (i) \(\mathrm{Rb}^{+},(\mathrm{j}) \mathrm{Sr}^{2+},(\mathrm{k}) \mathrm{Sn}^{2+}\) (1) \(\mathrm{Te}^{2-} \cdot(\mathrm{m}) \mathrm{Ba}^{2+}\) (n) \(\mathrm{Pb}^{2+}\) (o) \(\mathrm{In}^{3+},(\mathrm{p}) \mathrm{Tl}^{+}\) (a) \(\mathrm{Tl}^{3+}\)

Short Answer

Expert verified
The ground-state configurations are derived by adding or removing electrons based on ion charges.

Step by step solution

01

Understand Electron Configuration

Electron configuration represents the distribution of electrons in an atom or ion. You need to know the neutral atom's electron configuration and then adjust it for the ions by adding or removing electrons based on the charge of the ion.
02

Electron Configuration of Neutral Atoms

Before adjusting for the ions, write down the electron configurations for the neutral atoms: \(\mathrm{Li}: 1s^2 2s^1\), \(\mathrm{H}: 1s^1\), \(\mathrm{N}: 1s^2 2s^2 2p^3\), \(\mathrm{F}: 1s^2 2s^2 2p^5\), \(\mathrm{S}: 1s^2 2s^2 2p^6 3s^2 3p^4\), \(\mathrm{Al}: 1s^2 2s^2 2p^6 3s^2 3p^1\), \(\mathrm{Se}: [Ar] 3d^{10} 4s^2 4p^4\), \(\mathrm{Br}: [Ar] 3d^{10} 4s^2 4p^5\), \(\mathrm{Rb}: [Kr] 5s^1\), \(\mathrm{Sr}: [Kr] 5s^2\), \(\mathrm{Sn}: [Kr] 4d^{10} 5s^2 5p^2\), \(\mathrm{Te}: [Kr] 4d^{10} 5s^2 5p^4\), \(\mathrm{Ba}: [Xe] 6s^2\), \(\mathrm{Pb}: [Xe] 4f^{14} 5d^{10} 6s^2 6p^2\), \(\mathrm{In}: [Kr] 4d^{10} 5s^2 5p^1\), \(\mathrm{Tl}: [Xe] 4f^{14} 5d^{10} 6s^2 6p^1\).
03

Adjust for Ion Charges

Modify the electron configurations based on the charges of the ions. Lose electrons for positive charges and gain electrons for negative charges:
04

Write Configurations for Each Ion

- \(\mathrm{Li}^+: 1s^2\) (lost 1 electron) - \(\mathrm{H}^-: 1s^2\) (gained 1 electron) - \(\mathrm{N}^{3-}: 1s^2 2s^2 2p^6\) (gained 3 electrons) - \(\mathrm{F}^-: 1s^2 2s^2 2p^6\) (gained 1 electron) - \(\mathrm{S}^{2-}: 1s^2 2s^2 2p^6 3s^2 3p^6\) (gained 2 electrons) - \(\mathrm{Al}^{3+}: 1s^2 2s^2 2p^6\) (lost 3 electrons) - \(\mathrm{Se}^{2-}: [Kr] 4d^{10} 5s^2 5p^6\) (gained 2 electrons) - \(\mathrm{Br}^-: [Kr] 4d^{10} 5s^2 5p^6\) (gained 1 electron) - \(\mathrm{Rb}^+: [Kr]\) (lost 1 electron) - \(\mathrm{Sr}^{2+}: [Kr]\) (lost 2 electrons) - \(\mathrm{Sn}^{2+}: [Kr] 4d^{10} 5s^2 5p^0\) (lost 2 electrons) - \(\mathrm{Te}^{2-}: [Xe]\) (gained 2 electrons) - \(\mathrm{Ba}^{2+}: [Xe]\) (lost 2 electrons) - \(\mathrm{Pb}^{2+}: [Xe] 4f^{14} 5d^{10} 6s^2\) (lost 2 electrons) - \(\mathrm{In}^{3+}: [Kr] 4d^{10}\) (lost 3 electrons) - \(\mathrm{Tl}^+: [Xe] 4f^{14} 5d^{10} 6s^2 6p^0\) (lost 1 electron) - \(\mathrm{Tl}^{3+}: [Xe] 4f^{14} 5d^{10} 6s^2\) (lost 3 electrons)

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Key Concepts

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

Ground-State Electron Configurations
Ground-state electron configurations describe the arrangement of electrons in the lowest possible energy level for an atom. This concept is fundamental in chemistry because it helps predict the properties and reactivity of elements. In a neutral atom, electrons fill energy levels in order of increasing energy, starting from the lowest. An easy way to determine this is by using the Aufbau principle, which suggests that electrons occupy orbitals starting with the lowest energy level. The sequence follows the format of filling up the 1s, 2s, 2p, 3s, 3p, and so on, keeping in mind the Pauli Exclusion Principle and Hund's Rule. For example, lithium (Li) in its ground state has an electron configuration of 1s虏 2s鹿, indicating that the first two electrons fill the 1s orbital, and the third electron resides in the higher energy 2s orbital.
Ion Electron Configuration
When atoms gain or lose electrons, they form ions with electron configurations differing from their neutral counterparts. For example, an atom that loses an electron will form a cation, while an atom that gains an electron will form an anion. The electron configuration of ions is crucial because it can influence processes like chemical bonding and ionic stability. To determine the electron configuration of an ion, you start with the ground-state configuration of the atom and then either remove or add electrons as needed. For instance,
  • For ewlineewlinehydrogen anion (\(\text{H}^{-}\)), start with hydrogen's electron configuration 1s鹿 and add an electron, resulting in 1s虏.
  • Similarly, lithium cation (\(\text{Li}^+\)) loses one electron, reducing its configuration from 1s虏 2s鹿 to 1s虏.
Understanding these configurations helps clarify how atoms interact and bond at the molecular level.
Adjusting for Ion Charge
Adjusting for ion charge involves modifying the electron configuration of an atom based on whether it forms a positive or negative ion. Positive ions, or cations, have fewer electrons than their neutral atoms, while negative ions, or anions, have more. For cations like aluminum (\(\text{Al}^{3+}\)), you remove electrons: Al starts as 1s虏 2s虏 2p鈦� 3s虏 3p鹿. This changes to 1s虏 2s虏 2p鈦� after removing three electrons for the +3 charge. Anions, like the fluoride ion (\(\text{F}^-\)), result from gaining electrons. Fluorine starts as 1s虏 2s虏 2p鈦�, which becomes 1s虏 2s虏 2p鈦� after gaining an electron. Consider the charge: a + 2 charge means losing two electrons, while a - 2 charge means gaining two.
Neutral Atom Configuration
The neutral atom configuration is essential as it forms the baseline before any electrons are added or removed when forming ions. This configuration shows how electrons are distributed in an atom without considering any ions or charges. Understanding the neutral atom configuration makes it easier to predict how the atom will change when forming ions. For example, sulfur (\(\text{S}\)) as a neutrally charged atom has the configuration 1s虏 2s虏 2p鈦� 3s虏 3p鈦�. This indicates that sulfur has 16 electrons distributed across its orbitals. Recognizing this neutral state assists in envisioning the potential changes needed to achieve stability through ion formation. By understanding the beginning neutral state, we can anticipate the rules governing electron addition or removal that result in cations or anions.

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Most popular questions from this chapter

You are given four substances: a fuming red liquid, a dark metallic-looking solid, a pale-yellow gas, and a yellow-green gas that attacks glass. You are told that these substances are the first four members of Group 7A, the halogens. Name each one.

Arrange the following in order of increasing first ionization energy: \(\mathrm{Na}, \mathrm{Cl}, \mathrm{Al}, \mathrm{S},\) and \(\mathrm{Cs}\).

A student is given samples of three elements, \(X, Y,\) and \(\mathrm{Z}\), which could be an alkali metal, a member of Group 4A, or a member of Group 5A. She makes the following observations: Element \(\mathrm{X}\) has a metallic luster and conducts electricity. It reacts slowly with hydrochloric acid to produce hydrogen gas. Element \(Y\) is a light yellow solid that does not conduct electricity. Element \(Z\) has a metallic luster and conducts electricity. When exposed to air, it slowly forms a white powder. A solution of the white powder in water is basic. What can you conclude about the elements from these observations?

The ionization energy of a certain element is \(412 \mathrm{~kJ} / \mathrm{mol}\). When the atoms of this element are in the first excited state, however, the ionization energy is only \(126 \mathrm{~kJ} / \mathrm{mol}\). Based on this information, calculate the wavelength of light emitted in a transition from the first excited state to the ground state.

Write a balanced equation for the preparation of (a) molecular oxygen, (b) ammonia, (c) carbon dioxide, (d) molecular hydrogen, (e) calcium oxide. Indicate the physical state of the reactants and products in each equation.
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