Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{P}^{3-}\) (c) \(Z r^{4+}\) (d) \(\mathrm{Ru}^{3+}\), (e) \(\mathrm{As}^{3-},(\mathrm{f}) \mathrm{Ag}^{+}\)

Short Answer

Expert verified
The electron configurations of the given ions are as follows: (a) \(\mathrm{Cd}^{2+}\): [Kr]\(\,4d^{10}\) (not a noble-gas configuration) (b) \(\mathrm{P}^{3-}\): [Ne]\(\,3s^2\,3p^6\) (noble-gas configuration, matches Ar) (c) \(Z r^{4+}\): [Kr] (noble-gas configuration) (d) \(\mathrm{Ru}^{3+}\): [Kr]\(\,4d^3\) (not a noble-gas configuration) (e) \(\mathrm{As}^{3-}\): [Ar]\(\,4s^2\,3d^{10}\,4p^6\) (noble-gas configuration, matches Kr) (f) \(\mathrm{Ag}^{+}\): [Kr]\(\,4d^{10}\) (not a noble-gas configuration) The ions with noble-gas configurations are \(\mathrm{P}^{3-}\), \(Z r^{4+}\), and \(\mathrm{As}^{3-}\).

Step by step solution

01

(a) Electron Configuration of \(\mathrm{Cd}^{2+}\)

: 1. For Neutral \(\mathrm{Cd}\): \(\mathrm{Cd}\) (Cadmium) has an atomic number of 48. Therefore, an uncharged \(\mathrm{Cd}\) atom has 48 protons and 48 electrons. The electron configuration of \(\mathrm{Cd}\) is [Kr]\(\,5s^2\,4d^1$$\,0\). 2. For \(\mathrm{Cd}^{2+}\) ion: \(\mathrm{Cd}^{2+}\) has lost 2 electrons. Therefore, the electron configuration becomes [Kr]\(\,4d^1$$\,0\).
02

(b) Electron Configuration of \(\mathrm{P}^{3-}\)

: 1. For Neutral \(\mathrm{P}\): \(\mathrm{P}\) (Phosphorus) has an atomic number of 15. Therefore, an uncharged \(\mathrm{P}\) atom has 15 protons and 15 electrons. The electron configuration of \(\mathrm{P}\) is [Ne]\(\,3s^2\,3p^3\). 2. For \(\mathrm{P}^{3-}\) ion: \(\mathrm{P}^{3-}\) has gained 3 electrons. Therefore, the electron configuration becomes [Ne]\(\,3s^2\,3p^6\).
03

(c) Electron Configuration of \(Z r^{4+}\)

: 1. For Neutral \(Z r\): \(Z r\) (Zirconium) has an atomic number of 40. Therefore, an uncharged \(Z r\) atom has 40 protons and 40 electrons. The electron configuration of \(Z r\) is [Kr]\(\,5s^2\,4d^2\). 2. For \(Z r^{4+}\) ion: \(Z r^{4+}\) has lost 4 electrons. Therefore, the electron configuration becomes [Kr].
04

(d) Electron Configuration of \(\mathrm{Ru}^{3+}\)

: 1. For Neutral \(\mathrm{Ru}\): \(\mathrm{Ru}\) (Ruthenium) has an atomic number of 44. Therefore, an uncharged \(\mathrm{Ru}\) atom has 44 protons and 44 electrons. The electron configuration of \(\mathrm{Ru}\) is [Kr]\(\,5s^2\,4d^6\). 2. For \(\mathrm{Ru}^{3+}\) ion: \(\mathrm{Ru}^{3+}\) has lost 3 electrons. Therefore, the electron configuration becomes [Kr]\(\,4d^3\).
05

(e) Electron Configuration of \(\mathrm{As}^{3-}\)

: 1. For Neutral \(\mathrm{As}\): \(\mathrm{As}\) (Arsenic) has an atomic number of 33. Therefore, an uncharged \(\mathrm{As}\) atom has 33 protons and 33 electrons. The electron configuration of \(\mathrm{As}\) is [Ar]\(\,4s^2\,3d^1$$\,0\,4p^3\). 2. For \(\mathrm{As}^{3-}\) ion: \(\mathrm{As}^{3-}\) has gained 3 electrons. Therefore, the electron configuration becomes [Ar]\(\,4s^2\,3d^1$$\,0\,4p^6\).
06

(f) Electron Configuration of \(\mathrm{Ag}^{+}\)

: 1. For Neutral \(\mathrm{Ag}\): \(\mathrm{Ag}\) (Silver) has an atomic number of 47. Therefore, an uncharged \(\mathrm{Ag}\) atom has 47 protons and 47 electrons. The electron configuration of \(\mathrm{Ag}\) is [Kr]\(\,5s^2\,4d^9\). 2. For \(\mathrm{Ag}^{+}\) ion: \(\mathrm{Ag}^{+}\) has lost 1 electron. Therefore, the electron configuration becomes [Kr]\(\,4d^1$$\,0\). Now, let's identify which ions have noble-gas configurations: - \(\mathrm{Cd}^{2+}\): [Kr]\(\,4d^1$$\,0\) (No) - \(\mathrm{P}^{3-}\): [Ne]\(\,3s^2\,3p^6\) (Yes, matches \(\mathrm{Ar}\)) - \(Z r^{4+}\): [Kr] (Yes) - \(\mathrm{Ru}^{3+}\): [Kr]\(\,4d^3\) (No) - \(\mathrm{As}^{3-}\): [Ar]\(\,4s^2\,3d^1$$\,0\,4p^6\) (Yes, matches \(\mathrm{Kr}\)) - \(\mathrm{Ag}^{+}\): [Kr]\(\,4d^1$$\,0\) (No) So, the ions with noble-gas configurations are \(\mathrm{P}^{3-}\), \(Z r^{4+}\), and \(\mathrm{As}^{3-}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ions
An ion is an atom or a group of atoms that has a net electric charge due to the loss or gain of one or more electrons. There are two primary types of ions:
  • Cations, which are positively charged ions formed by the loss of electrons. For example, a neutral cadmium (Cd) atom losing two electrons becomes a \( \mathrm{Cd}^{2+} \) ion.
  • Anions, which are negatively charged ions formed by the gain of electrons. For example, a neutral phosphorus (P) atom gaining three electrons becomes a \( \mathrm{P}^{3-} \) ion.
Ions move towards electrodes in a solution during electrolysis, and this movement is a key part of electrical conductivity. Understanding the formation of ions is crucial, as it helps predict how elements react chemically, form compounds, and influence materials'electrical properties in different environments. Additionally, knowing the charge of an ion is essential in balancing chemical equations, a fundamental skill in chemistry.
Noble Gas Configuration
In the realm of chemistry, achieving a noble gas configuration is a pivotal concept. Noble gases like helium (He), neon (Ne), and argon (Ar) have a complete outer electron shell, making them highly stable and chemically inert.Atoms strive to reach a noble gas configuration because it represents a state of maximum stability, usually having eight electrons in their outer shell, known as the Octet Rule. When ions form, they often gain or lose electrons to emulate the electron configuration of the nearest noble gas.For instance:
  • The \( \mathrm{P}^{3-} \) ion gains three electrons to have the same electron configuration as argon, \([\mathrm{Ne}]3s^2 3p^6\).
  • The \( \mathrm{Zr}^{4+} \) ion loses four electrons to achieve the noble gas configuration of krypton, \([\mathrm{Kr}]\).
  • \( \mathrm{As}^{3-} \) achieves a noble-gas configuration of krypton after gaining three electrons, \([\mathrm{Ar}]4s^2 3d^{10} 4p^6\).
Achieving noble gas configuration reduces an atom's energy level and makes it energetically favorable. This concept is fundamental for understanding atom stability in various chemical environments.
Chemical Elements
Chemical elements are the fundamental building blocks of matter. Each element is defined by the number of protons in its nucleus, known as the atomic number, which determines the element's identity and its placement in the periodic table. The periodic table is an organized map that categorizes elements based on their similar properties and recurring trends in chemical behavior.Every element is unique in terms of its properties and applications, for example:
  • Cadmium (Cd): A metal used in batteries and pigments, Cd forms a \( \mathrm{Cd}^{2+} \) ion by losing two electrons.
  • Phosphorus (P): A non-metal essential in biological systems, gaining electrons to form a \( \mathrm{P}^{3-} \) ion.
  • Zirconium (Zr): Known for its corrosion resistance, Zr forms \( \mathrm{Zr}^{4+} \) ion when it loses electrons.
  • Arsenic (As): Used in alloys and semiconductors, gains electrons forming \( \mathrm{As}^{3-} \) ion.
Understanding the properties of elements is essential for grasping the complexities of matter and chemical reactions. It forms the basis for discovering new materials and applications in technology and medicine. Recognizing how an element behaves in its neutral state compared to its ionized form is a fundamental concept in chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Based on data in Table 8.2 , estimate (within \(30 \mathrm{~kJ} / \mathrm{mol}\) ) the lattice energy for (a) LiBr, (b) CsBr, (c) \(\mathrm{CaCl}_{2}\)

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

The following three Lewis structures can be drawn for \(\mathrm{N}_{2} \mathrm{O}:\) \(\mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{O}}: \longleftrightarrow: \mathrm{N}^{*}-\mathrm{N} \equiv \mathrm{O}: \longleftrightarrow: \ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{O}}:\) (a) Using formal charges, which of these three resonance forms is likely to be the most important? (b) The \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(1.12 \mathrm{~A}\), slightly longer than a typical \(\mathrm{N} \equiv \mathrm{N}\) bond; and the \(\mathrm{N}-\mathrm{O}\) bond length is \(1.19 \AA\), slightly shorter than a typical \(\mathrm{N}=\mathrm{O}\) bond. (See Table \(8.5 .\) ) Rationalize these observations in terms of the resonance structures shown previously and your conclusion for part (a).

Which of these elements are unlikely to form covalent bonds: \(\mathrm{S}, \mathrm{H}, \mathrm{K}, \mathrm{Ar},\) Si? Explain your choices.

Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free