/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Write electron configurations fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write electron configurations for a. the cations \(\mathrm{Mg}^{2+}, \mathrm{K}^{+}\), and \(\mathrm{Al}^{3+}\). b. the anions \(\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}\), and \(\mathrm{Te}^{2-}\).

Short Answer

Expert verified
The electron configurations for the given ions are: a. Cations: - \(\mathrm{Mg}^{2+}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{K}^{+}\): \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\] - \(\mathrm{Al}^{3+}\): \[1s^{2} 2s^{2} 2p^{6}\] b. Anions: - \(\mathrm{N}^{3-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{O}^{2-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{F}^{-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{Te}^{2-}\): \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6}\]

Step by step solution

01

1. Determine the original electron configuration of the neutral atom

For each ion, find the atomic number (number of protons, and in neutral state - also of electrons) in the periodic table and write the electron configuration accordingly: - \(\mathrm{Mg}\) (magnesium): atomic number 12; electron configuration: \[1s^{2} 2s^{2} 2p^{6} 3s^{2}\] - \(\mathrm{K}\) (potassium): atomic number 19; electron configuration: \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}\] - \(\mathrm{Al}\) (aluminum): atomic number 13; electron configuration: \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}\]
02

2. Account for the gained or lost electrons due to the charge

Cations have a positive charge, which means electrons have been lost: - \(\mathrm{Mg}^{2+}\): loses 2 electrons - \(\mathrm{K}^{1+}\): loses 1 electron - \(\mathrm{Al}^{3+}\): loses 3 electrons
03

3. Write the electron configuration

Now we can write the electron configurations for the cations: - \(\mathrm{Mg}^{2+}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{K}^{+}\): \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\] - \(\mathrm{Al}^{3+}\): \[1s^{2} 2s^{2} 2p^{6}\] b. For the anions:
04

1. Determine the original electron configuration of the neutral atom

As before, find the atomic number and electron configuration for the neutral atoms: - \(\mathrm{N}\) (nitrogen): atomic number 7; electron configuration: \[1s^{2} 2s^{2} 2p^{3}\] - \(\mathrm{O}\) (oxygen): atomic number 8; electron configuration: \[1s^{2} 2s^{2} 2p^{4}\] - \(\mathrm{F}\) (fluorine): atomic number 9; electron configuration: \[1s^{2} 2s^{2} 2p^{5}\] - \(\mathrm{Te}\) (tellurium): atomic number 52; electron configuration: \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{4}\]
05

2. Account for the gained or lost electrons due to the charge

Anions have a negative charge, which means electrons have been gained: - \(\mathrm{N}^{3-}\): gains 3 electrons - \(\mathrm{O}^{2-}\): gains 2 electrons - \(\mathrm{F}^{-}\): gains 1 electron - \(\mathrm{Te}^{2-}\): gains 2 electrons
06

3. Write the electron configuration

Now we can write the electron configurations for the anions: - \(\mathrm{N}^{3-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{O}^{2-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{F}^{-}\): \[1s^{2} 2s^{2} 2p^{6}\] - \(\mathrm{Te}^{2-}\): \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6}\]

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 91Ó°ÊÓ!

Key Concepts

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

Cation Electron Configuration
Understanding the electron configuration of cations is foundational in chemistry, especially when dealing with ionic compounds. A cation is an atom that has lost one or more electrons, gaining a positive charge as a result. This loss alters the electron configuration, which is the arrangement of electrons around the nucleus. For instance, magnesium, in its neutral state, has 12 electrons with the configuration 1s^{2} 2s^{2} 2p^{6} 3s^{2}. As a cation Mg^{2+}, it loses two electrons making the new configuration 1s^{2} 2s^{2} 2p^{6}.

This simplified configuration shows that the outermost electrons, which are the most energetically accessible, were removed. Likewise, potassium as K^{+} loses an electron from the 4s orbital changing from 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} to 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}. When writing cation configurations, always remove electrons starting from the outermost shell. It's essential to understand this when examining ionic bonds and reactivity of elements.
Anion Electron Configuration
In contrast to cations, anions are atoms that have gained extra electrons, resulting in a negative charge. The electron configurations of anions are characterized by adding electrons to the existing structure. For nitrogen, with a neutral configuration of 1s^{2} 2s^{2} 2p^{3}, adding three electrons to form N^{3-} transforms it to 1s^{2} 2s^{2} 2p^{6}, which is the configuration of the noble gas neon. Similarly, an oxygen atom gains two electrons to achieve the electron configuration of neon, and fluorine gains one.

Tellurium is a bit more involved due to its position in the periodic table. As Te^{2-}, it gains two electrons to complete its p orbital, resulting in the configuration 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6}. When learning about anions, it's crucial to note that the additional electrons populate the lowest available energy level. These configurations are key to predicting anions' chemical behavior and their role in creating ionic bonds.
Periodic Table Notation
The periodic table notation for electron configurations offers a roadmap to understanding element properties. It organizes elements based on increasing atomic number and recurring chemical properties. Electron configuration notation involves using numbers to represent energy levels, letters (s, p, d, f) for subshells, and superscripts for the number of electrons in those subshells.

For instance, the notation 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} represents potassium with one electron in its outermost 4s orbital. This style of notation is not only concise but also indicative of an element's potential chemical behaviors and its place within the periodic table. As you explore the table, patterns emerge—elements within the same group usually have similar valence electron configurations, hinting at similar chemical properties. Thus, mastering periodic table notation is indispensable for anyone delving into the intricacies of 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

Without using Fig. \(8.3\), predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c. \(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Describe the type of bonding that exists in the \(\mathrm{F}_{2}(g)\) molecule. How does this type of bonding differ from that found in the \(\mathrm{HF}(g)\) molecule? How is it similar?

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl}\) c. \(\mathrm{Sr}\) and \(\mathrm{E}\) b. \(\mathrm{Na}\) and \(\mathrm{O}\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Which member of the following pairs would you expect to be more energetically stable? Justify each choice. a. \(\mathrm{NaBr}\) or \(\mathrm{NaBr}_{2}\) b. \(\mathrm{ClO}_{4}\) or \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{SO}_{4}\) or \(\mathrm{XeO}_{4}\) d. \(\mathrm{OF}_{4}\) or \(\mathrm{SeF}_{4}\)

Use the following data (in \(\mathrm{kJ} / \mathrm{mol}\) ) to estimate \(\Delta H\) for the reaction \(\mathrm{S}^{-}(g)+\mathrm{e}^{-} \rightarrow \mathrm{S}^{2-}(g)\). Include an estimate of uncertainty. $$ \begin{array}{cl} \mathrm{S}(s) \longrightarrow \mathrm{S}(g) & \Delta H=277 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{S}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{-}(g) & \Delta H=-200 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Assume that all values are known to \(\pm 1 \mathrm{~kJ} / \mathrm{mol}\).

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.