/*! 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 71 What monatomic ions would you ex... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 71

What monatomic ions would you expect potassium \((Z=19)\) and bromine \((Z=35)\) to form?

Short Answer

Expert verified
Potassium forms \(\text{K}^+\) and bromine forms \(\text{Br}^-\).

Step by step solution

01

Determine the Electron Configuration

Find the electron configuration of potassium and bromine. For potassium \((Z=19)\), the electron configuration is \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \]. For bromine \((Z=35)\), the electron configuration is \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \].
02

Determine the Tendency to Gain or Lose Electrons

Potassium, with one electron in its 4s orbital, will readily lose one electron to achieve a stable noble gas configuration (like argon, which has \[1s^2 2s^2 2p^6 3s^2 3p^6 \]). Bromine, with seven electrons in its 4p orbital, will tend to gain one electron to complete its octet, achieving a stable noble gas configuration (like krypton, which has \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 \]).
03

Identify the Monatomic Ions Formed

When potassium loses one electron, it becomes a cation with a charge of \(+1\), forming \(\text{K}^+\). When bromine gains one electron, it becomes an anion with a charge of \(-1\), forming \(\text{Br}^-\).

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 91影视!

Key Concepts

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

electron configuration
Electron configuration describes the distribution of electrons in an atom's orbitals. It's key to understanding how atoms interact and bond. Electrons occupy different energy levels or shells around a nucleus. These shells (s, p, d, f) fill based on increasing energy.
For potassium \((Z=19)\), the electron configuration is \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \]. This sequence shows how 19 electrons are arranged. They fill lower energy orbitals before higher ones.
Bromine \((Z=35)\) has a different sequence: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \]. Here, the electrons fill up to the 4th energy level, and the 4p subshell has 5 electrons.
noble gas configuration
Noble gas configuration is about achieving electron arrangements like those of noble gases. Noble gases (like helium, neon, argon) have very stable and complete electron shells. Atoms often react to reach similar stability.
Potassium (K) loses one electron to mimic argon \[ 1s^2 2s^2 2p^6 3s^2 3p^6 \]. This way, it achieves a stable noble gas configuration.
Bromine (Br), in contrast, gains one electron to have the same electron configuration as krypton \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 \]. It fills its 4p orbital to complete the octet.
cation and anion formation
Cations and anions form when atoms lose or gain electrons, respectively. This process helps atoms reach a more stable electron configuration.
Potassium (\text{K}), being a metal, loses one electron to form a cation: \[ \text{K}^+ \]. This makes it positively charged with a configuration like argon.
Bromine (\text{Br}), being a non-metal, gains one electron to form an anion: \[ \text{Br}^- \]. It becomes negatively charged while achieving a krypton-like configuration.
In summary, potassium forms \[ \text{K}^+ \] by losing an electron, and bromine forms \[ \text{Br}^- \] by gaining one. This gain or loss ensures both reach more stable electron arrangements.

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

Give the name and formula for the acid derived from each of the following anions: (a) hydrogen carbonate (b) \(\mathrm{IO}_{4}^{-}\) (c) cyanide (d) \(\mathrm{HS}^{-}\)

Give the systematic names for the formulas or the formulas for the names: (a) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} ;\) (b) ammonium perchlorate; (c) \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2} \cdot 3 \mathrm{H}_{2} \mathrm{O} ;\) (d) sodium nitrite.

Before the use of systematic names, many compounds had common names. Give the systematic name for each of the following: (a) Blue vitriol, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (b) Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) Oil of vitriol, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) Washing soda, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) (e) Muriatic acid, HCl (f) Epsom salt, \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\)

Boron trifluoride is used as a catalyst in the synthesis of organic compounds. When this compound is analyzed by mass spectrometry, several different \(1+\) ions form, including ions representing the whole molecule as well as molecular fragments formed by the loss of one, two, and three \(\mathrm{F}\) atoms. Given that boron has two naturally occurring isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B},\) and fluorine has one, \({ }^{19} \mathrm{~F}\), calculate the masses of all possible \(1+\) ions.

Given that the ions in \(\mathrm{LiF}\) and in \(\mathrm{MgO}\) are of similar size, which compound has stronger ionic bonding? Use Coulomb's law in your explanation.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation
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.