General, Organic, and Biological Chemistry - 4th edition
General, Organic, and Biological Chemistry - 4th edition
4th Edition
ISBN: 9781259883989
Author: by Janice Smith
Publisher: McGraw-Hill Education
Question
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Chapter 3, Problem 48P
Interpretation Introduction

(a)

Interpretation:

The cations & anions derived from Barium & Bromine should be combined in order to get an electrically neutralized ionic compound.

Concept introduction:

An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cationis derived from Barium (Ba), while anion is derived from Bromine (Br).

Expert Solution
Check Mark

Answer to Problem 48P

BaBr2

Explanation of Solution

Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.

Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.

So, in the given two elements metal is Barium (Ba). So cation is derived from Ba. The only cation derived from Ba is Ba2+. The cation is derived by removing 2 electrons from Ba.

Ba → Ba2+ + 2e (electrons are indicated as e)

Bromine is the non-metal which is forming anion, Br- / Bromide, by gaining 1 electron to a Br atom.

Br + e → Br-

From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.

As Ba2+ is having +2 positive charge, 2 Br- ions should becombined.

So the ionic compound form from Barium & Bromine is BaBr2.

Interpretation Introduction

(b)

Interpretation:

The cations & anions derived from Aluminum&Sulfur should be combined in order to get an electrically neutralized ionic compound.

Concept introduction:

An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Aluminum (Al), while anion is derived from Sulfur(S).

Expert Solution
Check Mark

Answer to Problem 48P

Al2S3

Explanation of Solution

Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.

Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.

In the given two elements metal is Aluminum (Al). So cation is derived from Al. The only cation derived from Al is Al3+. The cation is derived by removing 3 electrons from Al.

Al → Al3++ 3e (electrons are indicated as e)

Sulfur is the non-metal which is forming anion, S2-/Sulfide, by gaining 2 electrons.

S + 2e → S2-

From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.

Al3+ has a +3 charge. S2-/ Sulfide ion has a -2 charge. So S2-: Al3+ should be combined in a ratio of 3: 2, to maintain the charge neutrality. This is the first ratio that the both values are integers.

The above ratio is proven correct by the following simple calculation to balance charges.

  Charge on cations + Charge on anions = 0            (2Al3+) + (3 S2)= 0  

(Equals zero because the ionic compound should have no net charge)

  2(+3) + 3(2)      = 0(+6) + (6)              = 0

So in the ionic compound form from Aluminum & Sulfur there should be 2 Al3+s & 3 S2- s. So the compound is Al2S3

Interpretation Introduction

(c)

Interpretation:

The cations & anions derived from Manganese&Chlorine should be combined in order to get an electrically neutralized ionic compound.

Concept introduction:

An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Manganese (Mn), while anion is derived from Chlorine (Cl).

Expert Solution
Check Mark

Answer to Problem 48P

MnCl2

Explanation of Solution

Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.

Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.

So in the given two elements metal is Manganese (Mn). So cation is derived from Mn. Manganese has the ability to form several cations like Mn2+, Mn3+&Mn4+. Out of them the most stable cation is Mn2+. This cation is derived by removing 2 electrons from Mn.

Mn → Mn2+ + 2e (electrons are indicated as e)

Chlorine is the non-metal which is forming anion, Cl-/Chloride, by gaining 1 electron to a Cl atom.

Cl + e → Cl-

From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.

As Mn2+ is having +2 positive charge, 2 Cl- ions should becombined.

So, the ionic compound form from Manganese&Chlorine is MnCl2.

Interpretation Introduction

(d)

Interpretation:

The cations & anions derived from Zinc&Sulfur should be combined in order to get an electrically neutralized ionic compound.

Concept introduction:

An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Zinc (Zn), while anion is derived from Sulfur (S).

Expert Solution
Check Mark

Answer to Problem 48P

ZnS

Explanation of Solution

Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.

Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.

So in the given two elements metal is Zinc (Zn). So cation is derived from Zn. The most common & stable cation derived from Zn is Zn2+. The cation is derived by removing 2 electrons from Zn.

Zn → Zn2+ + 2e (electrons are indicated as e)

Sulfur is the non-metal which is forming anion, S2- / Sulfide, by gaining 2 electrons.

S + 2e → S2-

From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.

As both Zn2+& S2-is having +2 & -2 charge, one cation & one anion are combined to form the Compound ZnS.

So, the ionic compound form from Zinc&Sulfur is ZnS.

Interpretation Introduction

(e)

Interpretation:

The cations & anions derived from Magnesium&Fluorine should be combined in order to get an electrically neutralized ionic compound.

Concept introduction:

An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Magnesium (Mg), while anion is derived from Fluorine (F).

Expert Solution
Check Mark

Answer to Problem 48P

MgF2

Explanation of Solution

Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.

Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.

So in the given two elements metal is Magnesium (Mg). So cation is derived from Mg. The only cation derived from Mg is Mg2+. The cation is derived by removing 2 electrons from Mg.

Mg → Mg2+ + 2e (electrons are indicated as e)

Fluorine is the non-metal, which is forming anion, F- / Fluoride, by gaining 1 electron to a F atom.

F + e → F-

From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.

As Mg2+ is having +2 positive charge, 2 F- ions should becombined.

So the ionic compound form from Magnesium&Fluorine is MgF2.

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Chapter 3 Solutions

General, Organic, and Biological Chemistry - 4th edition

Ch. 3.2 - How many electrons and protons are contained in...Ch. 3.2 - Prob. 3.6PCh. 3.2 - Mn2+ is an essential nutrient needed for blood...Ch. 3.3 - Prob. 3.7PPCh. 3.3 - Prob. 3.8PPCh. 3.3 - Prob. 3.8PCh. 3.4 - Prob. 3.9PCh. 3.4 - Prob. 3.10PCh. 3.4 - Prob. 3.9PPCh. 3.4 - Prob. 3.10PPCh. 3.4 - Prob. 3.11PCh. 3.4 - Prob. 3.12PCh. 3.4 - Prob. 3.11PPCh. 3.5 - List four physical properties of ionic compounds.Ch. 3.6 - Prob. 3.12PPCh. 3.6 - Prob. 3.14PCh. 3.6 - Prob. 3.15PCh. 3.6 - Prob. 3.13PPCh. 3.6 - Using the charges on the ions that compose...Ch. 3 - Which formulas represent ionic compounds and which...Ch. 3 - Which formulas represent ionic compound and which...Ch. 3 - Which pairs of elements are likely to form ionic...Ch. 3 - Which pairs of elements are likely to form ionic...Ch. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Write the ion symbol for an atom with the given...Ch. 3 - How many protons and electrons are present in each...Ch. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - What species fits each description? a period 2...Ch. 3 - What species fits each description? a period 3...Ch. 3 - Give the ion symbol for each ion. sodium ion...Ch. 3 - Give the ion symbol for each ion. barium ion iron...Ch. 3 - What noble gas has the same electronic...Ch. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - How many electrons must be gained or lost by each...Ch. 3 - For each of the general electron-dot formulas for...Ch. 3 - Label each of the following elements or regions in...Ch. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - How many protons and electrons are contained in...Ch. 3 - How many protons and electrons are contained in...Ch. 3 - Identify the polyatomic anion (including its...Ch. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - What is the charge on the cation M in each of the...Ch. 3 - What is the charge on the anion Z in each of the...Ch. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Write the formula for the ionic compound formed...Ch. 3 - Prob. 58PCh. 3 - Write the formula for the ionic compound formed...Ch. 3 - Write the formula for the ionic compound formed...Ch. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Liquid nutritional shakes that contain protein,...Ch. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Prob. 72PCh. 3 - Prob. 73PCh. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - Prob. 86PCh. 3 - Energy bars contain ionic compounds that serve as...Ch. 3 - Prob. 88CP
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