Bundle: Chemistry & Chemical Reactivity, Loose-Leaf Version, 9th + OWLv2, 4 terms (24 Months) Printed Access Card
Bundle: Chemistry & Chemical Reactivity, Loose-Leaf Version, 9th + OWLv2, 4 terms (24 Months) Printed Access Card
9th Edition
ISBN: 9781305367425
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 21, Problem 13PS

(a)

Interpretation Introduction

Interpretation: The complete balanced equation should be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group 1A and group 2A and elements from group 3A to 8A are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    X+neXn

The metals of group 1A form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group 1A elements is +1. Similarly, group 2A elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than 2. If the difference in electronegativity is more than 2 then ionic compounds are formed.

(b)

Interpretation Introduction

Interpretation: The complete balanced equation should be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group 1A and group 2A and elements from group 3A to 8A are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    X+neXn

The metals of group 1A form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group 1A elements is +1. Similarly, group 2A elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than 2. If the difference in electronegativity is more than 2 then ionic compounds are formed.

(c)

Interpretation Introduction

Interpretation: The complete balanced equation should be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group 1A and group 2A and elements from group 3A to 8A are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    X+neXn

The metals of group 1A form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group 1A elements is +1. Similarly, group 2A elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than 2. If the difference in electronegativity is more than 2 then ionic compounds are formed.

(d)

Interpretation Introduction

Interpretation: The complete balanced equation should be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group 1A and group 2A and elements from group 3A to 8A are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  MMn++ne

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    X+neXn

The metals of group 1A form +1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group 1A elements is +1. Similarly, group 2A elements form +2 ions by losing two electrons and have an oxidation number of +2. The non-metal gains these electrons to form anions with 1 and 2 charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than 2. If the difference in electronegativity is more than 2 then ionic compounds are formed.

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Chapter 21 Solutions

Bundle: Chemistry & Chemical Reactivity, Loose-Leaf Version, 9th + OWLv2, 4 terms (24 Months) Printed Access Card

Ch. 21.8 - Prob. 4QCh. 21.8 - Prob. 3RCCh. 21.11 - Prob. 1QCh. 21.11 - Prob. 2QCh. 21 - Give examples of two basic oxides. Write equations...Ch. 21 - Prob. 2PSCh. 21 - Prob. 3PSCh. 21 - Prob. 4PSCh. 21 - Prob. 5PSCh. 21 - Prob. 6PSCh. 21 - For the product of the reaction you selected in...Ch. 21 - For the product of the reaction you selected in...Ch. 21 - Prob. 9PSCh. 21 - Prob. 10PSCh. 21 - Place the following oxides in order of increasing...Ch. 21 - Place the following oxides in order of increasing...Ch. 21 - Prob. 13PSCh. 21 - Prob. 14PSCh. 21 - Prob. 15PSCh. 21 - Prob. 16PSCh. 21 - Prob. 17PSCh. 21 - Prob. 18PSCh. 21 - Prob. 19PSCh. 21 - Prob. 20PSCh. 21 - Prob. 21PSCh. 21 - Write balanced equations for the reaction of...Ch. 21 - Prob. 23PSCh. 21 - (a) Write equations for the half-reactions that...Ch. 21 - When magnesium bums in air, it forms both an oxide...Ch. 21 - Prob. 26PSCh. 21 - Prob. 27PSCh. 21 - Prob. 28PSCh. 21 - Calcium oxide, CaO, is used to remove SO2 from...Ch. 21 - Prob. 30PSCh. 21 - Prob. 31PSCh. 21 - The boron trihalides (except BF3) hydrolyze...Ch. 21 - When boron hydrides burn in air, the reactions are...Ch. 21 - Prob. 34PSCh. 21 - Write balanced equations for the reactions of...Ch. 21 - Prob. 36PSCh. 21 - Prob. 37PSCh. 21 - Alumina, Al2O3, is amphoteric. Among examples of...Ch. 21 - Prob. 39PSCh. 21 - Prob. 40PSCh. 21 - Describe the structure of pyroxenes (see page...Ch. 21 - Describe how ultrapure silicon can be produced...Ch. 21 - Prob. 43PSCh. 21 - Prob. 44PSCh. 21 - Prob. 45PSCh. 21 - Prob. 46PSCh. 21 - Prob. 47PSCh. 21 - The overall reaction involved in the industrial...Ch. 21 - Prob. 49PSCh. 21 - Prob. 50PSCh. 21 - Prob. 51PSCh. 21 - Prob. 52PSCh. 21 - Prob. 53PSCh. 21 - Prob. 54PSCh. 21 - Prob. 55PSCh. 21 - Sulfur forms a range of compounds with fluorine....Ch. 21 - The halogen oxides and oxoanions are good...Ch. 21 - Prob. 58PSCh. 21 - Bromine is obtained from brine wells. The process...Ch. 21 - Prob. 60PSCh. 21 - Prob. 61PSCh. 21 - Halogens combine with one another to produce...Ch. 21 - The standard enthalpy of formation of XeF4 is 218...Ch. 21 - Draw the Lewis electron dot structure for XeO3F2....Ch. 21 - Prob. 65PSCh. 21 - Prob. 66PSCh. 21 - Prob. 67GQCh. 21 - Prob. 68GQCh. 21 - Consider the chemistries of the elements...Ch. 21 - When BCl3 gas is passed through an electric...Ch. 21 - Prob. 71GQCh. 21 - Prob. 72GQCh. 21 - Prob. 73GQCh. 21 - Prob. 74GQCh. 21 - Prob. 75GQCh. 21 - Prob. 76GQCh. 21 - Prob. 77GQCh. 21 - Prob. 78GQCh. 21 - Prob. 79GQCh. 21 - Prob. 80GQCh. 21 - Prob. 81GQCh. 21 - Prob. 83GQCh. 21 - Prob. 84GQCh. 21 - A Boron and hydrogen form an extensive family of...Ch. 21 - In 1774, C. Scheele obtained a gas by reacting...Ch. 21 - What current must be used in a Downs cell...Ch. 21 - The chemistry of gallium: (a) Gallium hydroxide,...Ch. 21 - Prob. 89GQCh. 21 - Prob. 90GQCh. 21 - Prob. 91GQCh. 21 - Prob. 92GQCh. 21 - Prob. 93ILCh. 21 - Prob. 94ILCh. 21 - Prob. 95ILCh. 21 - Prob. 96ILCh. 21 - Prob. 97ILCh. 21 - Prob. 98ILCh. 21 - Prob. 99SCQCh. 21 - Prob. 100SCQCh. 21 - Prob. 101SCQCh. 21 - Prob. 102SCQCh. 21 - Prob. 103SCQCh. 21 - Prob. 104SCQCh. 21 - Prob. 105SCQCh. 21 - Prob. 106SCQCh. 21 - Prob. 107SCQCh. 21 - Prob. 108SCQCh. 21 - Prob. 109SCQCh. 21 - Prob. 110SCQCh. 21 - Comparing the chemistry of carbon and silicon. (a)...Ch. 21 - Prob. 112SCQCh. 21 - Xenon trioxide, XeO3, reacts with aqueous base to...
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