Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Question
Chapter 21, Problem 18PS

(a)

Interpretation Introduction

Interpretation: The complete balanced equation has to 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.

(a)

Expert Solution
Check Mark

Answer to Problem 18PS

The complete balanced equation for the reaction of potassium and iodine is:

    2K(s)+I2(g)2KI(s)

Explanation of Solution

Potassium belongs to group 1A of periodic table and has an oxidation number of +1. Thus, it loses one electron to attain noble gas configuration.

    KK++e

This electron is gained by iodine to form an anion with one negative charge. Iodine belongs to halogen family and it has the oxidation number of 1.

    I2+e_2I

The number of electrons in both the equations is same. Thus an ionic compound is formed in which potassium has +1 charge and iodine bears 1 charge. Thus, the formula of the product is KI.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation Since iodine is present as I2 in the reaction, hence the stoichiometric coefficient of KI is 2 due to which potassium also has a stoichiometric coefficient of 2.

Thus, the overall balanced equation is:

    2K(s)+I2(g)2KI(s)

(b)

Interpretation Introduction

Interpretation: The complete balanced equation has to 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)

Expert Solution
Check Mark

Answer to Problem 18PS

The complete balanced equation for the reaction of barium and oxygen is:

    2Ba(s)+O2(g)2BaO(s)

Explanation of Solution

Barium belongs to group 2A of periodic table and thus has +2 oxidation number. Barium loses two electrons to attain noble gas configuration.

  BaBa2++2e

These two electrons are gained by the oxygen leading to the formation of an ionic compound. Oxygen belongs to the sulfur family and exists in -2 oxidation number.

  O2+2e2O2

The number of electrons in both the equations is same. Barium has a charge of +2 and oxygen has a charge of 2. Thus, the formula of the product formed is BaO.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since oxygen is present as O2 in the reaction, hence the stoichiometric coefficient of BaO is 2 due to which barium also has a stoichiometric coefficient of 2.

Thus, the overall balanced equation is:

    2Ba(s) + O2(g)2BaO(s)

(c)

Interpretation Introduction

Interpretation: The complete balanced equation has to 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)

Expert Solution
Check Mark

Answer to Problem 18PS

The complete balanced equation for the reaction of aluminium with sulfur is:

    16Al(s)+3S8(s)8Al2S3(s)

Explanation of Solution

Aluminium belongs to group 3A of the periodic table and has a highest oxidation number of +3. It loses three electrons to attain noble gas configuration of filled s and p subshells.

  AlAl3++3e

Sulphur belongs to oxygen family and exists in -2 oxidation number. These two electrons are gained by the sulphur leading to the formation of a product compound.

    S8+2e_8S2

The numbers of electrons are not same in both the equations. Aluminium bear charge and sulfur bears 2 charge.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since sulphur is present as S8 in the reaction and aluminium forms the compound with three sulphur atoms, hence the stoichiometric coefficients of Al is 16, S8 is 3 and Al2S3 is 8.

Thus, the overall balanced equation is:

    16Al(s)+3S8(s)8Al2S3(s)

(d)

Interpretation Introduction

Interpretation: The complete balanced equation has to 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)

Expert Solution
Check Mark

Answer to Problem 18PS

The complete balanced equation for the reaction of silicon with chlorine is:

    Si(s)+2Cl2(g)SiCl4(l)

Explanation of Solution

Silicon belongs to group 4A of the periodic table and has four electrons in the outermost shell. The electronic configuration of silicon is [Ne]3s23p2. Silicon can either gain four electrons or can lose its four electrons to form an ionic compound. However, a very high energy is required in both the cases. Thus, silicon forms covalent compounds.

The electronegativity difference between silicon and chlorine is less than 2. Thus, the compound formed is covalent compound by sharing of electrons between silicon and chlorine. The oxidation number of silicon is +4. The oxidation number of chlorine is 1. Thus, the compound formed is SiCl4.

The stoichiometric coefficients are multiplied with species to have equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since chlorine is present as Cl2 in the reaction and silicon forms compound with four chlorine atoms, thus the stoichiometric coefficient of Cl2 is 2.

Thus, the overall balanced equation is:

    Si(s)+2Cl2(g)SiCl4(l)

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Chapter 21 Solutions

Chemistry & Chemical Reactivity

Ch. 21.11 - The best catalysts used to accelerate the...Ch. 21.11 - Prob. 2.5ACPCh. 21 - Which of the following formulas is incorrect? (a)...Ch. 21 - The reaction of elemental phosphorus and excess...Ch. 21 - Like sulfur, selenium forms compounds in several...Ch. 21 - Prob. 4PSCh. 21 - Give examples of two basic oxides. Write equations...Ch. 21 - Prob. 6PSCh. 21 - Prob. 7PSCh. 21 - Prob. 8PSCh. 21 - Prob. 9PSCh. 21 - Prob. 10PSCh. 21 - For the product of the reaction you selected in...Ch. 21 - For the product of the reaction you selected in...Ch. 21 - Prob. 13PSCh. 21 - Prob. 14PSCh. 21 - Place the following oxides in order of increasing...Ch. 21 - Place the following oxides in order of increasing...Ch. 21 - Prob. 17PSCh. 21 - Prob. 18PSCh. 21 - Prob. 19PSCh. 21 - Prob. 20PSCh. 21 - Prob. 21PSCh. 21 - Prob. 22PSCh. 21 - Prob. 23PSCh. 21 - Prob. 24PSCh. 21 - Prob. 25PSCh. 21 - Prob. 26PSCh. 21 - Prob. 27PSCh. 21 - The compound Na2O2 consists of (a) two Na+ ions...Ch. 21 - Prob. 29PSCh. 21 - Write balanced equations for the reaction of...Ch. 21 - Prob. 31PSCh. 21 - (a) Write equations for the half-reactions that...Ch. 21 - Prob. 33PSCh. 21 - Prob. 34PSCh. 21 - When magnesium bums in air, it forms both an oxide...Ch. 21 - Prob. 36PSCh. 21 - Prob. 37PSCh. 21 - Prob. 38PSCh. 21 - Calcium oxide, CaO, is used to remove SO2 from...Ch. 21 - Prob. 40PSCh. 21 - Prob. 41PSCh. 21 - The element below aluminum in Group 3A is gallium,...Ch. 21 - Prob. 43PSCh. 21 - The boron trihalides (except BF3) hydrolyze...Ch. 21 - When boron hydrides burn in air, the reactions are...Ch. 21 - Prob. 46PSCh. 21 - Write balanced equations for the reactions of...Ch. 21 - Prob. 48PSCh. 21 - Prob. 49PSCh. 21 - Alumina, Al2O3, is amphoteric. Among examples of...Ch. 21 - Prob. 51PSCh. 21 - Prob. 52PSCh. 21 - Prob. 53PSCh. 21 - Silicon and oxygen form a six-membered ring in the...Ch. 21 - Describe the structure of pyroxenes (see page...Ch. 21 - Describe how ultrapure silicon can be produced...Ch. 21 - Prob. 57PSCh. 21 - Prob. 58PSCh. 21 - Prob. 59PSCh. 21 - Prob. 60PSCh. 21 - Prob. 61PSCh. 21 - Prob. 62PSCh. 21 - Prob. 63PSCh. 21 - The overall reaction involved in the industrial...Ch. 21 - Prob. 65PSCh. 21 - Prob. 66PSCh. 21 - Prob. 67PSCh. 21 - Prob. 68PSCh. 21 - Prob. 69PSCh. 21 - Which statement about oxygen is not true? (a)...Ch. 21 - Prob. 71PSCh. 21 - Prob. 72PSCh. 21 - Prob. 73PSCh. 21 - Sulfur forms a range of compounds with fluorine....Ch. 21 - Prob. 75PSCh. 21 - Which of the following statements is not correct?...Ch. 21 - The halogen oxides and oxoanions are good...Ch. 21 - Prob. 78PSCh. 21 - Bromine is obtained from brine wells. The process...Ch. 21 - Prob. 80PSCh. 21 - Prob. 81PSCh. 21 - Halogens combine with one another to produce...Ch. 21 - Prob. 83PSCh. 21 - Prob. 84PSCh. 21 - The standard enthalpy of formation of XeF4 is 218...Ch. 21 - Draw the Lewis electron dot structure for XeO3F2....Ch. 21 - Prob. 87PSCh. 21 - Prob. 88PSCh. 21 - Prob. 89GQCh. 21 - Prob. 90GQCh. 21 - Consider the chemistries of the elements...Ch. 21 - When BCl3 gas is passed through an electric...Ch. 21 - Prob. 93GQCh. 21 - Prob. 94GQCh. 21 - Prob. 95GQCh. 21 - Prob. 96GQCh. 21 - Prob. 97GQCh. 21 - Prob. 98GQCh. 21 - Prob. 99GQCh. 21 - Prob. 100GQCh. 21 - Prob. 101GQCh. 21 - Prob. 102GQCh. 21 - Prob. 103GQCh. 21 - Prob. 105GQCh. 21 - Prob. 106GQCh. 21 - A Boron and hydrogen form an extensive family of...Ch. 21 - In 1774, C. Scheele obtained a gas by reacting...Ch. 21 - The chemistry of gallium: (a) Gallium hydroxide,...Ch. 21 - Prob. 111GQCh. 21 - Prob. 112GQCh. 21 - Prob. 113GQCh. 21 - Prob. 114GQCh. 21 - Prob. 115ILCh. 21 - Prob. 116ILCh. 21 - Prob. 117ILCh. 21 - Prob. 118ILCh. 21 - Prob. 119ILCh. 21 - Prob. 120ILCh. 21 - Prob. 121SCQCh. 21 - Prob. 122SCQCh. 21 - Prob. 123SCQCh. 21 - Prob. 124SCQCh. 21 - Prob. 125SCQCh. 21 - Prob. 126SCQCh. 21 - Prob. 127SCQCh. 21 - Prob. 128SCQCh. 21 - Comparing the chemistry of carbon and silicon. (a)...Ch. 21 - Prob. 130SCQCh. 21 - Xenon trioxide, XeO3, reacts with aqueous base to...
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