Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Question
Book Icon
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.

Blurred answer
Students have asked these similar questions
5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3
Show work. don't give Ai generated solution.  How many carbons and hydrogens are in the structure?
13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c.…

Chapter 21 Solutions

Chemistry & Chemical Reactivity

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...
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning