Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 8, Problem 8.14E
(a)
Interpretation Introduction
Interpretation:
The species have to be divided into two isoelectronic groups.
(b)
Interpretation Introduction
Interpretation:
The strongest Lewis base and strongest reducing agent in each other has to be given.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The elements sodium, aluminum, and chlorine are in the same period.
(a) Which has the greatest electronegativity?
(b) Which of the atoms is smallest?
(c) Which is the largest possible oxidation state for each of these elements?
(d) Will the oxide of each element in the highest oxidation state (write its formula)
be acidic, basic, or amphoteric?
The elements sodium, aluminum, and chlorine are in the same period.(a) Which has the greatest electronegativity?(b) Which of the atoms is smallest?(c) Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine.(d) Will the oxide of each element be acidic, basic, or amphoteric?
Would you expect potassium perchlorate, KClO4(aq), in a concentrated acidic solution to act as an oxidizing agent or as a reducing agent?
Chapter 8 Solutions
Chemical Principles: The Quest for Insight
Ch. 8 - Prob. 8A.1ASTCh. 8 - Prob. 8A.1BSTCh. 8 - Prob. 8A.2ASTCh. 8 - Prob. 8A.2BSTCh. 8 - Prob. 8A.1ECh. 8 - Prob. 8A.2ECh. 8 - Prob. 8A.3ECh. 8 - Prob. 8A.4ECh. 8 - Prob. 8A.5ECh. 8 - Prob. 8A.6E
Ch. 8 - Prob. 8A.7ECh. 8 - Prob. 8A.8ECh. 8 - Prob. 8A.9ECh. 8 - Prob. 8A.10ECh. 8 - Prob. 8A.11ECh. 8 - Prob. 8A.12ECh. 8 - Prob. 8A.13ECh. 8 - Prob. 8A.14ECh. 8 - Prob. 8A.15ECh. 8 - Prob. 8A.16ECh. 8 - Prob. 8A.17ECh. 8 - Prob. 8A.18ECh. 8 - Prob. 8A.19ECh. 8 - Prob. 8A.20ECh. 8 - Prob. 8B.1ASTCh. 8 - Prob. 8B.1BSTCh. 8 - Prob. 8B.2ASTCh. 8 - Prob. 8B.2BSTCh. 8 - Prob. 8B.1ECh. 8 - Prob. 8B.2ECh. 8 - Prob. 8B.3ECh. 8 - Prob. 8B.4ECh. 8 - Prob. 8B.5ECh. 8 - Prob. 8B.6ECh. 8 - Prob. 8B.7ECh. 8 - Prob. 8B.8ECh. 8 - Prob. 8C.1ASTCh. 8 - Prob. 8C.1BSTCh. 8 - Prob. 8C.2BSTCh. 8 - Prob. 8C.1ECh. 8 - Prob. 8C.2ECh. 8 - Prob. 8C.3ECh. 8 - Prob. 8C.4ECh. 8 - Prob. 8C.5ECh. 8 - Prob. 8C.6ECh. 8 - Prob. 8D.1ASTCh. 8 - Prob. 8D.1BSTCh. 8 - Prob. 8D.2ASTCh. 8 - Prob. 8D.2BSTCh. 8 - Prob. 8D.1ECh. 8 - Prob. 8D.2ECh. 8 - Prob. 8D.3ECh. 8 - Prob. 8D.4ECh. 8 - Prob. 8D.5ECh. 8 - Prob. 8D.6ECh. 8 - Prob. 8D.7ECh. 8 - Prob. 8D.8ECh. 8 - Prob. 8E.1ASTCh. 8 - Prob. 8E.1BSTCh. 8 - Prob. 8E.2ASTCh. 8 - Prob. 8E.2BSTCh. 8 - Prob. 8E.1ECh. 8 - Prob. 8E.2ECh. 8 - Prob. 8E.3ECh. 8 - Prob. 8E.4ECh. 8 - Prob. 8E.5ECh. 8 - Prob. 8E.6ECh. 8 - Prob. 8E.7ECh. 8 - Prob. 8E.8ECh. 8 - Prob. 8F.1ASTCh. 8 - Prob. 8F.1BSTCh. 8 - Prob. 8F.2ASTCh. 8 - Prob. 8F.2BSTCh. 8 - Prob. 8F.1ECh. 8 - Prob. 8F.2ECh. 8 - Prob. 8F.3ECh. 8 - Prob. 8F.4ECh. 8 - Prob. 8F.5ECh. 8 - Prob. 8F.6ECh. 8 - Prob. 8G.1ASTCh. 8 - Prob. 8G.1BSTCh. 8 - Prob. 8G.2ASTCh. 8 - Prob. 8G.2BSTCh. 8 - Prob. 8G.1ECh. 8 - Prob. 8G.2ECh. 8 - Prob. 8G.3ECh. 8 - Prob. 8G.4ECh. 8 - Prob. 8G.5ECh. 8 - Prob. 8G.6ECh. 8 - Prob. 8G.7ECh. 8 - Prob. 8G.8ECh. 8 - Prob. 8G.9ECh. 8 - Prob. 8G.10ECh. 8 - Prob. 8H.1ASTCh. 8 - Prob. 8H.1BSTCh. 8 - Prob. 8H.2ASTCh. 8 - Prob. 8H.2BSTCh. 8 - Prob. 8H.1ECh. 8 - Prob. 8H.2ECh. 8 - Prob. 8H.3ECh. 8 - Prob. 8H.4ECh. 8 - Prob. 8H.5ECh. 8 - Prob. 8H.6ECh. 8 - Prob. 8H.7ECh. 8 - Prob. 8H.8ECh. 8 - Prob. 8H.10ECh. 8 - Prob. 8H.11ECh. 8 - Prob. 8H.12ECh. 8 - Prob. 8I.1ASTCh. 8 - Prob. 8I.1BSTCh. 8 - Prob. 8I.2ASTCh. 8 - Prob. 8I.2BSTCh. 8 - Prob. 8I.1ECh. 8 - Prob. 8I.2ECh. 8 - Prob. 8I.3ECh. 8 - Prob. 8I.5ECh. 8 - Prob. 8I.6ECh. 8 - Prob. 8I.7ECh. 8 - Prob. 8I.8ECh. 8 - Prob. 8I.9ECh. 8 - Prob. 8I.10ECh. 8 - Prob. 8I.11ECh. 8 - Prob. 8I.12ECh. 8 - Prob. 8I.13ECh. 8 - Prob. 8I.14ECh. 8 - Prob. 8I.15ECh. 8 - Prob. 8I.16ECh. 8 - Prob. 8J.1ASTCh. 8 - Prob. 8J.1BSTCh. 8 - Prob. 8J.1ECh. 8 - Prob. 8J.2ECh. 8 - Prob. 8J.3ECh. 8 - Prob. 8J.4ECh. 8 - Prob. 8J.5ECh. 8 - Prob. 8J.6ECh. 8 - Prob. 8J.7ECh. 8 - Prob. 8J.8ECh. 8 - Prob. 8.3ECh. 8 - Prob. 8.4ECh. 8 - Prob. 8.5ECh. 8 - Prob. 8.6ECh. 8 - Prob. 8.7ECh. 8 - Prob. 8.8ECh. 8 - Prob. 8.9ECh. 8 - Prob. 8.11ECh. 8 - Prob. 8.12ECh. 8 - Prob. 8.13ECh. 8 - Prob. 8.14ECh. 8 - Prob. 8.15ECh. 8 - Prob. 8.17ECh. 8 - Prob. 8.18ECh. 8 - Prob. 8.19ECh. 8 - Prob. 8.20ECh. 8 - Prob. 8.21ECh. 8 - Prob. 8.22ECh. 8 - Prob. 8.25ECh. 8 - Prob. 8.29ECh. 8 - Prob. 8.31CE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- (c) Suggest how the following boron species could be prepared, showing any intermediates. -B(OH)2 B(OH)2arrow_forward(a) One of the alkali metals reacts with oxygen to form a solidwhite substance. When this substance is dissolved in water,the solution gives a positive test for hydrogen peroxide, H2O2.When the solution is tested in a burner flame, a lilac-purpleflame is produced. What is the likely identity of the metal?(b) Write a balanced chemical equation for the reaction ofthe white substance with water.arrow_forward(a) The third row element with the largest first ionization energy. (b) The Group 3A element with the largest first ionization energy. (c) The Group 3A element with the largest atomic radius.arrow_forward
- (a) Which poisonous gas is evolved when white phosphorus is heated with Cone. NaOH solution? Write the chemical equation. (b) Write the formula of first noble gas compound prepared by N. Bartlett. What inspired N. Bartlett to prepare this compound? (c) Fluorine is a stronger oxidising agent than chlorine. Why? (d)Write one use of chlorine gas.arrow_forwardWould you expect an aqueous solution of manganese (VII) oxide to have a ph greater or less than 7.0? Justify your answerarrow_forwardBased on the information shown in the attached image, which of the following statements are CORRECT? (I) Hydrogen is the best reducing agent in acid solution. (II) Hydrogen is the best reducing agent in basic solution. (III) Oxygen is the best oxidizing agent in acid solution. (IV) Oxygen is the best oxidizing agent in basic solution.arrow_forward
- Each of the chemically active Period 2 elements forms sta-ble compounds that have bonds to fluorine.(a) What are the names and formulas of these compounds?(b) Does ΔEN increase or decrease left to right across the period?(c) Does percent ionic character increase or decrease left to right?(d) Draw Lewis structures for these compounds.arrow_forwardWhich do you expect to be the stronger oxidizing agent, NaBrO3 or NaClO3?arrow_forwardWhat is a disproportionation reaction, and which of thefollowing fit the description?(a) I₂(s) +KI(aq) →KI₃(aq)(b) 2ClO₂(g) +H₂O(l) →HClO₃(aq) +HClO₂(aq)(c) Cl₂(g)+ 2NaOH(aq)→ NaCl(aq)+ NaClO(aq)+ H₂O(l)(d) NH₄NO₂(s)→ N₂(g) +2H₂O(g)(e) 3MnO₄²⁻(aq)+ 2H₂O(l) →2MnO₄(aq) +MnO₂(s) +4OH(aq)(f) 3AuCl(s) →AuCl₃(s)+ 2Au(sarrow_forward
- Compounds such as NaBH₄, Al(BH₄)₃, and LiAlH₄ arecomplex hydrides used as reducing agents in many syntheses.(a) Give the oxidation state of each element in these compounds.(b) Write a Lewis structure for the polyatomic anion in NaBH₄,and predict its shape.arrow_forwardConsider the series of reactions to synthesize the alum (KAl(SO4 )2 · xH2O(s)) from the introduction. (a) Assuming an excess of the other reagents, from one mole of aluminum Al (s), how many moles of alum will be produced? (b) Assuming an excess of the other reagents, from one mole of potassium hydroxide KOH, how many moles of alum will be produced? (c) Assuming an excess of the other reagents, from one mole of sulfuric acid H2SO4 , how many moles of alum will be produced? (d) If you start the synthesis with 1.00 g of Al, 40.0 mL of 1.50 M KOH, and 20.0 mL of 9.00 M H2SO4 , which of the three will be the limiting reagent? (e) Assuming that the product is anhydrous (that there are no waters of hydration), calculate the theoretical yield of alum, in grams, based on the amounts of reagents in part (d). 3. Consider the nickel salt: (NH4 )2Ni(SO4 )2 ·y H2O (Ammonium Nickel Sulfate Hydrate), where y is the number of coordinated waters. (a) Assuming that the product is anhydrous (y = 0),…arrow_forwardConsider the elements O, Ba, Co, Be, Br, and Se. From this list,select the element that (a) is most electronegative, (b) exhibitsa maximum oxidation state of +7, (c) loses an electronmost readily, (d) forms π bonds most readily, (e) is a transitionmetal, (f) is a liquid at room temperature and pressure.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStax
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Liquids: Crash Course Chemistry #26; Author: Crash Course;https://www.youtube.com/watch?v=BqQJPCdmIp8;License: Standard YouTube License, CC-BY
Chemistry of Group 16 elements; Author: Ch-11 Chemical Engg, Chemistry and others;https://www.youtube.com/watch?v=5B1F0aDgL6s;License: Standard YouTube License, CC-BY