
(a)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(a)

Answer to Problem 100A
It is binary molecular compound.
Explanation of Solution
The given compound is
(b)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(b)

Answer to Problem 100A
It is an acid and binary ionic compound.
Explanation of Solution
The given compound is
(c)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(c)

Answer to Problem 100A
It is a base and ionic compound with polyatomic ion.
Explanation of Solution
The given compound is
(d)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(d)

Answer to Problem 100A
It is a binary ionic compound.
Explanation of Solution
The given compound is
(e)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(e)

Answer to Problem 100A
It is an acid and ionic compound with polyatomic ion.
Explanation of Solution
The given compound is
(f)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(f)

Answer to Problem 100A
It is a binary molecular compound.
Explanation of Solution
The given compound is
(g)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(g)

Answer to Problem 100A
It is a binary ionic compound.
Explanation of Solution
The given compound is
(h)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(h)

Answer to Problem 100A
It is an acid and ionic compound with polyatomic ion.
Explanation of Solution
The given compound is
(i)
Interpretation: The compounds are to be separated into five categories or more than one category.
Concept Introduction : Molecular compounds are formed by the sharing of electrons between non-metals. They are known as carbon compounds.
Ionic compounds are formed by the transfer of electrons between metal and non-metal.
Polyatomic ions are ions that have more than one atom. While naming polyatomic ions, the name of the less electronegative element is placed first, followed by the higher electronegative element. ‘−ate’ or ‘-ite’ is used as a suffix for anions.
(i)

Answer to Problem 100A
It is an ionic compound with polyatomic ion.
Explanation of Solution
The given compound is
Chapter 9 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- Iarrow_forwardDraw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward
- 4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)arrow_forward5. Consider the compounds shown below as ligands in coordination chemistry and identify their denticity; comment on their ability to form chelate complexes. (6 points) N N A B N N N IN N Carrow_forward
- 1. Use standard reduction potentials to rationalize quantitatively why: (6 points) (a) Al liberates H2 from dilute HCl, but Ag does not; (b) Cl2 liberates Br2 from aqueous KBr solution, but does not liberate C12 from aqueous KCl solution; c) a method of growing Ag crystals is to immerse a zinc foil in an aqueous solution of AgNO3.arrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1 1. PPh3 2. n-BuLi 3 2 • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Click and drag to start drawing a structure. Xarrow_forwardWhat is the missing reactant R in this organic reaction? N N H3O+ +R + • Draw the structure of R in the drawing area below. • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure. fmarrow_forward
- The product on the right-hand side of this reaction can be prepared from two organic reactants, under the conditions shown above and below the arrow. Draw 1 and 2 below, in any arrangement you like. 1+2 NaBH3CN H+ N Click and drag to start drawing a structure. 5arrow_forwardAssign this HSQC Spectrum ( please editing clearly on the image)arrow_forward(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY





