Glencoe Chemistry: Matter and Change, Student Edition
1st Edition
ISBN: 9780076774609
Author: McGraw-Hill Education
Publisher: MCGRAW-HILL HIGHER EDUCATION
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 7, Problem 51A
Interpretation Introduction
Interpretation:
The formation of sulfur ions needs to be explained.
Concept introduction:
The electronic configuration of atomic species is followed by set of rules defined by Aufbau’s principle, Hund’s Rule, Pauli-Exclusion principle. Each orbital is occupied by two electrons of opposite spins. As per Aufbau’s principle, electrons will occupy orbital in order of increasing energy. According to Hund’s Rule, orbitals of same energy will be filled with one electron each having the same spin.
Expert Solution & Answer
Answer to Problem 51A
Sulfur can take up two electrons to become an octet.
Explanation of Solution
Given that the orbital notation of sulfur is given as
Figure 1
From the orbital notation, it can be seen that the 3p orbital has vacancy for two electrons. In other words, it can take in two electrons. To get an octet configuration, elements generally give up or take in electrons to get a stable configuration. If sulfur take up electrons, it will achieve octet configuration and become stable.
Hence, sulfur will take up two electrons to form a stable configuration.
Conclusion
Sulfur can take up two electrons to become an octet.
Chapter 7 Solutions
Glencoe Chemistry: Matter and Change, Student Edition
Ch. 7.1 - Prob. 1SSCCh. 7.1 - Prob. 2SSCCh. 7.1 - Prob. 3SSCCh. 7.1 - Prob. 4SSCCh. 7.1 - Prob. 5SSCCh. 7.1 - Prob. 6SSCCh. 7.2 - Prob. 7PPCh. 7.2 - Prob. 8PPCh. 7.2 - Prob. 9PPCh. 7.2 - Prob. 10PP
Ch. 7.2 - Prob. 11PPCh. 7.2 - Prob. 12SSCCh. 7.2 - Prob. 13SSCCh. 7.2 - Prob. 14SSCCh. 7.2 - Prob. 15SSCCh. 7.2 - Prob. 16SSCCh. 7.2 - Prob. 17SSCCh. 7.2 - Prob. 18SSCCh. 7.3 - Prob. 19PPCh. 7.3 - Prob. 20PPCh. 7.3 - Prob. 21PPCh. 7.3 - Prob. 22PPCh. 7.3 - Prob. 23PPCh. 7.3 - Prob. 24PPCh. 7.3 - Prob. 25PPCh. 7.3 - Prob. 26PPCh. 7.3 - Prob. 27PPCh. 7.3 - Prob. 28PPCh. 7.3 - Prob. 29PPCh. 7.3 - Prob. 30PPCh. 7.3 - Prob. 31PPCh. 7.3 - Prob. 32PPCh. 7.3 - Prob. 33PPCh. 7.3 - Prob. 34SSCCh. 7.3 - Prob. 35SSCCh. 7.3 - Prob. 36SSCCh. 7.3 - Prob. 37SSCCh. 7.3 - Prob. 38SSCCh. 7.3 - Prob. 39SSCCh. 7.4 - Prob. 40SSCCh. 7.4 - Prob. 41SSCCh. 7.4 - Prob. 42SSCCh. 7.4 - Prob. 43SSCCh. 7.4 - Prob. 44SSCCh. 7.4 - Prob. 45SSCCh. 7 - How do positive ions and negative ions form?Ch. 7 - Prob. 47ACh. 7 - Prob. 48ACh. 7 - Prob. 49ACh. 7 - Prob. 50ACh. 7 - Prob. 51ACh. 7 - Prob. 52ACh. 7 - Prob. 53ACh. 7 - Prob. 54ACh. 7 - Prob. 55ACh. 7 - Prob. 56ACh. 7 - Prob. 57ACh. 7 - Prob. 58ACh. 7 - Prob. 59ACh. 7 - Prob. 60ACh. 7 - Prob. 61ACh. 7 - Prob. 62ACh. 7 - Prob. 63ACh. 7 - Prob. 64ACh. 7 - Prob. 65ACh. 7 - Prob. 66ACh. 7 - Prob. 67ACh. 7 - Prob. 68ACh. 7 - Prob. 69ACh. 7 - Prob. 70ACh. 7 - Prob. 71ACh. 7 - Prob. 72ACh. 7 - Prob. 73ACh. 7 - Prob. 74ACh. 7 - Prob. 75ACh. 7 - Prob. 76ACh. 7 - Prob. 77ACh. 7 - Prob. 78ACh. 7 - Prob. 79ACh. 7 - Prob. 80ACh. 7 - Prob. 81ACh. 7 - Prob. 82ACh. 7 - Prob. 83ACh. 7 - Prob. 84ACh. 7 - Prob. 85ACh. 7 - Prob. 86ACh. 7 - Prob. 87ACh. 7 - Prob. 88ACh. 7 - Prob. 89ACh. 7 - Prob. 90ACh. 7 - Prob. 91ACh. 7 - Prob. 92ACh. 7 - Prob. 93ACh. 7 - Prob. 94ACh. 7 - Prob. 95ACh. 7 - Prob. 96ACh. 7 - Prob. 97ACh. 7 - Prob. 98ACh. 7 - Prob. 99ACh. 7 - Prob. 100ACh. 7 - Prob. 101ACh. 7 - Prob. 102ACh. 7 - Prob. 103ACh. 7 - Prob. 104ACh. 7 - Prob. 105ACh. 7 - Prob. 106ACh. 7 - Prob. 107ACh. 7 - Prob. 108ACh. 7 - Prob. 109ACh. 7 - Prob. 110ACh. 7 - Prob. 111ACh. 7 - Prob. 112ACh. 7 - Prob. 113ACh. 7 - Prob. 114ACh. 7 - Prob. 115ACh. 7 - Prob. 116ACh. 7 - Prob. 117ACh. 7 - Prob. 118ACh. 7 - Prob. 119ACh. 7 - Prob. 120ACh. 7 - Prob. 121ACh. 7 - Prob. 122ACh. 7 - Prob. 123ACh. 7 - Prob. 124ACh. 7 - Prob. 125ACh. 7 - Prob. 126ACh. 7 - Prob. 127ACh. 7 - Prob. 128ACh. 7 - Prob. 129ACh. 7 - Prob. 130ACh. 7 - Prob. 131ACh. 7 - Prob. 132ACh. 7 - Prob. 133ACh. 7 - Prob. 134ACh. 7 - Prob. 1STPCh. 7 - Prob. 2STPCh. 7 - Prob. 3STPCh. 7 - Prob. 4STPCh. 7 - Prob. 5STPCh. 7 - Prob. 6STPCh. 7 - Prob. 7STPCh. 7 - Prob. 8STPCh. 7 - Prob. 9STPCh. 7 - Prob. 10STPCh. 7 - Prob. 11STPCh. 7 - Prob. 12STPCh. 7 - Prob. 13STPCh. 7 - Prob. 14STPCh. 7 - Prob. 15STPCh. 7 - Prob. 16STPCh. 7 - Prob. 17STPCh. 7 - Prob. 18STPCh. 7 - Prob. 19STPCh. 7 - Prob. 20STP
Additional Science Textbook Solutions
Find more solutions based on key concepts
Police Captain Jeffers has suffered a myocardial infarction. a. Explain to his (nonmedically oriented) family w...
Human Physiology: An Integrated Approach (8th Edition)
2. The three ropes in FIGURE EX6.2 are tied to a small, very light ring. Two of the ropes are anchored to wa...
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
1. ___ Mitosis 2. ___ Meiosis 3. __ Homologous chromosomes 4. __ Crossing over 5. __ Cytokinesis A. Cytoplasmic...
Microbiology with Diseases by Body System (5th Edition)
1. Why is the quantum-mechanical model of the atom important for understanding chemistry?
Chemistry: Structure and Properties (2nd Edition)
Fill in the blanks: The nose is to the mouth. The ankle is to the knee. The ring finger is to the inde...
Human Anatomy & Physiology (2nd Edition)
What percentage of Earths land surface do glaciers presently cover? ____________
Applications and Investigations in Earth Science (9th Edition)
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
- A block of zinc has an initial temperature of 94.2 degrees celcius and is immererd in 105 g of water at 21.90 degrees celcius. At thermal equilibrium, the final temperature is 25.20 degrees celcius. What is the mass of the zinc block? Cs(Zn) = 0.390 J/gxdegrees celcius Cs(H2O) = 4.18 J/gx degrees celcusarrow_forwardPotential Energy (kJ) 1. Consider these three reactions as the elementary steps in the mechanism for a chemical reaction. AH = -950 kJ AH = 575 kJ (i) Cl₂ (g) + Pt (s) 2C1 (g) + Pt (s) Ea = 1550 kJ (ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s) (iii) Cl (g) + CICO (g) → Cl₂CO (g) Ea = 2240 kJ Ea = 2350 kJ AH = -825 kJ 2600 2400 2200 2000 1800 1600 1400 1200 1000 a. Draw the potential energy diagram for the reaction. Label the data points for clarity. The potential energy of the reactants is 600 kJ 800 600 400 200 0 -200- -400 -600- -800- Reaction Progressarrow_forwardCan u help me figure out the reaction mechanisms for these, idk where to even startarrow_forward
- Hi, I need your help with the drawing, please. I have attached the question along with my lab instructions. Please use the reaction from the lab only, as we are not allowed to use outside sources. Thank you!arrow_forwardHi, I need your help i dont know which one to draw please. I’ve attached the question along with my lab instructions. Please use the reaction from the lab only, as we are not allowed to use outside sources. Thank you!arrow_forward5. Write the formation reaction of the following complex compounds from the following reactants: 6. AgNO₃ + K₂CrO₂ + NH₄OH → 7. HgNO₃ + excess KI → 8. Al(NO₃)₃ + excess NaOH →arrow_forward
- Indicate whether the product formed in the reaction exhibits tautomerism. If so, draw the structure of the tautomers. CO₂C2H5 + CH3-NH-NH,arrow_forwardDraw the major product of this reaction N-(cyclohex-1-en-1-yl)-1-(pyrrolidino) reacts with CH2=CHCHO, heat, H3O+arrow_forwardDraw the starting material that would be needed to make this product through an intramolecular Dieckmann reactionarrow_forward
- Draw the major product of this reaction. Nitropropane reacts + pent-3-en-2-one reacts with NaOCH2CH3, CH3CHOHarrow_forwardIndicate whether the product formed in the reaction exhibits tautomerism. If so, draw the structure of the tautomers. OC2H5 + CoHs-NH-NH,arrow_forwardExplain how substitutions at the 5-position of barbituric acid increase the compound's lipophilicity.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles 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 Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
The Bohr Model of the atom and Atomic Emission Spectra: Atomic Structure tutorial | Crash Chemistry; Author: Crash Chemistry Academy;https://www.youtube.com/watch?v=apuWi_Fbtys;License: Standard YouTube License, CC-BY