Concept explainers
(a)
Interpretation:
Lewis dot structure of CF4 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(a)
Answer to Problem 5E
Lewis dot structure of CF4 is given below.
. 
Explanation of Solution
C has 4 valence electrons. F has 7 valence electrons. Thus one carbon is covalently bonded with 4 F atoms. There are three lone pairs of electrons on each F atom. Accordingly Lewis structure is shown as follows:
(b)
Interpretation:
Lewis dot structure of CH3Cl must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(b)
Answer to Problem 5E
Lewis dot structure of CH3Cl is given below.
. 
Explanation of Solution
C has 4 valence electrons. Cl has 7 valence electrons. H has one valence electron. Thus one carbon is covalently bonded with 3 H atoms and 1 Cl. There are three lone pairs of electrons on Cl atom. Accordingly Lewis structure is shown as follows:
(c)
Interpretation:
Lewis dot structure of SiCl2H2 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(c)
Answer to Problem 5E
Lewis dot structure of SiCl2H2 is given below.
.
Explanation of Solution
Si has 4 valence electrons, Cl has 7 valence electrons and H has one valence electron. Thus one carbon is covalently bonded with 2 H atoms and 2 Cl atoms. There are three lone pairs of electrons on Cl atom. Accordingly Lewis structure is shown as follows:
(d)
Interpretation:
Lewis dot structure of CH3OH must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(d)
Answer to Problem 5E
Lewis dot structure of CH3OH is given below:
.
Explanation of Solution
C has 4 valence electrons. O has 6 valence electrons. H has one valence electron. Thus one carbon is covalently bonded with 3 H atoms and 1 O atom. The O atom is bonded with central carbon atom and one hydrogen atom. There are 2 lone pairs of electrons on O atom. Accordingly, Lewis structure is shown as follows:
(e)
Interpretation:
Lewis dot structure of HOCl must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(e)
Answer to Problem 5E
Lewis dot structure of HOCl is given below.
.
Explanation of Solution
O has 6 valence electrons, Cl has 7valence electrons and H has one valence electron. Thus one central O is covalently bonded with 1 H atoms and 1 Cl atom. There are 2 lone pairs of electrons on O atom and 3 lone pairs of electrons on Cl atom. Accordingly Lewis structure is shown as follows
(f)
Interpretation:
Lewis dot structure of CH3NH2 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule with the valence electrons shown as dots.
(f)
Answer to Problem 5E
Lewis dot structure of CH3NH2 is given below.
.
Explanation of Solution
C has 4 valence electrons, N has 5 valence electrons and H has one valence electron. Thus one central C is covalently bonded with 3 H atoms and 1 N atom. N atom is covalently bonded with central C atom and two hydrogen atoms. There are 2 lone pairs of electrons on N atom. Accordingly Lewis structure is shown as follows:
Chapter U2 Solutions
Living by Chemistry
Additional Science Textbook Solutions
Cosmic Perspective Fundamentals
Applications and Investigations in Earth Science (9th Edition)
Introductory Chemistry (6th Edition)
Biology: Life on Earth with Physiology (11th Edition)
Biology: Life on Earth (11th Edition)
Microbiology: An Introduction
- Phenol is the starting material for the synthesis of 2,3,4,5,6-pentachlorophenol, known al-ternatively as pentachlorophenol, or more simply as penta. At one time, penta was widely used as a wood preservative for decks, siding, and outdoor wood furniture. Draw the structural formula for pentachlorophenol and describe its synthesis from phenol.arrow_forward12 Mass Spectrometry (d) This unknown contains oxygen, but it does not show any significant infrared absorption peaks above 3000 cm . 59 100- BO 40 Relative Abundance M(102) - 15 20 25 30 35 40 45 50 5 60 65 70 75 80 85 90 95 100 105 mizarrow_forwardDraw a Haworth projection of a common cyclic form of this monosaccharide: H HO H HO H HO H H -OH CH2OH Click and drag to start drawing a structure. Х : Darrow_forward
- : Draw the structure of valylasparagine, a dipeptide made from valine and asparagine, as it would appear at physiological pH. Click and drag to start drawing a structure. P Darrow_forwardDraw the Haworth projection of α-L-mannose. You will find helpful information in the ALEKS Data resource. Click and drag to start drawing a structure. : ཊི Х Darrow_forwardDraw the structure of serine at pH 6.8. Click and drag to start drawing a structure. : d كarrow_forward
- Take a look at this molecule, and then answer the questions in the table below it. CH2OH H H H OH OH OH CH2OH H H H H OH H H OH H OH Is this a reducing sugar? yes α β ロ→ロ no ☑ yes Does this molecule contain a glycosidic bond? If you said this molecule does contain a glycosidic bond, write the symbol describing it. O no 0+0 If you said this molecule does contain a glycosidic bond, write the common names (including anomer and enantiomer labels) of the molecules that would be released if that bond were hydrolyzed. If there's more than one molecule, separate each name with a comma. ☐arrow_forwardAnswer the questions in the table below about this molecule: H₂N-CH₂ -C—NH–CH–C—NH–CH—COO- CH3 CH CH3 What kind of molecule is this? 0= CH2 C If you said the molecule is a peptide, write a description of it using 3-letter codes separated ☐ by dashes. polysaccharide peptide amino acid phospolipid none of the above Хarrow_forwardDraw a Haworth projection of a common cyclic form of this monosaccharide: CH₂OH C=O HO H H -OH H OH CH₂OH Click and drag to start drawing a structure. : ☐ Х S '☐arrow_forward
- Nucleophilic Aromatic Substitution 22.30 Predict all possible products formed from the following nucleophilic substitution reactions. (a) (b) 9 1. NaOH 2. HCI, H₂O CI NH₁(!) +NaNH, -33°C 1. NaOH 2. HCl, H₂Oarrow_forwardSyntheses 22.35 Show how to convert toluene to these compounds. (a) -CH,Br (b) Br- -CH3 22.36 Show how to prepare each compound from 1-phenyl-1-propanone. 1-Phenyl-1-propanone ہتی. Br. (b) Br (racemic) 22.37 Show how to convert ethyl benzene to (a) 2,5-dichlorobenzoic acid and (b) 2,4-dichlorobenzoic acid. 22.38 Show reagents and conditions to bring about the following conversions. (a) 9 NH2 8 CO₂H NH2 CO₂Et (d) NO2 NH2 S NH₂ NO2 CHS CHarrow_forwardive the major organic product(s) of each of the following reactions or sequences of reactions. Show all rant stereochemistry. [10 only] A. B. NaN3 1. LiAlH4, ether Br 2. H₂O CH3 HNO3 H₂/Pt H₂SO ethanol C. 0 0 CH3CC1 NaOH NHCCH AICI H₂O . NH₂ CH3CH2 N CH2CH3 + HCI CH₂CH 3 1. LIAIH, THE 2. H₂Oarrow_forward
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