(a)
Interpretation:
The Lewis structure for
Concept Introduction:
Lewis structure is also known as Lewis dot diagrams or electron dot structures. The bond between atoms and lone pairs of electrons that is present in the molecule. Lewis structure represents each atom and their position in structure using the chemical symbol. Excess electrons forms the lone pair are given by pair of dots, and are located next to the atom.
(a)

Explanation of Solution
Oxygen is in Group 6A and Chlorine is in Group 7A and the valence electrons present in the
The two chlorine atoms connect with one Oxygen atom through single bonds.
Chlorine atoms attain octet by adding six electrons as dots in pairs.
Complete the octet of the two chlorine atom uses
Put the last six electrons on Oxygen atom.
The correct Lewis structure of the
Oxygen has eight electrons four in the bonds and four as dots, hence the structure is complete.
Hence, the total number of electrons can be counted as
(b)
Interpretation:
The Lewis structure for
Concept Introduction:
Refer part (a).
(b)

Explanation of Solution
Hydrogen atom is from Group one A and Oxygen atom is from group 6A, hence the valence electrons are
Complete the octet of the two chlorine atom uses
The incomplete Lewis structure of
The correct Lewis structure of the
Hence, the total number of electrons can be counted as
(c)
Interpretation:
The Lewis structure for
Concept Introduction:
Refer part (a).
(c)

Explanation of Solution
The four Hydrogen atoms connect to boron with single bonds uses eight electrons. Boron atom is the central atom with the hydrogen atoms around it. So, the valence electron present in the
Boron atom must be the central atom with the four Hydrogen atoms bonded to it. Boron has eight electrons so, the structure is complete. Boron has eight electrons, and each Hydrogen atom has just two electrons, the structure clockwise, the total number of electrons can be counted
The correct Lewis structure of the
(d)
Interpretation:
The Lewis structure for
Concept Introduction:
Refer part (a).
(d)

Explanation of Solution
The four Hydrogen atoms connect to Phosphorous with single bonds uses eight electrons. Phosphorous atom is the central atom with the hydrogen atoms around it. So, the valence electrons present in the
Phosphorous atom must be the central atom with the four Hydrogen atoms bonded to it. Phosphorous has eight electrons so, the structure is complete. It has eight electrons, and each Hydrogen atom has just two electrons, the structure clockwise, the total number of electrons can be counted
The structure is a
The correct Lewis structure of the
(e)
Interpretation:
The Lewis structure for
Concept Introduction:
Refer part (a).
(e)

Explanation of Solution
The five chlorine atoms connect to Phosphorous with single bonds uses ten electrons. Phosphorous atom is the central atom with the Chlorine atoms around it.
The number of valence electrons present in
Chlorine atoms prefer making only one bond, and Phosphorous prefers to make three and five bonds. So use Phosphorous atom as central atom with the five Chlorine atoms around it.
Each Chlorine atom has three lone pair and one bond pair so it attains octet.
The correct Lewis structure of the
The total number of electrons can be counted as
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Chapter 6 Solutions
OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
- For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. དྲ。 ✗MgBr ? O CI Will the first product that forms in this reaction create a new C-C bond? Yes No • ? Will the first product that forms in this reaction create a new CC bond? Yes No × : ☐ Xarrow_forwardPredict the major products of this organic reaction: OH NaBH4 H ? CH3OH Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. ☐ : Sarrow_forwardPredict the major products of this organic reaction: 1. LIAIHA 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. X : ☐arrow_forward
- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
- give example for the following(by equation) a. Converting a water insoluble compound to a soluble one. b. Diazotization reaction form diazonium salt c. coupling reaction of a diazonium salt d. indacator properties of MO e. Diazotization ( diazonium salt of bromobenzene)arrow_forward2-Propanone and ethyllithium are mixed and subsequently acid hydrolyzed. Draw and name the structures of the products.arrow_forward(Methanesulfinyl)methane is reacted with NaH, and then with acetophenone. Draw and name the structures of the products.arrow_forward
- 3-Oxo-butanenitrile and (E)-2-butenal are mixed with sodium ethoxide in ethanol. Draw and name the structures of the products.arrow_forwardWhat is the reason of the following(use equations if possible) a.) In MO preperation through diazotization: Addition of sodium nitrite in acidfied solution in order to form diazonium salt b.) in MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at low pH c.) In MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at pH 4.5 d.) Avoiding not cooling down the reaction mixture when preparing the diazonium salt e.) Cbvcarrow_forwardA 0.552-g sample of an unknown acid was dissolved in water to a total volume of 20.0 mL. This sample was titrated with 0.1103 M KOH. The equivalence point occurred at 29.42 mL base added. The pH of the solution at 10.0 mL base added was 3.72. Determine the molar mass of the acid. Determine the Ka of the acid.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
