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Draw Lewis structures for the following compounds and

Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr ISBN: 9780321768414 33

Solution for problem 6P Chapter 1

Organic Chemistry | 8th Edition

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Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr

Organic Chemistry | 8th Edition

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Problem 6P

Draw Lewis structures for the following compounds and ions, showing appropriate formal charges.

Step-by-Step Solution:
Step 1 of 3

Solution 6P

Lewis formulas are written based on the number of electrons in the outermost energy level of the atoms involved. When atoms form covalent bond, they will share enough electrons so that each atom forms covalent bonds, they will share enough electrons so that each atom will have eight electrons in its outer energy level.

(a) :

First draw the skeletal structure as follows:

The total number of valence electrons is calculated as follows:

1O6= 6

1C4= 4

5H1= 5

____________

           = 15

So, there are 15 valence electrons.

Subtract 12 electrons to account for the bonds in the skeletal structure by leaving 3 electrons to distribute. Thus, 2 electrons are distributed as lone pairs on the oxygen atom.

In this structure, there is one electron less on the oxygen atom, which is not satisfied with the octet rule. So, we have to assign a formal charge on the oxygen atom.

Now, the Lewis structure shows that the oxygen atom shared three bonding pair of electrons and two nonbonding electrons. So, the formal charge oxygen is calculated by using the formula as follows:

Formal charge (FC) = Group number-Nonbonding electrons
                        -
(Shared electrons)

For oxygen, FC = 6-2-(6)

                   = + 1

Hence, by assigning the formal charges to the structure, the final Lewis structure is written as follows:

(b)

First draw the skeletal structure as follows:

The total number of valence electrons is calculated as follows:

1N5= 5

1 Cl7= 7

4H1 =  4

_____________

            = 16

So, there are 16 valence electrons.

Subtract 8 electrons to account for the bonds in the skeletal structure by leaving 8 electrons to distribute. Thus, 8 electrons are distributed as lone pairs on the chlorine atom. So, the Lewis structure is shown as follows:

Now, the Lewis structure shows that the nitrogen atom shared four bonding pair of electrons and chlorine has no shared pair of electrons. So, the formal charge for nitrogen and chlorine is calculated by using the formula as follows:

Formal charge (FC) = Group number-Nonbonding electrons
                         -
(Shared electrons)

For nitrogen, FC = 5-0 -(8)
                   = +1

For chlorine, FC = 7-8-(0)

                  = -1

Hence, by assigning the formal charges to the structure, the final Lewis structure is written as follows:

(c)  N CI

First draw the skeletal structure as follows:

The total number of valence electrons is calculated as follows:

1N5= 5

1Cl 7=7

4C4= 16

12H 1= 12

_______________

              = 40

So, there are 40 valence electrons.

Subtract 32 electrons to account for the bonds in the skeletal structure by leaving 8 electrons to distribute. Thus, 8 electrons are distributed as lone pairs on the chlorine atom. So, the Lewis structure is shown as follows:

Now, the Lewis structure shows that the nitrogen atom shared four bonding pair of electrons and chlorine has no shared pair of electrons. So, the formal charge for nitrogen and chlorine is calculated by using the formula as follows:

Formal charge(FC) = Group number- Nonbonding electrons
                        -
(Shared electrons)

For nitrogen, FC = 5 - 0 - (8)

                 = + 1

For chlorine, FC = 7-8 -(0)

                  =-1        

Hence, by assigning the formal charges to the structure, the final Lewis structure is written as follows:

(d) Na :

First draw the skeletal structure as follows:

The total number of valence electrons is calculated as follows:

1O6= 6

1Na1= 1

1C4= 4

3H1= 3

_____________

           =14

So, there are 14 valence electrons.

Subtract 8 electrons to account for the bonds in the skeletal structure by leaving 6 electrons to distribute. Thus, 6 electrons are distributed as lone pairs on the oxygen atom. So, the Lewis structure is shown as follows:

Step 2 of 3

Chapter 1, Problem 6P is Solved
Step 3 of 3

Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

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Draw Lewis structures for the following compounds and