OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th
OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th
9th Edition
ISBN: 9781285185446
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 9, Problem 89CP

Cyanamide (H2NCN), an important industrial chemical, is produced by the following steps:

Chapter 9, Problem 89CP, Cyanamide (H2NCN), an important industrial chemical, is produced by the following steps: Calcium , example  1

Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics:

Chapter 9, Problem 89CP, Cyanamide (H2NCN), an important industrial chemical, is produced by the following steps: Calcium , example  2

a. Write Lewis structures for NCN2, H2NCN, dicyandiarnide, and melamine, including resonance structures where appropriate.

b. Give the hybridization of the C and N atoms in each species.

c. How many σ bonds and how many π bonds are in each species?

d. Is the ring in melamine planar?

e. There are three different C—N bond distances in dicyandiamide, NCNC(NH2)2, and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

(a)

Expert Solution
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Interpretation Introduction

Interpretation: The reactions leading to the formation of dicyandiamide and melamine are given. The answers to the various questions on basis of these are to be determined.

Concept introduction: The hybridization of an atom can be obtained by finding its steric number. The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

To determine: The Lewis structures for NCN2-,H2NCN , dicyandiamide and melamine, along with their resonance structures.

Explanation of Solution

The electronic configurations of the elements present are,

H=1s1C=1s22s22p2N=1s22s22p3

The number of valence electrons present in carbon is four and in nitrogen are five. Hydrogen shows the presence of only one electron.

The total number of valence electrons in NCN2 ,

2N+C+2e=((2×5)+4+2)e=16e

The predicted Lewis structures for NCN2 are,

OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th, Chapter 9, Problem 89CP , additional homework tip  1

Figure 1

The total number of valence electrons in H2NCN ,

2H+2N+C=(2+(2×5)+4)e=16e

The predicted Lewis structures for H2NCN are,

OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th, Chapter 9, Problem 89CP , additional homework tip  2

Figure 2

The total number of valence electrons in dicyandiamide,

4H+4N+2C=((4×1)+(4×5)+(2×4))e=32e

The predicted Lewis structures for dicyandiamide are,

OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th, Chapter 9, Problem 89CP , additional homework tip  3

Figure 3

The total number of valence electrons in melamine,

6H+6N+3C=((6×1)+(6×5)+(3×4))e=48e

The predicted Lewis structures for melamine are,

OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl's Chemistry, 9th, Chapter 9, Problem 89CP , additional homework tip  4

Figure 4

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reactions leading to the formation of dicyandiamide and melamine are given. The answers to the various questions on basis of these are to be determined.

Concept introduction: The hybridization of an atom can be obtained by finding its steric number. The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

To determine: The hybridization of the carbon and nitrogen atoms for each molecule.

Explanation of Solution

The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

The hybridization can hence be obtained by calculating the value of the steric number for an atom.

For NCN2- ,

In the given structure,

  • The nitrogen atom is sp2 hybridized (steric number 3 ) and the carbon atom is sp hybridized (steric number 2 ).

For H2NCN ,

The resonance structure (II) is more stable.

In the given structure,

  • One nitrogen atom is sp3 hybridized (steric number 4 ), The other nitrogen atom is sp hybridized (steric number 2 ) and the carbon atom is sp hybridized (steric number 2 ).

For melamine,

In the given structure,

  • One nitrogen atom present in the ring is sp2 hybridized (steric number 3 ), The other nitrogen atom present in the NH2 group is sp3 hybridized (steric number 4 ) and the carbon atom present is sp2 hybridized (steric number 3 ).

For dicyandiamide,

In the given structure,

  • One carbon atom is sp hybridized and the other is sp2 hybridized.
  • Four nitrogen atoms are present,

One nitrogen atom present is sp2 hybridized (steric number 3 ).

One nitrogen atom present is sp hybridized (steric number 2 ).

The remaining two nitrogen atoms are sp3 hybridized (steric number 4 ).

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reactions leading to the formation of dicyandiamide and melamine are given. The answers to the various questions on basis of these are to be determined.

Concept introduction: The hybridization of an atom can be obtained by finding its steric number. The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

To determine: The number of sigma (σ) and pi (π) bonds present in the structure of the given molecules.

Answer to Problem 89CP

The ring present in the case of melamine is planar.

Explanation of Solution

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

For NCN2- ,

In the structure for the given compound,

  • 2 sigma bonds are present between atoms.
  • The presence of 2 pi bonds is observed.

For H2NCN ,

In the structure for the given compound,

  • 4 sigma bonds are present between atoms.
  • The presence of 2 pi bonds is observed.

For melamine,

In the structure for the given compound,

  • 15 sigma bonds are present between atoms.
  • The presence of 3 pi bonds is observed.

For dicyandiamide,

In the structure for the given compound,

  • 9 sigma bonds are present between atoms.
  • The presence of 3 pi bonds is observed.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reactions leading to the formation of dicyandiamide and melamine are given. The answers to the various questions on basis of these are to be determined.

Concept introduction: The hybridization of an atom can be obtained by finding its steric number. The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

To determine: A justification on the planar nature of the ring present in the structure of melamine.

Explanation of Solution

All the atoms involved in the formation of a ring in the structure of the melamine ring have the same sp2 hybridization. The π electrons are delocalized over the six atoms present. Hence, this ring is considered planar.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The reactions leading to the formation of dicyandiamide and melamine are given. The answers to the various questions on basis of these are to be determined.

Concept introduction: The hybridization of an atom can be obtained by finding its steric number. The sum of the numbers of atoms bonded to the required atom and the number of lone pairs the atom has is known as the steric number.

If the steric number is 4 , the atom is sp3 hybridized.

If the steric number is 3 , the atom is sp2 hybridized.

If the steric number is 2 , the atom is sp hybridized.

Sigma bonds are the single bonds present between atoms in a compound. The second bonding that occurs between the atoms is known as the pi bonding.

To determine: The most important resonance structure for dicyandiamide.

Answer to Problem 89CP

The structure with the least formal charge is the most stable.

Explanation of Solution

 Explanation:

Formula

Formal charge =V[N+B2]

Where,

  • V is the number of valence electrons of the neutral atom.
  • N is the number of non-bonding valence electrons.
  • B is the total number of shared electrons.

For structure (I) for dicyandiamide,

For the hydrogen atom,

Formal charge =1[0+(12×2)]=0

For the single bonded nitrogen atom,

Formal charge =5[2+(12×6)]=0

For the double bonded central nitrogen atom,

Formal charge =5[0+(12×8)]=1

For the double bonded terminal nitrogen atom,

Formal charge =5[4+(12×4)]=1

For the carbon atoms,

Formal charge =4[0+(12×8)]=0

For structure (II) for dicyandiamide,

For the hydrogen atom,

Formal charge =1[0+(12×2)]=0

For the single bonded nitrogen atom,

Formal charge =5[2+(12×6)]=0

For the double bonded central nitrogen atom,

Formal charge =5[2+(12×6)]=0

For the triple bonded terminal nitrogen atom,

Formal charge =5[2+(12×6)]=0

For the double bonded carbon atom,

Formal charge =4[0+(12×8)]=0

For the triple bonded carbon atom,

Formal charge =4[0+(12×8)]=0

The resonance structure (II) has a lesser formal charge. Hence, irt is the more stable structure of dicyandiamide.

Conclusion

The resonance structure having the least value of formal charge is termed as the most stable resonance structure of a molecule. The most important resonance structure for dicyandiamide is the structure (II).

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