Chemistry (Instructor's)
Chemistry (Instructor's)
10th Edition
ISBN: 9781305957787
Author: ZUMDAHL
Publisher: CENGAGE L
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 9, Problem 95CP

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

Chapter 9, Problem 95CP, 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 95CP, 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
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 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,

Chemistry (Instructor's), Chapter 9, Problem 95CP , 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,

Chemistry (Instructor's), Chapter 9, Problem 95CP , 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,

Chemistry (Instructor's), Chapter 9, Problem 95CP , 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,

Chemistry (Instructor's), Chapter 9, Problem 95CP , 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 95CP

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 95CP

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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Which is the most appropriate Lewis structure for the "interhalogen" compound chlorine monofluoride, CIF, and how many lone pairs of electrons does each atom have? If we combine two atoms with a total of 14 valence electrons, how many valence electrons will there be? How many electrons does an atom need to make an octet? Octets cannot be applied to halogens :CH-F: CI-F:: :CI=F:
7. Ethylene dichloride (C2H4C12) is used to make vinyl chloride (C2H3C1), which, in turn, is used to manufacture the plastic poly(vinyl chloride) (PVC), found in piping, siding, floor tiles, and toys. Write the Lewis structures of ethylene dichloride and vinyl chloride. Classify the bonds as covalent or polar.
Phosphorus pentachloride, a key industrial compound with annual world production of about 2107kg, is used to make other compounds. It reacts with sulfur dioxide to produce phos-phorus oxychloride (POCl₃) and thionyl chloride (SOCl₂).Draw a Lewis structure, and name the molecular shape of each product.

Chapter 9 Solutions

Chemistry (Instructor's)

Ch. 9 - Which is the more correct statement: The methane...Ch. 9 - Compare and contrast the MO model with the local...Ch. 9 - What are the relationships among bond order, bond...Ch. 9 - The molecules N2 and CO are isoelectronic but...Ch. 9 - Do lone pairs about a central atom affect the...Ch. 9 - In the hybrid orbital model, compare and contrast ...Ch. 9 - In the molecular orbital mode l, compare and...Ch. 9 - Why are d orbitals sometimes used to form hybrid...Ch. 9 - The atoms in a single bond can rotate about the...Ch. 9 - As compared with CO and O2, CS and S2 are very...Ch. 9 - Compare and contrast bonding molecular orbitals...Ch. 9 - What modification to the molecular orbital model...Ch. 9 - Why does the molecular orbital model do a better...Ch. 9 - The three NO bonds in NO3 are all equivalent in...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - The space-filling models of ethane and ethanol are...Ch. 9 - The space-filling models of hydrogen cyanide and...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - For each of the following molecules, write the...Ch. 9 - For each of the following molecules or ions that...Ch. 9 - Prob. 35ECh. 9 - The allene molecule has the following Lewis...Ch. 9 - Indigo is the dye used in coloring blue jeans. The...Ch. 9 - Urea, a compound formed in the liver, is one of...Ch. 9 - Biacetyl and acetoin are added to margarine to...Ch. 9 - Many important compounds in the chemical industry...Ch. 9 - Two molecules used in the polymer industry are...Ch. 9 - Hot and spicy foods contain molecules that...Ch. 9 - One of the first drugs to be approved for use in...Ch. 9 - Minoxidil (C9H15N15O) is a compound produced by...Ch. 9 - Consider the following molecular orbitals formed...Ch. 9 - Sketch the molecular orbital and label its type (...Ch. 9 - Which of the following are predicted by the...Ch. 9 - Which of the following are predicted by the...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - Consider the following electron configuration:...Ch. 9 - Using the molecular orbital model to describe the...Ch. 9 - A Lewis structure obeying the octet rule can be...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - In which of the following diatomic molecules would...Ch. 9 - In terms of the molecular orbital model, which...Ch. 9 - Show how two 2p atomic orbitals can combine to...Ch. 9 - Show how a hydrogen 1s atomic orbital and a...Ch. 9 - Use Figs. 4-54 and 4-55 to answer the following...Ch. 9 - Acetylene (C2H2) can be produced from the reaction...Ch. 9 - Describe the bonding in NO+, NO, and NO, using...Ch. 9 - Describe the bonding in the O3 molecule and the...Ch. 9 - Describe the bonding in the CO32 ion using the...Ch. 9 - Draw the Lewis structures, predict the molecular...Ch. 9 - The antibiotic thiarubin-A was discovered by...Ch. 9 - Two structures can be drawn for cyanuric acid: a....Ch. 9 - Give the expected hybridization for the molecular...Ch. 9 - Vitamin B6 is an organic compound whose deficiency...Ch. 9 - Aspartame is an artificial sweetener marketed...Ch. 9 - Prob. 73AECh. 9 - The three most stable oxides of carbon are carbon...Ch. 9 - Complete the following resonance structures for...Ch. 9 - Prob. 77AECh. 9 - The transport of O2 in the blood is carried out by...Ch. 9 - Using molecular orbital theory, explain why the...Ch. 9 - Describe the bonding in the first excited state of...Ch. 9 - Using an MO energy-level diagram, would you expect...Ch. 9 - Show how a dxz. atomic orbital and a pz, atomic...Ch. 9 - What type of molecular orbital would result from...Ch. 9 - Consider three molecules: A, B, and C. Molecule A...Ch. 9 - Draw the Lewis structures for TeCl4, ICl5, PCl5,...Ch. 9 - A variety of chlorine oxide fluorides and related...Ch. 9 - Pelargondin is the molecule responsible for the...Ch. 9 - Complete a Lewis structure for the compound shown...Ch. 9 - Which of the following statements concerning SO2...Ch. 9 - Consider the molecular orbital electron...Ch. 9 - Place the species B2+ , B2, and B2 in order of...Ch. 9 - Consider the following computer-generated model of...Ch. 9 - Cholesterol (C27liu;O) has the following...Ch. 9 - Cyanamide (H2NCN), an important industrial...Ch. 9 - A flask containing gaseous N2 is irradiated with...Ch. 9 - Values of measured bond energies may vary greatly...Ch. 9 - Use the MO model to explain the bonding in BeH2....Ch. 9 - Prob. 101CPCh. 9 - Arrange the following from lowest to highest...Ch. 9 - Use the MO model to determine which of the...Ch. 9 - Given that the ionization energy of F2 is 290...Ch. 9 - Carbon monoxide (CO) forms bonds to a variety of...Ch. 9 - Prob. 106CPCh. 9 - As the bead engineer of your starship in charge of...Ch. 9 - Determine the molecular structure and...Ch. 9 - Although nitrogen trifluoride (NF3) is a thermally...
Knowledge Booster
Background pattern image
Chemistry
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY