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Chapter 9: Problem 81

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \end{aligned} $$ Cyanamid 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: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Short Answer

Expert verified
The short answer for this exercise is: a. Lewis structures: 1. NCN虏鈦�: N鈮鈮虏鈦� 2. H鈧侼CN: H鈧侼鈮=N 3. Dicyandiamide: H鈧侼-C=N-C-NH鈧� 4. Melamine: \[ \begin{array}{cccccc} & & {\mathrm{NH}_{2}}\\ &-&{\mathrm{CN}_{2}-}\\ - & {\circlearrowright} & {\circlearrowright} & {\circlearrowright} \\ &{\mathrm{CN}_{2}-}\\ & & {\mathrm{NH}_{2}}\\ \end{array} \] b. Hybridization: 1. NCN虏鈦�: N (sp), C (sp), N (sp) 2. H鈧侼CN: N (sp鲁), C (sp), N (sp虏) 3. Dicyandiamide: N (sp鲁), C (sp虏), N (sp虏), C (sp虏), N (sp鲁) 4. Melamine: N (sp鲁), C (sp虏) c. Sigma (蟽) and Pi (蟺) bonds: 1. In NCN虏鈦�: 2 蟽 bonds and 4 蟺 bonds 2. In H鈧侼CN: 4 蟽 bonds and 2 蟺 bonds 3. In dicyandiamide: 8 蟽 bonds and 2 蟺 bonds 4. In melamine: 12 蟽 bonds and 3 蟺 bonds d. Planarity of the ring in melamine: The ring is planar. e. Most important resonance structure of dicyandiamide: \[ \mathrm{H}_{2} \mathrm{N} - \mathrm{C} - \mathrm{N} - \mathrm{C} - \mathrm{N} \mathrm{H}_{2} \]

Step by step solution

01

Lewis Structures

: For each compound, we will first count the valence electrons, then arrange the atoms, and finally distribute the electrons in pairs to complete the octet rule. Consider resonance structures where appropriate. 1. NCN虏鈦�: Total valence electrons = 5 (from N) + 4 (from C) + 5 (from N) + 2 (for 2- charge) = 16. The Lewis structure is: \[ \mathrm{N} \equiv \mathrm{C} \equiv \mathrm{N}^{2-} \] 2. H鈧侼CN (Cyanamide): Total valence electrons = 2 (from 2H) + 5 (from N) + 4 (from C) + 5 (from N) = 16. The Lewis structure is: \[ \mathrm{H}_{2} \mathrm{N} \equiv \mathrm{C} = \mathrm{N} \] 3. Dicyandiamide: The molecular formula is H鈧凜鈧侼鈧�. Total valence electrons = 4 (from 4H) + 8 (from 2C) + 20 (from 4N) = 32. The Lewis structure is: \[ \mathrm{H}_{2} \mathrm{N} - \mathrm{C} = \mathrm{N} - \mathrm{C} - \mathrm{N} \mathrm{H}_{2} \] 4. Melamine: The molecular formula is C鈧僅鈧哊鈧�. Total valence electrons = 6 (from 6H) + 12 (from 3C) + 30 (from 6N) = 48. The Lewis structure is: \[ \begin{array}{cccccc} & & {\mathrm{NH}_{2}}\\ &-&{\mathrm{CN}_{2}-}\\ \boldsymbol{-} & {\circlearrowright} & {\circlearrowright} & {\circlearrowright} \\ &{\mathrm{CN}_{2}-}\\ & & {\mathrm{NH}_{2}}\\ \end{array} \] b.
02

Hybridization of Carbon and Nitrogen atoms

: 1. In NCN虏鈦�: N (sp), C (sp), N (sp) 2. In H鈧侼CN: N (sp鲁), C (sp), N (sp虏) 3. In dicyandiamide: N (sp鲁), C (sp虏), N (sp虏), C (sp虏), N (sp鲁) 4. In melamine: N (sp鲁), C (sp虏) c.
03

Sigma (蟽) and Pi (蟺) bonds in each species

: 1. In NCN虏鈦�: 2 蟽 bonds and 4 蟺 bonds 2. In H鈧侼CN: 4 蟽 bonds and 2 蟺 bonds 3. In dicyandiamide: 8 蟽 bonds and 2 蟺 bonds 4. In melamine: 12 蟽 bonds and 3 蟺 bonds d.
04

Planarity of the ring in melamine

: The ring in melamine is composed of alternating Carbon (sp虏) and Nitrogen (sp鲁) atoms. The sp虏 hybridization of Carbon atoms and the presence of a continuous 蟺 electron cloud around the Carbon and Nitrogen atoms in the ring make it planar. e.
05

Most important resonance structure of dicyandiamide

: Since the molecule is nonlinear and has three different C-N bond distances, the most important resonance structure should have higher electron density around the central Carbon atom with single bonds to both Nitrogen atoms. This effectively increases its bond length as compared to double-bonded C-N. Thus, the most important resonance structure is: \[ \mathrm{H}_{2} \mathrm{N} - \mathrm{C} - \mathrm{N} - \mathrm{C} - \mathrm{N} \mathrm{H}_{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structures
Lewis structures are a simple way to represent molecules and compounds visually, showing the arrangement of atoms and their valence electrons. They help us understand the bonding between atoms and the distribution of electrons.

**Steps to Draw Lewis Structures:**
  • Count the total number of valence electrons. This includes adding extra electrons for negative charges and subtracting for positive charges.
  • Arrange the atoms, placing the least electronegative atom in the center (excluding hydrogen).
  • Distribute the electrons as pairs to satisfy the octet rule (or duet rule for hydrogen).
  • Consider multiple resonance structures if different arrangements can be drawn.
For example, for NCN2-, the 16 valence electrons allow for a triple bond between the nitrogen and carbon atoms: \[ \mathrm{N} \equiv \mathrm{C} \equiv \mathrm{N}^{2-} \]This representation shows the linear arrangement of the atoms and helps visualize the molecule's structure.
Molecular Hybridization
Molecular hybridization explains how atomic orbitals mix to form new hybrid orbitals used in bonding. This concept helps predict molecular geometry and bonding properties, crucial in understanding molecular interactions.

**Types of Hybridization:**
  • **sp Hybridization:** Combines one s and one p orbital, forming two equivalent sp hybrid orbitals. Found in linear molecules like NCN虏鈦�, where the carbon atom exhibits sp hybridization.
  • **sp虏 Hybridization:** Involves one s and two p orbitals, creating three sp虏 hybrid orbitals. Common in trigonal planar molecules. Observed in melamine for carbon atoms.
  • **sp鲁 Hybridization:** Mixes one s and three p orbitals, resulting in four sp鲁 hybrid orbitals. Seen in tetrahedral molecules like ammonia, affecting the nitrogen atoms in H鈧侼CN.
Each type of hybridization dictates different angles and shapes, influencing the overall molecular geometry and properties.
Sigma and Pi Bonds
Chemical bonds in molecules can be categorized into sigma (蟽) and pi (蟺) bonds. Understanding these helps in analyzing molecular stability and reactivity.

**Sigma (蟽) Bonds:**
  • Formed by head-on overlapping of orbitals, creating a bond along the axis between atoms.
  • Typically stronger than pi bonds, allowing for free rotation around the bond axis.

**Pi (蟺) Bonds:**
  • Produced by the side-to-side overlap of p orbitals, above and below the sigma bond axis.
  • Weaker than sigma bonds, restricting rotation and often leading to multiple bonds.
In the example of cyanamide, H鈧侼CN, there are 4 sigma bonds and 2 pi bonds. This understanding helps us scrutinize the compound's structural features, as seen in the multiple bonds and rotations within the molecules.
Resonance Structures
Resonance structures are multiple ways to depict the same molecule where electrons can be differently distributed, providing insight into a molecule's stability and reactivity.

**Key Points About Resonance:**
  • Resonance structures are not real molecules themselves; rather, they show possible electron configurations.
  • The true structure of a molecule is a hybrid of all possible resonance structures, resulting in increased stability.
For instance, in dicyandiamide, various resonance structures are drawn with different distributions of single and double bonds between carbon and nitrogen atoms. Determining the most significant resonance form involves identifying structures with the most stable electron arrangements. The most contributing structure usually has the least charge separation and the most optimized bonding.

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Most popular questions from this chapter

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-}\), and NO using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C} \equiv \mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43 .\) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide \(\left(\mathrm{CO}_{2}\right)\), and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right)\). The spacefilling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{aligned} \mathrm{NCl}_{3}(g) & \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta H &=375 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) & \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & \Delta H &=158 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)
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