Chapter 4, Problem 115CP

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

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 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.

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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^2$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ 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 ${\text{NCN}}^{\text{2-}},{\text{H}}_{\text{2}}\text{NCN}$, dicyandiamide and melamine, along with their resonance structures.

Explanation of Solution

The electronic configurations of the elements present are,

$\begin{array}{l}\text{H}=1{\text{s}}^1\\ \text{C}=1{\text{s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^2\\ \text{N}=1{\text{s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}\end{array}$

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 ${\text{NCN}}^{2-}$,

$\text{2N}+\text{C}+2{\text{e}}^{-}=\left(\left(2\times 5\right)+4+2\right){\text{e}}^{-}=16{\text{e}}^{-}$

The predicted Lewis structures for ${\text{NCN}}^{2-}$ are,

Figure 1

The total number of valence electrons in ${\text{H}}_2\text{NCN}$,

$\text{2H+2N}+\text{C}=\left(2+\left(2\times 5\right)+4\right){\text{e}}^{-}=16{\text{e}}^{-}$

The predicted Lewis structures for ${\text{H}}_2\text{NCN}$ are,

Figure 2

The total number of valence electrons in dicyandiamide,

$\text{4H+4N}+2\text{C}=\left(\left(4\times 1\right)+\left(4\times 5\right)+\left(2\times 4\right)\right){\text{e}}^{-}=32{\text{e}}^{-}$

The predicted Lewis structures for dicyandiamide are,

Figure 3

The total number of valence electrons in melamine,

$\text{6H+6N}+3\text{C}=\left(\left(6\times 1\right)+\left(6\times 5\right)+\left(3\times 4\right)\right){\text{e}}^{-}=48{\text{e}}^{-}$

The predicted Lewis structures for melamine are,

Figure 4

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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^2$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ 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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^2$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ hybridized.

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

For $NC{N}^{2-}$,

In the given structure,

• The nitrogen atom is ${\text{sp}}^{\text{2}}$ hybridized (steric number $3$) and the carbon atom is $\text{sp}$ hybridized (steric number $2$).

For $H_{2}NCN$,

The resonance structure (II) is more stable.

In the given structure,

• One nitrogen atom is ${\text{sp}}^3$ hybridized (steric number $4$), The other nitrogen atom is $\text{sp}$ hybridized (steric number $2$) and the carbon atom is $\text{sp}$ hybridized (steric number $2$).

For melamine,

In the given structure,

• One nitrogen atom present in the ring is ${\text{sp}}^2$ hybridized (steric number $3$), The other nitrogen atom present in the $-{\text{NH}}_{\text{2}}$ group is ${\text{sp}}^3$ hybridized (steric number $4$) and the carbon atom present is ${\text{sp}}^2$ hybridized (steric number $3$).

For dicyandiamide,

In the given structure,

• One carbon atom is $\text{sp}$ hybridized and the other is ${\text{sp}}^{\text{2}}$ hybridized.
• Four nitrogen atoms are present,

One nitrogen atom present is ${\text{sp}}^2$ hybridized (steric number $3$).

One nitrogen atom present is $\text{sp}$ hybridized (steric number $2$).

The remaining two nitrogen atoms are ${\text{sp}}^{\text{3}}$ hybridized (steric number $4$).

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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^{\text{2}}$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ 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 $(\sigma )$ and pi $(\pi )$ bonds present in the structure of the given molecules.

Answer to Problem 115CP

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 $NC{N}^{2-}$,

In the structure for the given compound,

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

For $H_{2}NCN$,

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.

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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^{\text{2}}$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ 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 ${\text{sp}}^{\text{2}}$ hybridization. The $\text{π}$ electrons are delocalized over the six atoms present. Hence, this ring is considered planar.

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 ${\text{sp}}^{\text{3}}$ hybridized.

If the steric number is $3$, the atom is ${\text{sp}}^{\text{2}}$ hybridized.

If the steric number is $2$, the atom is $\text{sp}$ 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 115CP

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

Explanation of Solution

 Explanation:

Formula

Formal charge $=\text{V}-\left[\text{N}\text{+}\frac{\text{B}}{2}\right]$

Where,

• $\text{V}$ is the number of valence electrons of the neutral atom.
• $\text{N}$ is the number of non-bonding valence electrons.
• $\text{B}$ is the total number of shared electrons.

For structure (I) for dicyandiamide,

For the hydrogen atom,

Formal charge $=1-\left[0+\left(\frac{1}{2}\times 2\right)\right]=0$

For the single bonded nitrogen atom,

Formal charge $=5-\left[2+\left(\frac{1}{2}\times 6\right)\right]=0$

For the double bonded central nitrogen atom,

Formal charge $=5-\left[0+\left(\frac{1}{2}\times 8\right)\right]=1$

For the double bonded terminal nitrogen atom,

Formal charge $=5-\left[4+\left(\frac{1}{2}\times 4\right)\right]=-1$

For the carbon atoms,

Formal charge $=4-\left[0+\left(\frac{1}{2}\times 8\right)\right]=0$

For structure (II) for dicyandiamide,

For the hydrogen atom,

Formal charge $=1-\left[0+\left(\frac{1}{2}\times 2\right)\right]=0$

For the single bonded nitrogen atom,

Formal charge $=5-\left[2+\left(\frac{1}{2}\times 6\right)\right]=0$

For the double bonded central nitrogen atom,

Formal charge $=5-\left[2+\left(\frac{1}{2}\times 6\right)\right]=0$

For the triple bonded terminal nitrogen atom,

Formal charge $=5-\left[2+\left(\frac{1}{2}\times 6\right)\right]=0$

For the double bonded carbon atom,

Formal charge $=4-\left[0+\left(\frac{1}{2}\times 8\right)\right]=0$

For the triple bonded carbon atom,

Formal charge $=4-\left[0+\left(\frac{1}{2}\times 8\right)\right]=0$

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).