Chapter 1, Problem 1.37P

Citric acid is responsible for the tartness of citrus fruits, especially lemons and limes.

a. What is the molecular formula of citric acid?

b. How many lone pairs are present?

c. Draw a skeletal structure.

d. How many $s{p}^2$ hybridized carbon atoms are present?

e. What orbitals are used to form each indicated bond $\left(\left[1\right]-\left[4\right]\right)$?

Interpretation Introduction

(a)

Interpretation: The molecular formula of citric acid is to be stated.

Concept introduction: In ball-and-stick model, each colored ball represents a specific atom and each stick represents a bond. In this model, each black ball represents $\text{C}$ atoms, each gray ball represents $\text{H}$ atoms, and each red ball represents $\text{O}$ atoms.

Answer to Problem 1.37P

The molecular formula of citric acid is ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{7}}$.

Explanation of Solution

The given ball-and-stick model of citric acid is,

Figure 1

In ball-and-stick model, each colored ball represents a specific atom and each stick represents a bond. In this model, each black ball represents $\text{C}$ atoms, each gray ball represents $\text{H}$ atoms, and each red ball represents $\text{O}$ atoms.

In the above model,

• There are seven red balls. Thus, there are seven $\text{O}$ atoms.

• There are six black balls. Thus, there are six $\text{C}$ atoms.

• There are eight grey balls. Thus, there are eight $\text{H}$ atoms.

Hence, the molecular formula of citric acid is ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{7}}$.

Conclusion

The molecular formula of citric acid is ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{7}}$.

Interpretation Introduction

(b)

Interpretation: The number of lone pairs present in citric acid is to be stated.

Concept introduction: In a compound or molecule, the lone pairs represent number of unshared electrons on atom. An atom may or may not have unshared electrons. For example, carbon and hydrogen atoms have no lone pair but each oxygen atom has two lone pairs.

Answer to Problem 1.37P

There are total $14$ lone pairs in citric acid.

Explanation of Solution

The molecular formula of citric acid is ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{7}}$. Carbon and hydrogen atoms have no lone pair, but each oxygen atom has two lone pairs. There are seven oxygen atoms in citric acid. Thus, there are total $14$ lone pairs $\left(7\times 2=14\right)$ in citric acid.

Conclusion

There are total $14$ lone pairs in citric acid.

Interpretation Introduction

(c)

Interpretation: A skeletal structure of citric acid is to be drawn.

Concept introduction: A ball-and-stick model is converted into skeletal structure by replacing black ball with $\text{C}$, gray ball with $\text{H}$, and red ball with $\text{O}$. Omit the $\text{H}$ atom on carbon, but not in the case of heteroatom.

Answer to Problem 1.37P

A skeletal structure of citric acid is,

Explanation of Solution

In ball-and-stick model each colored ball represents a specific atom and each stick represents a bond. A ball-and-stick model is converted into skeletal structure by replacing black ball with $\text{C}$, gray ball with $\text{H}$, red ball with $\text{O}$, and blue ball with $\text{N}$. Omit the $\text{H}$ atom on carbon, but not in the case of heteroatom.

A skeletal structure of citric acid is shown in Figure 2.

Figure 2

Conclusion

In ball-and-stick model each colored ball represents a specific atom and each stick represents a bond.

Interpretation Introduction

(d)

Interpretation: The number of $s{p}^2$ hybridized carbon in citric acid is to be stated.

Concept introduction: According to the rule of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

Answer to Problem 1.37P

There are three $s{p}^2$ hybridized carbon atoms in citric acid.

Explanation of Solution

The Lewis structure of citric acid is,

Figure 3

According to the rules of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

The $s{p}^2$ hybridized carbon atoms (surrounded by three groups) are shown in Figure 4.

Figure 4

Thus, there are three $s{p}^2$ hybridized carbon atoms in citric acid.

Conclusion

There are three $s{p}^2$ hybridized carbon atoms in citric acid.

Interpretation Introduction

(e)

Interpretation: The orbitals that are used to form each indicated bond is to be stated.

Concept introduction: According to the rule of hybridization, an atom that is surrounded with two groups is $sp$ hybridized, an atom that is surrounded with three groups is $s{p}^2$ hybridized, and an atom that is surrounded with four groups is $s{p}^3$ hybridized.

Answer to Problem 1.37P

Bond $\left[1\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{O}}_{s{p}^2}$ hybridized orbitals. Bond $\left[2\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ hybridized orbitals. Bond $\left[3\right]$ is formed by ${\text{O}}_{s{p}^3}-{\text{H}}_{1s}$ orbitals. Bond $\left[4\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{O}}_{s{p}^3}$ hybridized orbitals.

Explanation of Solution

Bond [1] represents $\text{C}=\text{O}$ bond, in which both carbon atom and oxygen atom are $s{p}^2$ hybridized. Thus, it is formed by ${\text{C}}_{s{p}^2}-{\text{O}}_{s{p}^2}$ hybridized orbitals.

Bond $\left[2\right]$ represents bonding between the carbon atom of carbonyl group $\left(\text{C}=\text{O}\right)$, and the carbon atom of $-{\text{CH}}_{\text{2}}-$ group. The carbon atom of carbonyl group is $s{p}^2$ hybridized and the carbon atom of $-{\text{CH}}_{\text{2}}-$ group is $s{p}^3$ hybridized.

Thus, $\left[2\right]$ is formed by ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ hybridized orbitals.

Bond $\left[3\right]$ represents, $\text{O}-\text{H}$ bond. This bond is formed through $s{p}^3$ hybridized orbital of oxygen and $1s$ orbital of hydrogen.

Thus, $\left[3\right]$ is formed by ${\text{O}}_{s{p}^3}-{\text{H}}_{1s}$ orbitals.

Bond $\left[4\right]$ represents, $\text{C}-\text{O}$ bond, in which both carbon atom and oxygen atom are $s{p}^3$ hybridized. Thus, $\left[4\right]$ is formed by ${\text{C}}_{s{p}^3}-{\text{O}}_{s{p}^3}$ hybridized orbitals.

Conclusion

The number of surrounded group around any atom predicts the hybridization of that atom, which is further helpful to predict the orbitals involve in bond formation.