Chapter 8, Problem 8.119QP

(a)

Interpretation Introduction

Interpretation: The given ions are to be ranked in order of increasing nitrogen-oxygen bond lengths and increasing bond energies.

Concept introduction: The bond length is the distance between the nuclei of atoms and bond order is defined as the total number of bonds present in a molecule. When the bond order is high, the bond length is shorter because of the stronger attraction between the atoms. Bond length shows inverse relation with the bond energy.

To determine: The order of increasing nitrogen-oxygen bond lengths for the given ions.

Answer to Problem 8.119QP

Solution

The order of increasing nitrogen-oxygen bond lengths for the given ions is ${\text{NO}}^{+}{\text{<NO}}_2^{-}{\text{<NO}}_3^{-}$.

Explanation of Solution

Explanation

The order of nitrogen-oxygen bond lengths is predicted by calculating the bond order of the given ions. The bond order is inversely proportional to the bond length in a given ion or a molecule. So, more is the bond order value less is the bond strength.

The given ions are ${\text{NO}}_2^{-}$, ${\text{NO}}^{+}$ and ${\text{NO}}_3^{-}$. Nitrogen and oxygen belongs to $15$ and $16$ groups of the periodic table and has five and six valence electrons respectively. Therefore, $\text{NO}$ molecule has $5+6=11$ total valence electrons.

By losing one electron it forms ${\text{NO}}^{+}$ ion having $10$ valence electrons. It consists of three electron pair and one $\text{N}-\text{O}$ bond. The structure of ${\text{NO}}^{+}$ ion is,

Figure 1

The average bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}^{+}$ ion is calculated by using the formula,

$\text{Average}\text{bond}\text{order}=\frac{\text{Total}\text{number}\text{of}\text{electron}\text{pair}\text{in}\text{N}-\text{O}\text{bonds}}{\text{Total}\text{number}\text{of}\text{N}-\text{O}\text{bonds}}$

Substitute the values of number of electron pair and $\text{N}-\text{O}$ bonds in the above formula to calculate the bond order of ${\text{NO}}^{+}$ ion.

$\begin{array}{c}\text{Average}\text{bond}\text{order}=\frac{3}{1}\\ =3\end{array}$

The bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}^{+}$ ion is $3$.

The ${\text{NO}}_2^{-}$ ion has one nitrogen and two oxygen atoms.

Nitrogen and oxygen belongs to $15$ and $16$ groups of the periodic table and has five and six valence electrons respectively. The total valence electrons in ${\text{NO}}_2^{-}$ ion is calculated as,

$\begin{array}{c}1\text{N}+2\text{O}+\text{charge}\text{on}\text{ion}=1\left(5\right)+2\left(6\right)+1\\ =5+12+1\\ =18\end{array}$

The structure of ${\text{NO}}_2^{-}$ ion is,

Figure 2

The ${\text{NO}}_2^{-}$ ion has three electron pair and two $\text{N}-\text{O}$ bonds. The average bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}_2^{-}$ ion is calculated by using the formula,

$\text{Average}\text{bond}\text{order}=\frac{\text{Total}\text{number}\text{of}\text{electron}\text{pair}\text{in}\text{N}-\text{O}\text{bonds}}{\text{Total}\text{number}\text{of}\text{N}-\text{O}\text{bonds}}$

Substitute the values of number of electron pair and $\text{N}-\text{O}$ bonds in the above formula to calculate the bond order of ${\text{NO}}_2^{-}$ ion.

$\begin{array}{c}\text{Average}\text{bond}\text{order}=\frac{3}{2}\\ =1.5\end{array}$

The bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}_2^{-}$ ion is $1.5$.

The ${\text{NO}}_3^{-}$ ion has one nitrogen and three oxygen atoms.

Nitrogen and oxygen belongs to $15$ and $16$ groups of the periodic table and has five and six valence electrons respectively. The total valence electrons in ${\text{NO}}_3^{-}$ ion is calculated as,

$\begin{array}{c}1\text{N}+3\text{O}+\text{charge}\text{on}\text{ion}=1\left(5\right)+3\left(6\right)+1\\ =5+18+1\\ =24\end{array}$

The structure of ${\text{NO}}_3^{-}$ ion is,

Figure 3

The ${\text{NO}}_3^{-}$ ion has four electron pair and three $\text{N}-\text{O}$ bonds. The average bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}_3^{-}$ ion is calculated by using the formula,

$\text{Average}\text{bond}\text{order}=\frac{\text{Total}\text{number}\text{of}\text{electron}\text{pair}\text{in}\text{N}-\text{O}\text{bonds}}{\text{Total}\text{number}\text{of}\text{N}-\text{O}\text{bonds}}$

Substitute the values of number of electron pair and $\text{N}-\text{O}$ bonds in the above formula to calculate the bond order of ${\text{NO}}_3^{-}$ ion.

$\begin{array}{c}\text{Average}\text{bond}\text{order}=\frac{4}{3}\\ =1.3\end{array}$

The bond order of $\text{N}-\text{O}$ bond in ${\text{NO}}_3^{-}$ ion is $1.3$.

The increasing order of bond order for the given ions is ${\text{NO}}_3^{-}<{\text{NO}}_2^{-}<{\text{NO}}^{+}$. As the bond order is inversely proportional to the bond length, the increasing order of bond lengths for the given ions is ${\text{NO}}^{+}{\text{<NO}}_2^{-}{\text{<NO}}_3^{-}$.

(b)

Interpretation Introduction

To determine: The order of increasing bond energies for the given ions.

Answer to Problem 8.119QP

Solution

The order of increasing bond energies for the given ions is ${\text{NO}}_3^{-}<{\text{NO}}_2^{-}<{\text{NO}}^{+}$.

Explanation of Solution

Explanation

The bond energy shows inverse relation to the bond length that is more is the bond length lesser is the amount of energy required to break the bond. Bond energy is directly proportional to the bond order in a given molecule or ion.

Hence, the increasing order of bond energies for the given ions is ${\text{NO}}_3^{-}<{\text{NO}}_2^{-}<{\text{NO}}^{+}$.

Conclusion

1. a. The order of increasing $\text{N}-\text{O}$ bond lengths for the given ions is ${\text{NO}}^{+}{\text{<NO}}_2^{-}{\text{<NO}}_3^{-}$.
2. b. The order of increasing bond energies for the given ions is ${\text{NO}}_3^{-}<{\text{NO}}_2^{-}<{\text{NO}}^{+}$.