Problem 11 - Chemistry: Principles and Reactions

Question

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), when bubbled into water can form nitric acid. Its skeleton structure has no \(\mathrm{N}-\mathrm{N}\) or \(\mathrm{O}-\mathrm{O}\) bonds. Write its Lewis structure.

Step-by-Step Solution

Answer

Answer: The Lewis structure for dinitrogen pentoxide (N2O5) is: O O \\ // N - N - O // \\ O O

Step 1: Count the total number of valence electrons

The first step to draw the Lewis structure is to determine the valence electrons for each atom. Nitrogen (N) has 5 valence electrons and Oxygen (O) has 6 valence electrons. Since there are 2 Nitrogen atoms and 5 Oxygen atoms, the total valence electrons are: (2 × 5) + (5 × 6) = 10 + 30 = 40 valence electrons.

Step 2: Place the least electronegative atoms in the center of the structure

Nitrogen is less electronegative than Oxygen, so we place the two Nitrogen atoms at the center.

Step 3: Arrange the remaining atoms around the central atoms and form bonds

Place the five Oxygen atoms around the two Nitrogen atoms, with one Oxygen atom bound to each Nitrogen atom, and a double bond between one Nitrogen atom and another Oxygen atom. Form single bonds between the Nitrogen atoms and the Oxygen atoms like in a chain.

Step 4: Distribute the remaining valence electrons as lone pairs around each atom

Now that we have the basic skeleton structure, we need to distribute the remaining valence electrons as lone pairs around the atoms. We have used 8 electrons to form the bonds (2 electrons per bond). That leaves 40 - 8 = 32 electrons to be distributed as lone pairs. First, complete the octet for the terminal Oxygen atoms by adding six electrons (3 pairs) around each of these atoms. So, four terminal Oxygen atoms will use 24 electrons in total. Now, we have 8 electrons left. Place the remaining 8 electrons (4 pairs) as lone pairs on both Nitrogen atoms and the non-terminal Oxygen atom. Each atom will have an octet configuration.

Step 5: Check the formal charge of each atom

Once the Lewis structure is complete, verify that the formal charge of every atom is as close to zero as possible. For Nitrogen atoms, the formal charge = (5 - [3 non-bonding electrons + 5/2 bonding electrons]) = 0. For the terminal Oxygen atoms, the formal charge = (6 - [6 non-bonding electrons + 2/2 bonding electrons]) = 0. For the non-terminal Oxygen atom, the formal charge = (6 - [4 non-bonding electrons + 4/2 bonding electrons]) = 0. As all formal charges are equal to 0, our Lewis structure of N2O5 is correct. So, the Lewis structure of dinitrogen pentoxide (N2O5) is: O O \\ // N - N - O // \\ O O

Key Concepts

Lewis structures

Understanding the Lewis structures is essential for visualizing the distribution of electrons in a molecule. It graphically represents the individual atoms, the bonds between them, and the lone pairs of electrons. A Lewis structure adheres to certain rules to provide a clear diagram of how atoms bond and what the electron configuration is around each atom.

When drawing a Lewis structure, we start by counting valence electrons which are the electrons available for bonding in the outer shell. Next, we arrange the atoms to form the most likely structure, typically positioning the least electronegative elements in the center. After all single bonds are formed (each bond consisting of two electrons), we distribute the remaining valence electrons as lone pairs around the atoms to fulfill their octet requirement, which is the tendency to be surrounded by eight electrons.

Lewis structures also help in predicting molecular geometry, the reactivity of the compound, and its physical properties. Since they don’t represent the actual shapes of molecules, they work best for two-dimensional representations and simplify the electron interactions for educational purposes.

Valence electrons

Valence electrons are the electrons located in the outermost shell of an atom. They are significant since they are involved in bond formation and in chemical reactions. Determining the number of valence electrons is the cornerstone of drawing Lewis structures, as they indicate how many electrons can be shared or paired.

Nitrogen, with an atomic number of 7, has 5 valence electrons, while Oxygen, with an atomic number of 8, has 6 valence electrons. To determine the total number of valence electrons in a compound like dinitrogen pentoxide (N₂O₅), we need to tally the valence electrons of all the atoms involved - resulting in an overall count of 40 valence electrons for this molecule. These electrons are then used to create bonds and form lone pairs around the atoms in the Lewis structure, ensuring each atom achieves a stable configuration.

Formal charge calculation

The concept of formal charge is used to estimate the electric charge distribution within a molecule. The formal charge is calculated for each atom in a Lewis structure to facilitate the lowest energy configuration for the molecule. The formula to calculate the formal charge is:
\[ \text{Formal Charge} = (\text{Valence Electrons}) - (\text{Non-bonding Electrons}) - \frac{\text{Bonding Electrons}}{2} \]

For a stable molecule, the sum of formal charges for all atoms should be equal to the charge of the molecule itself. In our case with dinitrogen pentoxide, all atoms have a formal charge of zero, which indicates that the Lewis structure represents a stable arrangement of electrons. Calculating the formal charge is crucial when determining the best structure, especially for molecules with resonance structures or where multiple configurations are possible.