Problem 41 - Chemistry and Chemical Reactivity

Question

Two resonance structures are possible for \(\mathrm{NO}_{2}^{-} .\) Draw these structures and then find the formal charge on each atom in each resonance structure. If an \(\mathrm{H}^{+}\) ion is attached to \(\mathrm{NO}_{2}^{-}\) (to form the acid \(\mathrm{HNO}_{2}\) ), does it attach to \(\mathrm{O}\) or N?

Step-by-Step Solution

Answer

In NO2^-, H+ attaches to O, where lone pairs can stabilize the added hydrogen.

Step 1: Draw Lewis Structure for NO2^-

The first step is to draw the Lewis structure for the nitrate ion (\( \text{NO}_2^- \)). Start by counting the total number of valence electrons. Nitrogen (N) has 5 valence electrons, each oxygen (O) has 6 valence electrons, and the extra negative charge adds 1 more electron, totaling 18 valence electrons. Draw the structure with N in the center, connected to two oxygens.

Step 2: Assign Electrons to Each Atom

Connect N to each O with a single bond, using 4 electrons, leaving 14 valence electrons. Fill the octet of both oxygens with the remaining electrons, and place any leftover electrons on the nitrogen. In this configuration, N shares 4 electrons with each O, forming a double bond with one O, fulfilling the octet rule.

Step 3: Identify Resonance Structures

The possible resonance structures occur because either oxygen can form a double bond with the nitrogen. Draw the second structure, moving the double bond to the other oxygen, while ensuring that both forms obey the octet rule. Each structure is a resonance form of \( \text{NO}_2^- \).

Step 4: Calculate Formal Charge of Atoms

For each resonance structure, calculate the formal charge: Formal charge = (Valence electrons) - (Non-bonding electrons) - (Bonding electrons/2). Apply this to both N and O atoms. For \( \text{NO}_2^- \), one O will have a formal charge of 0 (with double bond), the other O will have a formal charge of -1 (with single bond), and N will have a formal charge of +1.

Step 5: Analyze Proton Attachment

Consider where a \( \text{H}^+ \) ion would attach in the acid \( \text{HNO}_2 \). Since the oxygens have lone pairs and a negative formal charge (better equipped for bonding with positive ions), the hydrogen attaches to one of the oxygen atoms. Thus, the oxygen with a lone pair becomes a hydroxyl group \( \text{OH} \), forming the structure of \( \text{HNO}_2 \).

Key Concepts

Lewis Structures

In chemistry, Lewis structures are diagrams that show the bond connections between atoms in a molecule, as well as the lone pairs of electrons that may exist. These structures are crucial for visualizing the arrangement of atoms and the flow of electrons in any chemical compound. For example, when constructing the Lewis structure for the nitrate ion (\(\text{NO}_2^- \)), we first calculate the total number of valence electrons. Nitrogen contributes 5, each oxygen contributes 6, and an additional electron is added due to the negative charge, making a total of 18 valence electrons.

• Arrange the atoms with nitrogen in the center since it is less electronegative than oxygen.
• Connect nitrogen to each oxygen atom with a single bond initially, which uses 4 electrons. This leaves us with 14 electrons to distribute.
• Complete the octet of each oxygen atom by adding remaining electrons, and any excess should be placed on the central nitrogen atom, while maintaining octet rules.

Formal Charge

To gain a deeper insight into the stability and potential reactivity of molecules, we calculate the formal charge. The formal charge is a technique to allocate charge among the atoms in a molecule based on the number of valence electrons. The formula used is:\[\text{Formal Charge} = \text{(Valence Electrons)} - \text{(Non-bonding Electrons)} - \frac{1}{2}\text{(Bonding Electrons)}\]For the resonance structures of \(\text{NO}_2^- \),it's critical to assign a formal charge:

• On nitrogen: Calculate using its 5 valence electrons minus the sum of non-bonding electrons and half of the shared bonding electrons.
• On each oxygen: Each will have different formal charges depending on single or double bonds; usually, oxygen with one double bond gets a formal charge of 0, while the other oxygen will carry a formal charge of -1.
• This results in nitrogen having a formal charge of +1, and the distribution effectively shows potential sites for further chemical interaction.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a pivotal role in the formation of chemical bonds. They determine how atoms interact with each other and how stable the bonds they form will be. Knowing the count of valence electrons is the foundational step in drawing Lewis structures:

• For nitrogen, with an atomic number of 7, we find 5 valence electrons in its outer shell.
• Each oxygen atom has 6 valence electrons (atomic number 8), but the entire molecule \(\text{NO}_2^- \)gains an additional electron due to its excess negative charge, accounting for the total 18 in the structure.
• Using these valence electrons efficiently helps form stable compounds by satisfying the octet rule, which aims for each atom to have eight electrons in its outer shell, similar to noble gases.

Proton Attachment in Acids

Acid formation involves the attachment of a proton (\(H^+\))to a molecule or ion, and is guided by the principle of stability in the resulting compound. In the case of forming the acid \(\text{HNO}_2\)from \(\text{NO}_2^- \):

• Identify the most likely site for attachment, which is usually the atom with the highest concentration of negative charge or with lone pair electrons available for bonding.
• Oxygens in \(\text{NO}_2^- \)have a negative formal charge and lone pairs, making them ideal for proton binding.
• Therefore, when \(H^+\)is added, it attaches to an oxygen, turning that oxygen into a hydroxyl group (\(\text{OH}\)). This reaction forms stable \(\text{HNO}_2\), providing insight into how proton transfer occurs in acid-base chemistry.