Question

Consider the formate ion, \(\mathrm{HCO}_{2}^{-},\) which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ton. The \(\mathrm{H}\) and the two O atoms are bonded to the central \(\mathrm{C}\) atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2} ?\)

Short answer

The best Lewis structures for the formate ion, \(\mathrm{HCO}_{2}^{-}\), are two resonance structures with an intermediate C-O bond order of 1.5. As a result, the C-O bond length in the formate ion would be expected to be longer than those in \(\mathrm{CO}_{2}\) due to the lower bond order (1.5) compared to the double bonds in \(\mathrm{CO}_{2}\) (2).

Step-by-step solution

(a) Draw the best Lewis structure(s) for the formate ion

First, we need to determine the total number of valence electrons in the formate ion, \(\mathrm{HCO}_{2}^{-}\). Using the periodic table, we find: - H has 1 valence electron - C has 4 valence electrons - O has 6 valence electrons (and we have two O atoms) - The negative charge (-1) Adding all the electrons, we have a total of \(1+4+(2\times6)-1=18\) valence electrons. Now, let's draw the Lewis structure. We know that the \(\mathrm{C}\) atom is at the center with \(\mathrm{H}\) and two \(\mathrm{O}\) atoms bonded to it. 1. Place the central atom with single bonds to the surrounding atoms: \[ \mathrm{C - H} \] \[\mathrm{|} \] \[ \mathrm{C - O \ \ O} \] 2. Distribute the remaining valence electrons as lone pairs to complete the octets for the surrounding atoms: - H atom is satisfied with 2 electrons since it's sharing a single bond with C - We have 14 valence electrons remaining to distribute on C and O atoms \[ \mathrm{H - C} \] \[ \ \mathrm{|}\ \ \] \[ \mathrm{O : - C - O} \] \[\ \ \ \ \ \ \mathrm{: \ \ :} \] 3. Central carbon has 6 electrons surrounding it (2 single bonds with H and O, and 1 double bond with another O atom). It needs two more electrons to complete its octet. So, we will add another bond between C and one of the O atoms, resulting in the final Lewis structure: \[ \mathrm{H - C = O} \] \[\ \ \ \mathrm{|} \ \ \] \[ \mathrm{O : - C} \] \[\ \ \mathrm{: \ \ :} \]

(b) Are resonance structures needed to describe the structure?

We can examine the possibility of a resonance structure by shifting the double bond between C and O atoms. Original structure: \[ \mathrm{H - C = O} \] \[\ \ \ \mathrm{|} \ \ \] \[ \mathrm{O : - C} \] \[\ \ \mathrm{: \ \ :} \] Possible resonance structure: \[ \mathrm{H - C -\ O} \] \[ \ \ \mathrm{|}\ \ \] \[\mathrm{O : - C = O} \] \[\ \ \ \ \ \ \mathrm{: \ \ :} \] Since both structures satisfy the octet rule for the central C atom, we can have resonance structures for the formate ion. The true structure of the formate ion is actually an average of these two resonance structures, where the C-O bond is not a single or double bond, but an intermediate between them (1.5 bonds).

(c) Predict if the C-O bond lengths in the formate ion would be longer or shorter relative to those in CO₂

In carbon dioxide, \(\mathrm{CO}_2\), there are two double bonds between the carbon atom and both oxygen atoms: \[ O = C = O \] In the formate ion, the true structure is an average of resonance structures, and the C-O bond order is 1.5 rather than 1(in single bond) or 2(in double bond). Therefore, the bond length of C-O in the formate ion would be intermediate between a single bond and a double bond. C-O bond length typically decreases as bond order increases, so the C-O bond length in the formate ion (1.5) would be expected to be longer than those in \(\mathrm{CO}_{2}\) (2).

Key concepts

Formate Ion

The formate ion, denoted as \( \mathrm{HCO_2^-} \), is an interesting species formed when a proton \( \mathrm{H^+} \) is removed from formic acid. The ion's central component is a carbon atom bonded to two oxygen atoms and one hydrogen atom. To understand its structure better, we look at the distribution of electrons around these atoms.

Each atom contributes a certain number of valence electrons. Hydrogen contributes 1, carbon contributes 4, and each oxygen contributes 6 electrons, considering there are two oxygens in the ion. Due to the negative charge on the formate ion, we add an extra electron to account for this, leading to a total count of 18 valence electrons.

The Lewis structure is a visual representation of how these electrons are used to form bonds between the atoms: a carbon atom centrally connects to both oxygens and hydrogen. This exercise emphasizes the importance of full octet satisfaction, especially for carbon, which strives to reach 8 electrons around it by forming additional bonds as necessary.

Resonance Structures

Resonance plays a crucial role in explaining the electronic structure of many molecules, including the formate ion. In chemistry, resonance structures are different Lewis structures that depict the same molecule and indicate the delocalization of electrons.

For the formate ion \( \mathrm{HCO_2^-} \), two resonance structures can be drawn by alternating which oxygen atom forms a double bond with the central carbon atom.

• In one resonance structure, the double bond is between the carbon and one oxygen, while the other oxygen is singly bonded.
• In the alternate structure, the double bond shifts to the other oxygen.

The resonance hybrid, or the true form of the ion, is a blend of these configurations. In essence, this delocalization results in each \( \mathrm{C-O} \) bond having a bond order of 1.5, rather than a single or double bond. This electron movement allows the molecule to achieve greater stability, showcasing the powerful concept of resonance structures.

Bond Length Prediction

Predicting bond lengths is an insightful way to understand molecular structure. In our context, the formate ion features bonds influenced by resonance. The average bond order for the \( \mathrm{C-O} \) bonds in the formate ion being 1.5 derives a middle ground between standard single and double bonds.

To contextualize, consider carbon dioxide \( \mathrm{CO_2} \), where carbon forms double bonds with each oxygen. These bond lengths are relatively short due to the double bond nature, which increases electron sharing and bond strength.

In comparison, the bond length in \( \mathrm{HCO_2^-} \) will be longer than typical double bonds but shorter than single bonds due to the resonance-induced bond order of 1.5.

Generally, increased bond order denotes shorter bonds, as seen in CO₂. However, for formate, the presence of resonance and its resulting impact mean bonds are more flexible in length yet not distinctly as strong as standard double bonds.