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In the formate ion, \(\mathrm{HCO}_{2}^{-}\) , the carbon atom is the central atom with the other three atoms attached to it. (a) Draw a Lewis structure for the formate ion. (b) What hybridization is exhibited by the C atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

Short Answer

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The formate ion (\(\mathrm{HCO}_{2}^{-}\)) has the following properties: (a) Its Lewis structure has the carbon atom in the center with a single bond to hydrogen, a single bond to one oxygen, and a double bond to another oxygen, and the oxygen atoms have two lone pairs. (b) The carbon atom exhibits sp2 hybridization. (c) There are two equivalent resonance structures for the ion. (d) There are 2 electrons in the π system of the ion.

Step by step solution

01

Draw the Lewis structure for the formate ion

First, we need to find the total number of valence electrons in the formate ion. For hydrogen (H), there is 1 valence electron; carbon (C) has 4 valence electrons; oxygen (O) has 6 valence electrons. Additionally, the ion has a negative charge, which means it has an extra valence electron. So, the total number of valence electrons is: \(1 + 4 + 6 + 6 + 1 = 18\). Now, we will draw the Lewis structure: 1. Place the least electronegative atom (C) in the center. 2. Place the remaining atoms (H and two O) around the central atom. 3. Connect the atoms with single bonds (each using 2 electrons). 4. Distribute the remaining electrons (12 left) as lone pairs to complete the octets of the outer atoms. The Lewis structure for formate ion is: ``` O || C - O - H | O- ``` with two lone pairs on each oxygen atom.
02

Identify the hybridization of the carbon atom

To determine the hybridization of the carbon atom, we need to count the number of electron groups (bonding pairs and lone pairs) around it. In the formate ion, carbon forms three single bonds - one with hydrogen and two with oxygen atoms. There are no lone pairs on the carbon atom. So, there are three electron groups in total, which corresponds to an sp2 hybridization.
03

Determine the multiple equivalent resonance structures for the ion

Different resonance structures can be formed by moving the π electrons between the carbon and oxygen atoms. In the formate ion, there exists 2 equivalent resonance structures: ``` O || C - O - H | O- ``` and ``` O- | C - O - H || O ``` These two structures have the same bond arrangements but differ in the location of the π bond.
04

Calculate the number of electrons in the π system of the ion

In the formate ion, there is a double bond between the carbon and one of the oxygen atoms. A double bond consists of one σ bond and one π bond. The π bond is formed by the overlap of the \(p_z\) orbitals of carbon and oxygen, which contains 2 π electrons. Therefore, the formate ion has a total of 2 electrons in its π system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structure
The Lewis structure is a graphical representation of the valence electrons in molecules or ions, which illustrates how these electrons are arranged around the atoms and how the atoms are bonded to each other. In the case of the formate ion electron count is a necessary first step. With a total of 18 valence electrons to work with, carbon is placed in the center being the least electronegative atom, with the hydrogen and oxygen atoms arranged around it. Single bonds are drawn between the central carbon atom and the other atoms, using 2 electrons each, with the remaining electrons distributed as lone pairs on the oxygen atoms to complete their octets. It is essential to understand that in such structures, the atoms aim to fulfill the octet rule, gaining a stable electron configuration.
Visualizing this structure is crucial for understanding the molecular geometry and reactivity of the formate ion.
sp2 Hybridization
Hybridization in chemistry refers to the concept of mixing atomic orbitals to form new hybrid orbitals that can better explain the molecular geometry of atoms within a molecule. In the formate ion, we determine the hybridization of the carbon atom by counting the number of electron groups around it. For carbon, three single bonds mean there are three electron groups, which corresponds to an sp2 hybridization. This means that one s orbital and two p orbitals mix to form three sp2 hybrid orbitals, each occupied by one electron group.

In the context of formate ion chemistry, recognizing that the carbon atom is sp2 hybridized helps us predict that the atoms bonded to it will lie on a plane, with bond angles approximately 120 degrees apart, characterizing the molecule's overall structure.
Resonance Structures
Resonance structures are a set of two or more Lewis structures that collectively describe the electron distribution in a molecule where the atomic arrangement remains constant, but the electron positions may differ. The formate ion has two equivalent resonance structures, which can be depicted by shifting the position of the pi (π) bond between the carbon and the oxygen atoms while keeping the sigma (σ) bonds intact.

The existence of resonance structures has profound implications for the stability and reactivity of the formate ion. It indicates that the actual structure is a hybrid of the resonance forms, suggesting delocalization of electrons across the molecule. This delocalization contributes to the stability of the ion, making it an essential concept to grasp when studying molecular chemistry.
Pi (π) Electron System
The pi (π) electron system of a molecule includes the electrons found in π bonds, which are formed by the sideways overlap of p orbitals on adjacent atoms. In the formate ion, the double bond between the carbon and one of the oxygen atoms contains one π bond. A π bond is a region of electron density located above and below the plane of the nuclei of the bonding atoms, contributing 2 π electrons to the ion's electron system.

Understanding the π electron system is vital for learning about the reactivity and properties of molecules. π electrons are often the electrons involved in chemical reactions, particularly those that occur via mechanisms that involve electron shifts, such as in the case of resonance structures. The π system in the formate ion allows for the delocalization of electrons, which is a key factor in determining the ion's stability and its ability to engage in chemical reactions.

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Most popular questions from this chapter

Draw a picture that shows all three 2\(p\) orbitals on one atom and all three 2\(p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of 2\(p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of 2\(p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of 2\(p\) orbitals?

Indicate whether each statement is true or false. (a) \(s\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is 100\(\%\) for finding an electron at the nucleus in a \(\pi^{*}\) orbital. (c) Antibonding orbitals are higher in energy than bonding orbitals (if all orbitals are created from the same atomic orbitals). (d) Electrons cannot occupy an antibonding orbital.

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}^{-} .\) (a) Draw the Lewis structure of the azide ion that minizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central Natom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},\) is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

Consider a molecule with formula \(\mathrm{AX}_{3}\) . Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{3}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle increases from \(100^{\circ}\) to \(120^{\circ}\)

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