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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 \(\mathrm{C}\) atom? (c) Are there multiple equivalent resonance structures for the ion? (d) Which of the atoms in the ion have \(p_{\pi}\) orbitals? (e) How many electrons are in the \(\pi\) system of the ion?

Short Answer

Expert verified
The Lewis structure for the formate ion, HCO2-, has a central carbon atom bonded to one hydrogen atom and two oxygen atoms, with one of the oxygen atoms forming a double bond: O || H - C - O - The central carbon atom exhibits sp2 hybridization, and there are multiple resonance structures for the ion. The carbon atom and the two oxygen atoms have pπ orbitals, and there are 2 electrons in the π system of the ion.

Step by step solution

01

Draw the Lewis structure of the formate ion

To draw the Lewis structure, start by counting the total number of valence electrons present in the ion. For the formate ion, HCO2-, we have: - 1 Hydrogen atom: 1 valence electron - 1 Carbon atom: 4 valence electrons - 2 Oxygen atoms: 6 valence electrons each (total of 12) - 1 extra electron from the negative charge This gives us a total of 18 valence electrons to distribute. Place the Carbon atom at the center and connect it to one Hydrogen atom and two Oxygen atoms. Distribute the remaining electrons among the atoms, starting with the more electronegative Oxygen atoms, to complete their octets. The Lewis structure for the formate ion will be: O || H - C - O -
02

Determine the hybridization of the central carbon atom

The hybridization of an atom is determined by the number of electron domains (regions of electron density) surrounding it. In the case of the central Carbon atom in the formate ion, we have three electron domains: - one single bond to Hydrogen - one single bond to Oxygen - one double bond to Oxygen (we can consider this as one electron domain) Since there are three electron domains, the hybridization for the Carbon atom is sp2.
03

Check for resonance structures

Resonance structures can occur when there is the possibility to distribute the π electrons differently without changing the placement of the atoms. In the formate ion, we have a double bond between the Carbon and one of the Oxygen atoms. We can draw an equivalent structure by moving the double bond to the other Oxygen atom: O || H - C - O - and O || H - C - O - Since we can draw more than one equivalent Lewis structure, there are multiple resonance structures for the formate ion.
04

Identify atoms with pπ orbitals

Atoms involved in π systems, like double bonds or resonance, have pπ orbitals. In the formate ion, the central Carbon atom and the two Oxygen atoms are involved in the double bond and resonance. Therefore, these three atoms have pπ orbitals.
05

Count the number of electrons in the π system

The π system in the formate ion consists of a double bond between the Carbon atom and one of the Oxygen atoms (it can be on either side due to resonance). In a double bond, there are 2 π electrons. So, there are 2 electrons in the π system of the formate ion.

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Key Concepts

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

Lewis Structure
Understanding the arrangement of atoms in a molecule begins with the Lewis structure, which represents atoms and their valence electrons through symbols and dots. To draw the Lewis structure of the formate ion, \( \mathrm{HCO_2}^- \), manage the total count of 18 valence electrons — one from hydrogen, four from carbon, six each from the two oxygens, and an extra one due to the negative charge. Place carbon at the center, connecting to a hydrogen and two oxygens. Electrons are first paired around oxygen atoms to complete their octets, which are preferred due to their high electronegativity. The resulting structure shows a single bond between carbon and hydrogen, and a double bond between carbon and one oxygen, with the remaining oxygen single-bonded to carbon, carrying a negative charge.

Correct representation of the Lewis structure ensures clear visualization and is crucial for exploring further molecular properties, including hybridization, resonance, and pi systems.
Hybridization
Hybridization explains the mixing of atomic orbitals to form new hybrid orbitals, which influences molecular geometry and bonding properties. In the formate ion, the carbon atom's hybridization is determined by the number of 'electron domains' or regions of electron density it possesses. With one single bond to hydrogen, another to oxygen, and a double bond to the second oxygen, carbon demonstrates three domains and thus an sp2 hybridization. This implies the formation of three sp2 hybrid orbitals that lie in a plane with 120-degree angles between them, accommodating the sigma bonds, and one unhybridized p orbital that participates in the pi bond formation.
Resonance Structures
Resonance structures depict alternate ways of arranging electrons in molecules with conjugated pi systems, offering a more complete picture of the electronic distribution. The formate ion demonstrates resonance as the pi electrons of the double bond can be located between carbon and either oxygen atom without altering the atoms' positions. Essential to understand, neither individual resonance form is actually accurate on its own; it's the concept of resonance stabilization—the blending of these forms—that accurately reflects the ion's true structure.
p\(\pi\) Orbitals
P\(\pi\) orbitals are involved in pi bond formation and are present in atoms that have regions of overlapping p orbitals. In the case of the formate ion, both oxygen atoms and the central carbon atom contain p\(\pi\) orbitals due to their involvement in the pi bond system. These orbitals extend above and below the plane of the atoms creating the pi bond through lateral overlap, which is essential for the resonance stability of the ion.
\(\pi\) Electron System
The \(\pi\) electron system includes the delocalized electrons in pi bonds or lone pairs that can contribute to resonance. In the formate ion, there are 2 \(\pi\) electrons found within the double bond shared between carbon and oxygen, forming part of the ion's resonating structure. Despite resinous depiction suggesting the double bond could be with either oxygen, there are always only 2 \(\pi\) electrons contributing to the system. Understanding the electron count in the \(\pi\) system is fundamental in predicting the molecule's reactivity and the distribution of electron density.

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Most popular questions from this chapter

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure \(9.46\), how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a 1 s orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\)-axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

Predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{CCl}_{4}\), (b) \(\mathrm{NH}_{3}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{XeF}_{4}\), (e) \(\mathrm{CH}_{3} \mathrm{Br}\), (f) \(\mathrm{GaH}_{3}\).

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6}\), is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\mathrm{SiH}_{4}\), (b) \(\mathrm{PF}_{3}\), (c) \(\mathrm{HBr}\), (d) \(\mathrm{HCN}\), (e) \(\mathrm{SO}_{2}\).

The molecule shown here is difluoromethane \(\left(\mathrm{CH}_{2} \mathrm{~F}_{2}\right)\), which is used as a refrigerant called \(R-32\). (a) Based on the structure, how many electron domains surround the \(\mathrm{C}\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) From the carbon atom toward a hydrogen atom? [Sections \(9.2\) and 9.3]

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