Chapter 3: Problem 99
Consider the bonding in \(\mathrm{CH}_{2}=\mathrm{CHCHO}\). (a) Draw the most important Lewis structure. Include all nonzero formal charges. (b) Identify the composition of the bonds and the hybridization of each lone pair-for example, by writing \(\sigma\left(H 1 s, C 2 \mathrm{sp}^{2}\right)\).
Short Answer
Expert verified
The most important Lewis structure has a double bond between the first and second carbons and between the third carbon and the oxygen, with a formal positive charge on the third carbon and a negative charge on the oxygen. The bonds are composed of \(\sigma\) and \(\pi\) types, with carbon atoms being sp2 or sp hybridized.
Step by step solution
01
Determine the Total Number of Valence Electrons
Count the valence electrons from all the atoms present in the compound. Carbon has 4 valence electrons, oxygen has 6, and each hydrogen has 1. Since the formula is \( \mathrm{CH}_2=\mathrm{CHCHO} \), the total is \(4(3) + 6 + 1(2) = 20\) valence electrons.
02
Sketch the Skeleton Structure
Arrange the atoms to show which atoms are connected to which, with carbon atoms in the center. The structure should be H2C=CH-CHO, depicting a double bond between the first and second carbon atoms and a double bond between the third carbon and the oxygen.
03
Distribute Electrons To Achieve Full Octets
Place pairs of electrons between atoms to form chemical bonds and around the outer atoms to fill their octets. The oxygen atom will have two lone pairs to complete its octet.
04
Include Nonzero Formal Charges
Calculate the formal charges for atoms that do not possess a complete octet or have more than an octet worth of sharing. In this case, the carbon atom involved in the double bond with oxygen will have a positive charge, and the oxygen will have a negative charge.
05
Identify the Composition of the Bonds
Identify each bond type between the atoms. The single bonds, such as those between carbon and hydrogen, are \(\sigma\) bonds. The double bonds consist of one sigma and one pi bond (\(\sigma\text{ and }\pi\)).
06
Determine Hybridization and Lone Pair Composition
Assign hybridizations: for sp2 hybridized carbons, the bonds are sigma(sp2, sp2) for C-C single bonds, sigma(sp2, s) for C-H, and for double bonds sigma(sp2, sp2) and pi(p, p). For the sp hybridized carbon (C bonded to O), it's sigma(sp, sp2) with the remaining C, and sigma(sp, s) with the H. The O has sigma(sp, sp2) and two lone pairs (nondirectional)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial in determining how an element will react chemically. They are available to form chemical bonds with other atoms to create molecules. The number of valence electrons can be derived from the group number of the elements in the periodic table – for instance, carbon is in group 14, typically possessing 4 valence electrons, while oxygen is in group 16, with 6 valence electrons.
In our exercise, to determine the valence electrons for (mathrm{CH}_2=mathrm{CHCHO}), it's essential to tally up these outer electrons from each atom, which is pivotal in sketching the initial Lewis structure. For (mathrm{CH}_2=mathrm{CHCHO}), there are 3 carbons (each contributing 4 electrons), 2 hydrogens (1 electron each), and 1 oxygen (6 electrons), totaling 20 valence electrons that will be used to create bonds in the molecule.
In our exercise, to determine the valence electrons for (mathrm{CH}_2=mathrm{CHCHO}), it's essential to tally up these outer electrons from each atom, which is pivotal in sketching the initial Lewis structure. For (mathrm{CH}_2=mathrm{CHCHO}), there are 3 carbons (each contributing 4 electrons), 2 hydrogens (1 electron each), and 1 oxygen (6 electrons), totaling 20 valence electrons that will be used to create bonds in the molecule.
Chemical Bonding
Chemical bonding involves the joining of atoms to form new substances; this occurs through interactions involving valence electrons. There are three primary types of chemical bonds: ionic, covalent, and metallic. In our exercise, we are focused on covalent bonds, where atoms share pairs of valence electrons.
To illustrate covalent bonding in (mathrm{CH}_2=mathrm{CHCHO}), we distribute the 20 valence electrons to form bonds. Each line represents a bond, equivalent to a pair of shared electrons. Single bonds or sigma bonds ((sigma)) are drawn between most atoms, but there is also a double bond between carbon atoms and between carbon and oxygen, each double bond consisting of both a sigma and a pi bond ((sigma) and (pi)).
To illustrate covalent bonding in (mathrm{CH}_2=mathrm{CHCHO}), we distribute the 20 valence electrons to form bonds. Each line represents a bond, equivalent to a pair of shared electrons. Single bonds or sigma bonds ((sigma)) are drawn between most atoms, but there is also a double bond between carbon atoms and between carbon and oxygen, each double bond consisting of both a sigma and a pi bond ((sigma) and (pi)).
Formal Charges
A formal charge is a tool for estimating the charge distribution within a molecule. It is calculated by taking the number of valence electrons in an isolated atom, subtracting the electrons in lone pairs, and subtracting half the electrons in bonds.
For (mathrm{CH}_2=mathrm{CHCHO}), the formal charges help us to decide how electron pairs are to be distributed around the structure. The formal charge calculation resulted in a positive charge on the carbon bonded to the oxygen (since it lacks a full octet) and a negative charge on the oxygen (since it is involved in a double bond and also has two lone pairs), indicating that charge is not completely shared equally.
For (mathrm{CH}_2=mathrm{CHCHO}), the formal charges help us to decide how electron pairs are to be distributed around the structure. The formal charge calculation resulted in a positive charge on the carbon bonded to the oxygen (since it lacks a full octet) and a negative charge on the oxygen (since it is involved in a double bond and also has two lone pairs), indicating that charge is not completely shared equally.
Hybridization
Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds. Hybrid orbitals contribute to the molecular geometry of a compound. There are several types of hybridization, like sp, sp2, and sp3.
In (mathrm{CH}_2=mathrm{CHCHO}), the central carbon atoms are sp2 hybridized due to the trigonal planar arrangement around these carbons, while the carbon bonded to the oxygen is sp hybridized, reflecting a linear arrangement. This influences how sigma bonds are formed; for instance, in sp2 hybridized carbons, sigma bonds are formed with other sp2 or s orbitals, and for sp hybridized carbon, sigma bonds are formed with sp2 or s orbitals, as outlined in the exercise solution.
In (mathrm{CH}_2=mathrm{CHCHO}), the central carbon atoms are sp2 hybridized due to the trigonal planar arrangement around these carbons, while the carbon bonded to the oxygen is sp hybridized, reflecting a linear arrangement. This influences how sigma bonds are formed; for instance, in sp2 hybridized carbons, sigma bonds are formed with other sp2 or s orbitals, and for sp hybridized carbon, sigma bonds are formed with sp2 or s orbitals, as outlined in the exercise solution.
Sigma and Pi Bonds
Sigma ((sigma)) and pi ((pi)) bonds are types of covalent bonds. A sigma bond is the strongest type of covalent bond, formed by the end-to-end overlap of atomic orbitals, while pi bonds are formed by the side-to-side overlap of two p-orbitals. Structures that contain double or triple bonds have both sigma and pi bonds.
In the molecule (mathrm{CH}_2=mathrm{CHCHO}), the double bonds are made up of one sigma bond, which forms the bond's basic framework, and one pi bond, which grants the ability for rotation or constraint. Pi bonds are less strong than sigma bonds and are responsible for the addition of chemical reactions due to their relative accessibility.
In the molecule (mathrm{CH}_2=mathrm{CHCHO}), the double bonds are made up of one sigma bond, which forms the bond's basic framework, and one pi bond, which grants the ability for rotation or constraint. Pi bonds are less strong than sigma bonds and are responsible for the addition of chemical reactions due to their relative accessibility.