One of the followving compounds does not exist. Use Lewis structures to identify that compound. (a) \(\mathrm{C}_{2} \mathrm{H}_{2} ;\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\); (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{8}\).

Short Answer

Expert verified
Compound (d) \(\mathrm{C}_{2} \mathrm{H}_{8}\) does not exist because it violates the octet rule for carbon.

Step by step solution

01

Recall the valency of carbon and hydrogen

Remember that carbon has 4 valence electrons and seeks to make 4 bonds to complete its octet, while hydrogen has 1 valence electron and can form only 1 bond.
02

Draw the Lewis structure for each compound

(a) For \(\mathrm{C}_{2} \mathrm{H}_{2}\), each carbon forms a triple bond with the other carbon, using 6 of its valence electrons, and each forms a single bond with a hydrogen atom. It satisfies the octet rule for both carbons. (b) For \(\mathrm{C}_{2} \mathrm{H}_{4}\), each carbon forms a double bond with the other carbon, using 4 of its valence electrons, and two single bonds with hydrogen atoms. It also satisfies the octet rule for both carbons. (c) For \(\mathrm{C}_{2} \mathrm{H}_{6}\), each carbon forms a single bond with the other carbon, using 2 of its valence electrons, and makes three single bonds with hydrogen atoms. Again, it satisfies the octet rule for both carbons.
03

Attempt to draw the Lewis structure for the remaining compound

(d) For \(\mathrm{C}_{2} \mathrm{H}_{8}\), we try to draw a structure where each carbon forms a single bond with the other carbon and makes four single bonds with hydrogen atoms. However, this results in each carbon exceeding the octet rule, as they would end up with 5 bonds or 10 valence electrons, which is not possible for carbon.
04

Identify the compound that does not exist

Compound (d) \(\mathrm{C}_{2} \mathrm{H}_{8}\) cannot exist because it requires carbon to form 5 bonds, which violates the octet rule.

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Key Concepts

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

Chemical Bonding
Chemical bonding is the force that holds atoms together in compounds. Atoms bond to achieve a more stable electron configuration, often akin to that of noble gases. There are three primary types of chemical bonds: ionic, covalent, and metallic.

Ionic bonds form between metals and nonmetals, where electrons are transferred from one atom to another. Covalent bonds, on the other hand, involve the sharing of electron pairs between nonmetals. This sharing allows each atom to reach the electron configuration of a noble gas. In a metallic bond, electrons are free to move around, which explains the conductive properties of metals.

Understanding how atoms bond enables us to predict the structures and properties of compounds. For example, the strong double bond in oxygen gas (O_2 ) makes it less reactive under normal conditions compared to ozone (O_3 ) with a weaker bond order. When we use Lewis structures, we represent these bonds with lines (a single line for a single bond, two for a double bond, etc.). In the exercise, we examined different hydrocarbons and how carbon bonds with hydrogen to form stable molecules.
Valence Electrons
Valence electrons are the electrons in the outermost shell of an atom and play a pivotal role in determining an element's chemical properties, including its ability to bond with other atoms. These electrons are the ones involved in forming bonds.

Each element has a characteristic number of valence electrons that influences its bonding behavior. For example, carbon has four valence electrons and can form four covalent bonds. Hydrogen, with one valence electron, can form one covalent bond. In the exercise, we used this knowledge to deduce the number of bonds each carbon and hydrogen can make in hydrocarbons.

It’s important to remember that in Lewis structures, valence electrons are depicted as dots surrounding the elements’ symbols, with shared pairs (or bonds) represented as lines. This visualization helps to understand how atoms interact to form molecules and the resulting molecular structure.
Octet Rule
The octet rule is a chemical rule of thumb that reflects the observation that atoms of main-group elements tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas.

The octet rule explains why atoms form the types of ions and molecules that they do. For example, atoms will share, donate, or receive electrons to achieve a stable octet. However, there are exceptions to the octet rule; for instance, hydrogen is stable with two electrons (a duet), and transition metals may not always follow the rule due to their unique electron configurations.

In the given exercise, this rule helped us determine why (C_{2}H_{8}) couldn't exist. Carbon cannot form five bonds without violating the octet rule, as it would result in having ten valence electrons. Lewis structures visually enforce the octet rule by showing that each carbon atom can only accommodate eight electrons through bonding.

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Most popular questions from this chapter

Write the Lewis structures that contribute to the resonance hybrid of the guanadinium ion, \(\mathrm{C}\left(\mathrm{NH}_{2}\right)_{3}{ }^{+}\).

In 1999, Karl Christe synthesized and characterized a salt that contained the \(\mathrm{N}_{5}{ }^{+}\)cation, in which the five \(\mathrm{N}\) atoms are connected in a long chain. This cation is the first all-nitrogen species to be isolated in more than 100 years. Draw the most important Lewis structure for this ion, including all equivalent resonance structures. Calculate the formal charges on all atoms.

Write the Lewis structure for each of the following molecules or ions and give the number of electrons about the central atom: (a) \(\mathrm{SiF}_{6}^{2-}\); (b) \(\mathrm{IF}_{7}\); (c) \(\mathrm{ClF}_{3} ;\) (d) \(\mathrm{BrF}_{2}^{+}\).

The perchlorate ion, \(\mathrm{ClO}_{4}^{-}\), is described by resonance structures. (a) Draw the Lewis structures that contribute to the resonance hyhrid and identify the most plawsible Lewis structures by using formal charge arguments. (b) The average length of a single \(\mathrm{Cl}-\mathrm{O}\) bond is 172 pm and that of a double \(\mathrm{Cl}=\mathrm{O}\) bond can be estimated at \(140 \mathrm{pm}\). The Cl-O bond length in the perchlorate ion is found experimentally to be \(144 \mathrm{pm}\) for all four bonds. Identify the most plausible Lewis structures of the perchlorate ion from these experimental data. (c) What is the oxidation mumber of chlorine in the perchlorate ion? Identify the most plausible Lewis structure by using the coxidation number, assuming that lone pairs belong to the atom to which they are attached but that all electrons shared in a bond helong to the atom of the more negative element. (d) Are these three approaches consistent? Explain why or why not.

The following species have the same number of electrons: Cd, \(\mathrm{In}^{+}\), and \(\mathrm{Sn}^{2+}\). (a) Write the electron configurations for each species. Are they the same or different? Explain. (b) How many unpaired electrons, if any, are present in each species? (c) What neutral atom, if any, has the same electron configuration as that of \(\operatorname{In}^{3+}\) ?

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