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Chapter 4: Problem 13

By considering only valence electrons in each case, draw Lewis structures for the following molecules: (a) methane \(\left(\mathrm{CH}_{4}\right)\); (b) bromomethane \(\left(\mathrm{CH}_{3} \mathrm{Br}\right)\), (c) ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ;\) (d) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} ,\mathrm{OH}\right)\); (e) ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\).

Short Answer

Expert verified
Draw the skeleton, distribute bonds with electrons, complete octets. Total e鈦� should match initial counts. E.g., CH鈧�, all single bonds; C鈧侶鈧�, double bond between C atoms.

Step by step solution

01

Determine Total Valence Electrons

First, we need to determine the total number of valence electrons for each molecule. - Methane (CH鈧�): C has 4 valence e鈦�, H has 1 valence e鈦� 脳 4 H = 4 + 4 = 8 e鈦�. - Bromomethane (CH鈧傿r): C has 4 valence e鈦�, H has 1 e鈦� 脳 3, Br has 7 e鈦� = 4 + 3 + 7 = 14 e鈦�. - Ethane (C鈧侶鈧�): Each C has 4 e鈦� 脳 2, H has 1 e鈦� 脳 6 = 8 + 6 = 14 e鈦�. - Ethanol (C鈧侶鈧匫H): Each C has 4 e鈦� 脳 2, H has 1 e鈦� 脳 6, O has 6 e鈦� = 8 + 6 + 6 = 20 e鈦�. - Ethene (C鈧侶鈧�): Each C has 4 e鈦� 脳 2, H has 1 e鈦� 脳 4 = 8 + 4 = 12 e鈦�.
02

Draw the Skeleton Structure

Arrange the atoms in a way that they are typically bonded, usually with carbon atoms at the center, and hydrogen atoms and halogens or other elements at the periphery. Also, place any functional groups appropriately. - Methane: C in the center bonded to 4 H. - Bromomethane: C in the center bonded to 3 H and 1 Br. - Ethane: Two C bonded together; each C bonded to 3 H. - Ethanol: Two C bonded together; one C bonded to 3 H, the other C bonded to 2 H and to OH. - Ethene: Two C bonded together with a double bond; each C bonded to 2 H.
03

Distribute Valence Electrons as Bonds

Each bond constitutes 2 electrons. Start by creating single bonds (lines) between all connected atoms according to the skeleton structure and verify the total bonds equal the total valence electrons calculated. - Methane: 4 single bonds (C-H). - Bromomethane: 3 single bonds (C-H) and 1 (C-Br). - Ethane: 6 single bonds (similar C-H bonds) and 1 (C-C). - Ethanol: Similar to Ethane, with an additional bond for O-H. - Ethene: 4 single bonds (C-H) and 1 double bond (C=C).
04

Complete Octets and Account for Hydrogen

Ensure each atom follows the octet rule (except hydrogen, which obeys the duet rule, needing only 2 electrons). - Methane, Bromomethane, Ethane, Ethanol: Check C with 4 bonds and other atoms until they obey their rules. - Ethene: Each C with its double bond and 2 H also satisfies the octet and duet rule respectively.
05

Confirm Structures

Ensure that all valence electrons are accounted for, satisfy the bonding requirements, and any remaining electrons are placed around atoms like Br or O as lone pairs. - Methane needs no lone pairs; others distribute any remaining electrons as lone pairs on atoms like Br in bromomethane or O in ethanol, ensuring these atoms achieve necessary electron configurations.

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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 play a key role in chemical reactions and bond formation. These electrons are typically involved in forming bonds and determining the chemical properties of elements. In Lewis structures, valence electrons are represented as dots around the elemental symbols.
For example, carbon has four valence electrons, commonly forming bonds by sharing its electrons with others to achieve a stable configuration. For hydrogen, there is only one valence electron, which easily bonds with other atoms to fill its outer shell.
  • Methane (CH鈧�): Carbon (4 valence e鈦�) shares electrons with four hydrogen atoms (1 valence e鈦� each), resulting in a total of 8 valence electrons.
  • Bromomethane (CH鈧傿r): Carbon bonds to three hydrogen atoms and one bromine atom, resulting in a total of 14 valence electrons.
  • Ethane (C鈧侶鈧�) and Ethanol (C鈧侶鈧匫H): These molecules involve carbon-carbon bonds and lead to shared valence electrons among multiple atoms.
  • Ethene (C鈧侶鈧�): Features a double bond between two carbons, resulting in additional shared pairs of electrons.
Octet Rule
The octet rule is a principle stating that atoms tend to form bonds until they have eight electrons in their valence shell, resembling the electron configuration of a noble gas. It is essential for predicting how atoms bond and form molecules.
In Lewis structures, we use the octet rule to ensure atoms are stable with appropriate electron sharing:
  • Carbon generally follows the octet rule, achieving eight electrons by sharing.
  • Hydrogen is the exception, as it follows the duet rule, needing only two electrons to be stable.
  • Molecules like Methane and Ethane perfectly align with the octet rule, with carbon atoms forming four bonds.
  • In molecules like Ethene, the double bond between carbon atoms helps satisfy the octet rule, enabling the shared electrons of each bond to reach eight surrounding an atom.
Chemical Bonding
Chemical bonding refers to the process by which atoms or groups of atoms combine to form molecules. It primarily involves interactions between valence electrons. There are different types of chemical bonds, but covalent bonds are most relevant for the molecules discussed:
  • Covalent Bonds: These occur when two atoms share one or more pairs of valence electrons. In Methane and Ethane, the carbon atoms form covalent bonds with hydrogen and other carbon atoms, sharing electrons to reach stability.
  • Bromomethane features both carbon-hydrogen covalent bonds and a carbon-bromine bond, where electrons are shared.
  • In molecules like Ethene, a double covalent bond is present between two carbon atoms, sharing two pairs of electrons to increase bonding strength.
Molecular Geometry
Molecular geometry involves the three-dimensional arrangement of atoms in a molecule, which influences physical and chemical properties. It is based on electron pair repulsion, where electrons around a central atom arrange themselves as far apart as possible.
  • Tetrahedral Shape: Methane (CH鈧�) is an example with a tetrahedral geometry, as the carbon bonds equally in a 3D shape with hydrogen atoms spaced at 109.5掳 angles.
  • Trigonal Pyramidal: For Bromomethane (CH鈧傿r), the central atom still retains a tetrahedral form but appears as a trigonal pyramidal structure due to the bromine atom being larger and occupying more space.
  • Linear and Planar Structures: Ethene (C鈧侶鈧�) features a trigonal planar geometry due to the double bond, creating a flat shape as the carbon atoms and attached hydrogens align in a plane.
  • V-shaped or Bent Structures: In Ethanol, the presence of the OH group leads to a bent molecular shape due to the lone pairs on oxygen, affecting bond angles and causing deviations from a perfect tetrahedral shape.

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Most popular questions from this chapter

State what you understand by each of the following: (a) covalent radius, (b) bond dissociation enthalpy and (c) the standard enthalpy of atomization of an element.

How many Lewis structures can you draw for \(\mathrm{C}_{2}\) ? In the text, we considered only one structure; suggest why the other structure(s) that you have drawn is (are) unreasonable. [Hint: Think about the distribution of the bonding electrons in space.]

Comment on the following observations. (a) An \(\mathrm{O}-\mathrm{O}\) bond is weaker than an \(\mathrm{S}-\mathrm{S}\) bond, but \(\mathrm{O}=\mathrm{O}\) is stronger than \(\mathrm{S}=\mathrm{S}\) (see Table 4.5 ). (b) There is evidence for an unstable \(\mathrm{Be}_{2}\) species with a bond enthalpy of \(10 \mathrm{kJ} \mathrm{mol}^{-1}\) and bond distance of \(245 \mathrm{pm}\) (c) \(P \equiv P\) bonds are weaker than \(N \equiv N\) or \(C \equiv C,\) but stronger than \(\mathrm{As} \equiv \mathrm{As}\) (d) The covalent radius for \(P\) appropriate for a triple bond is \(94 \mathrm{pm},\) but in \(\mathrm{P}_{4},\) the \(\mathrm{P}-\mathrm{P}\) bond distances are \(221 \mathrm{pm}\)

The bond dissociation enthalpy of \(\mathrm{Na}_{2}\) is \(74 \mathrm{kJ} \mathrm{mol}^{-1}\). Do you expect the bond enthalpy in \(\left[\mathrm{Na}_{2}\right]^{+}\) to be less than, equal to or greater than \(74 \mathrm{kJ} \mathrm{mol}^{-1} ?\) Rationalize your answer.

Construct a complete MO diagram for the formation of \(\mathrm{Li}_{2}\) showing the involvement of both the core and valence electrons of the two lithium atoms. Determine the bond order in \(\mathrm{Li}_{2}\). Does the inclusion of the core electrons affect the value of the bond order?
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