Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\). Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f .\) Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

Short Answer

Expert verified
The Lewis structures of acrylonitrile and methyl methacrylate are H2C=CH-C≡N and H3C-C(=O)-O-CH2=C(CH3)2, respectively. In acrylonitrile, bond angles a, b, and c are 120°, 120°, and 180°, respectively. In methyl methacrylate, bond angles d, e, and f are 109.5°, 120°, and 120°, respectively. All carbon atoms in acrylonitrile are sp² hybridized. In methyl methacrylate, the terminal methyl carbon is sp³ hybridized, the carbonyl carbon and the carbon connected to the ester group are both sp² hybridized. In acrylonitrile, all 5 atoms lie in the same plane. Acrylonitrile has 3 σ and 4 π bonds, while methyl methacrylate has 8 σ and 2 π bonds.

Step by step solution

01

Draw the Lewis structures of acrylonitrile and methyl methacrylate

Draw the structures showing the bonds between the atoms and the lone pairs of electrons. The Lewis structures for the two molecules are as follows: Acrylonitrile: H2C=CH-C≡N Methyl methacrylate: H3C-C(=O)-O-CH2=C(CH3)2
02

Determine bond angles

We need to determine the bond angles a through f for both molecules. For this, we will look at the bonds around the atoms in question: a) Angle H-C-H in acrylonitrile \[a = 120^{\circ}\] (since it is an sp² carbon) b) Angle H-C-C in acrylonitrile \[b = 120^{\circ}\] (since both carbons are sp² hybridized) c) Angle C-C-N in acrylonitrile \[c = 180^{\circ}\] (since the carbon and nitrogen are sp hybridized) d) Angle H-C-C in methyl methacrylate \[d=109.5^{\circ}\] (since it is an sp³ carbon) e) Angle O-C-O in methyl methacrylate \[e = 120^{\circ}\] (since it is sp² hybridized carbon) f) Angle H-C-C in methyl methacrylate \[f = 120^{\circ}\] (since it is an sp² hybridized carbon)
03

Identify hybridization

Now, we will analyze the hybridization of all the carbon atoms in the molecules: - In acrylonitrile, both carbons are sp² hybridized. - In methyl methacrylate, the terminal methyl carbon is sp³ hybridized, the carbonyl carbon is sp² hybridized, and the carbon connected to the ester group is also sp² hybridized.
04

Determine planar atoms in acrylonitrile

In acrylonitrile, the molecule is planar because all the atoms have either sp or sp² hybridizations. Therefore, all five atoms (2 carbons, 1 nitrogen, and 2 hydrogens) must lie in the same plane.
05

Identify σ and π bonds

Now, let's identify the number of σ and π bonds in the molecules: - In acrylonitrile, there are 3 σ bonds (C-H, C-C, and C-N) and 4 π bonds (2 C=C, and 2 C≡N) - In methyl methacrylate, there are 8 σ bonds (1 C-H, 3 O-C, 1 C=O, 1 O-CH2, and 2 CH2=C) and 2 π bonds (1 C=O and 1 C=C) So, in summary, acrylonitrile has 3 σ bonds and 4 π bonds, and methyl methacrylate has 8 σ bonds and 2 π bonds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lewis Structures
Understanding the basics of Lewis structures is crucial for predicting the properties and behavior of molecules. A Lewis structure is a diagram that represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist. In these structures, each dot represents an electron, and a line represents a bond between two atoms.

When drawing Lewis structures, balance is key. You should ensure that each atom, except for hydrogen, follows the octet rule, having eight electrons to fill its outer shell. Hydrogen is an exception as it can be stable with just two electrons. For the molecules in our exercise, the electrons are arranged to reflect the bonding pattern and molecular shape. By observing these drawings, one can predict the connectivity of the atoms and the distribution of electrons which acts as the link to understanding the molecule's geometry and reactivity.
Bond Angles
The bond angles of a molecule can greatly affect its shape and physical properties. These angles are determined by the spatial arrangement of bonds around a central atom which is influenced by the electrons' repulsion. In molecules where the central atom has sp² hybridization, such as acrylonitrile, the bond angles are approximately 120°. This corresponds to the trigonal planar arrangement of molecules.

Contrarily, a tetrahedral molecule, where the central atom has an sp³ hybridization, usually has bond angles close to 109.5°. These angles can provide insight into the spatial distribution of a molecule's atoms. Estimating bond angles allows students to better understand molecular shapes and the forces at play within the structure.
Hybridization
The concept of hybridization helps explain the observed shapes and bond angles in molecules. Hybridization occurs when atomic orbitals mix to form new, identical hybrid orbitals. This mixing directly influences molecular bonding and geometry.

For instance, in acrylonitrile and methyl methacrylate, the type of hybridization stipulates the bonding properties: an sp² hybridized atom will have one unhybridized p-orbital available for π bonding which is essential for double bonds. Similarly, an sp³ hybridized atom will form single bonds only. Recognizing hybridization patterns enables a student to predict not just shapes, but also the potential for chemical reactions.
Sigma and Pi Bonds
Breaking down bonds into sigma (σ) and pi (π) types offers deeper insight into how atoms in molecules connect. Sigma bonds are the single bonds that form between atoms. They allow for free rotation around the bond axis because they involve head-on overlapping of orbitals. In contrast, pi bonds are formed by the side-to-side overlap of atomic orbitals, typically p orbitals, which occur in double and triple bonds.

Pi bonds restrict the rotation about the bond axis, which contributes to the rigidity of the molecule's structure. By counting the sigma and pi bonds, as done in acrylonitrile and methyl methacrylate, students can better grasp the molecular complexity and the effects on the molecule’s behavior.
Molecular Geometry
The spatial arrangement of atoms in a molecule, or its molecular geometry, is determined by the electronic and steric interactions. The VSEPR (Valence Shell Electron Pair Repulsion) theory is a helpful tool for predicting molecular shapes based on the repulsion between electron pairs in the valence shell of the central atom.

Geometries like linear, trigonal planar, tetrahedral, and others are all derived from the specific electron pair repulsions. In both acrylonitrile and methyl methacrylate, the geometry affects the compound's polarity, reactivity, and physical properties. Grasping the principles of molecular geometry is vital for students to anticipate the behavior of different molecular structures in real-life chemical contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider three molecules: A, B, and C. Molecule A has a hybridization of \(s p^{3}\). Molecule \(\mathrm{B}\) has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule \(\mathrm{C}\) consists of two \(\sigma\) bonds and two \(\pi\) bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, \(\mathrm{B}_{2}\). a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2}\) ? b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part b) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{~kJ} / \mathrm{mol} $$ How much energy is needed to ionize \(1.5 \mathrm{~kg} \mathrm{~B}_{2}\) to the desired isoelectronic species?

An unusual category of acids known as superacids, which are defined as any acid stronger than \(100 \%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with \(\mathrm{SbF}_{5}\) produces the superacid \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared when \(2.93 \mathrm{~mL}\) anhydrous \(\mathrm{HF}\) (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free