/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 A carbene is a compound that has... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 75

A carbene is a compound that has a carbon bonded to two atoms and a lone pair remaining on the carbon. Many carbenes are very reactive. (a) Draw the Lewis structure for the simplest carbene, \(\mathrm{H}_{2} \mathrm{C}\). (b) Predict the length of the carbon-carbon bond you would expect if two \(\mathrm{H}_{2} \mathrm{C}\) molecules reacted with each other by a combination reaction.

Short Answer

Expert verified
The Lewis structure for \(\mathrm{H}_{2}\mathrm{C}\) is \(\mathrm{H}-\mathrm{C}: \mathrm{H}\). The expected \(\mathrm{C-C}\) bond length in the reaction product is around 154 pm.

Step by step solution

01

Understanding Carbenes

Carbenes are compounds with a carbon atom bonded to two other atoms and possessing a lone pair. In this exercise, we are considering the simplest carbene, \(\mathrm{H}_{2}\mathrm{C}\), where carbon is bonded to two hydrogen atoms and carries a lone pair of electrons.
02

Drawing the Lewis Structure for \(\mathrm{H}_{2}\mathrm{C}\)

Start by placing the carbon atom as the central atom. Attach two hydrogen atoms to the carbon. Since carbon forms four bonds or has an octet of electrons, for \(\mathrm{H}_{2}\mathrm{C}\), there are also two non-bonding electrons (a lone pair) on carbon to complete its octet. The Lewis structure is: \(\mathrm{H}-\mathrm{C}: \mathrm{H}\).
03

Reaction of Two Carbenes

When two \(\mathrm{H}_{2}\mathrm{C}\) molecules react, the lone pairs on each carbon atom can form a new \(\mathrm{C-C}\) bond. This type of reaction is known as a combination reaction.
04

Predicting the Carbon-Carbon Bond Length

After the combination reaction, the \(\mathrm{H}_{2}\mathrm{C}\) molecules form \(\mathrm{H}_{2}\mathrm{C}-\mathrm{CH}_{2}\) with a \(\mathrm{C-C}\) single bond. The typical carbon-carbon single bond length is approximately 154 pm (picometers). Thus, the expected \(\mathrm{C-C}\) bond length in the resulting molecule would be around 154 pm.

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 91影视!

Key Concepts

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

Lewis Structure
A Lewis structure is a simple way to represent the arrangement of electrons in a molecule. It helps visualize how atoms are bonded together and the presence of any lone pairs of electrons. For the simplest carbene, \(\mathrm{H}_{2}\mathrm{C}\), the Lewis structure shows carbon in the center. This carbon atom bonds with two hydrogen atoms and also carries a lone pair of electrons. This lone pair is essential for completing carbon's valence shell with eight electrons, following the octet rule.
  • The carbon atom is central, connected to two hydrogen atoms.
  • A lone pair of electrons sits on the carbon.
  • The structure is depicted as \(\mathrm{H}-\mathrm{C}:\mathrm{H}\).
Lewis structures are crucial for understanding molecular structure, as they provide insight into the bonding and electron pairs within a molecule.
Combination Reaction
A combination reaction involves two or more substances coming together to form a single product. In this exercise, two \(\mathrm{H}_{2}\mathrm{C}\) carbene molecules react by combining their lone pairs of electrons to form a new carbon-carbon bond.
This results in the production of a larger molecule, \(\mathrm{H}_{2}\mathrm{C}-\mathrm{CH}_{2}\).

  • Lone pairs from each carbon in \(\mathrm{H}_{2}\mathrm{C}\) facilitate the formation of a \(\mathrm{C-C}\) bond.
  • The molecules align such that carbon atoms are adjacent, allowing bond formation.
  • The result is a single, larger molecule with a new covalent bond.
Combination reactions are fundamental in chemical synthesis, as they allow simple molecules to come together and form complex structures.
Carbon-carbon bond length
The carbon-carbon bond length is a critical measurement in chemistry, as it influences the properties and stability of a molecule. When two \(\mathrm{H}_{2}\mathrm{C}\) react to form \(\mathrm{H}_{2}\mathrm{C}-\mathrm{CH}_{2}\), a single \(\mathrm{C-C}\) bond is created. The typical length for a carbon-carbon single bond in stable organic molecules is approximately 154 picometers (pm). This length can vary slightly based on the surrounding atoms and molecular environment, but 154 pm is a reliable average.

  • The bond originates from the overlap of \(sp^3\) hybrid orbitals on each carbon atom.
  • Stable \C-C\ single bonds form the backbone of many organic compounds.
  • The precise length can influence chemical reactivity and physical properties.
Understanding bond lengths is pivotal in predicting and explaining the behavior of organic and inorganic compounds in various environments.

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

Ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a very soluble salt in water. (a) Draw the Lewis structures of the ammonium and chloride ions. (b) Is there an \(\mathrm{N}-\mathrm{Cl}\) bond in solid ammonium chloride? (c) If you dissolve \(14 \mathrm{~g}\) of ammonium chloride in \(500.0 \mathrm{~mL}\) of water, what is the molar concentration of the solution? (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right)\), borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\right.\) which is analogous to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring), or boron trichloride?

A classmate of yours is convinced that he knows everything about electronegativity. (a) In the case of atoms X and Y having different electronegativities, he says, the diatomic molecule \(X-Y\) must be polar. Is your classmate correct? (b) Your classmate says that the farther the two atoms are apart in a bond, the larger the dipole moment will be. Is your classmate correct?

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Iodine has valence electrons and makes ___________ bond(s) in compounds. (b) Silicon has ____________ valence electrons and makes _________ bond(s) in compounds. (c) Phosphorus has ___________ valence electrons and makes ________ bond(s) in compounds. (d) Sulphur has __________ valence electrons and makes __________ bond(s) in compounds.

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: \((\mathbf{a}) \mathrm{PF}_{6}^{-},(\mathbf{b}) \mathrm{BeCl}_{2},(\mathbf{c}) \mathrm{NH}_{3},(\mathbf{d}) \mathrm{XeF}_{2} \mathrm{O}\) (the Xe is the central atom), \((\mathbf{e}) \mathrm{SO}_{4}^{2-} .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation
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.