Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 23

Use the localized electron model to describe the bonding in $\mathrm{H}_{2} \mathrm{CO}(\text { carbon is the central atom })$

Short Answer

Expert verified
In H₂CO (formaldehyde), carbon is the central atom and is double bonded to oxygen and single bonded to each hydrogen atom. The localized electron model demonstrates bonding through the Lewis structure, where valence electrons are distributed to satisfy the octet rule for all atoms involved. The Lewis structure of H₂CO is: H | H - C = O |

Step by step solution

01

Determine the number of valence electrons

First, we need to determine the number of valence electrons for each atom in the molecule. Hydrogen has 1 valence electron, carbon has 4 valence electrons, and oxygen has 6 valence electrons. The molecule has 2 hydrogen atoms, 1 carbon atom, and 1 oxygen atom. Therefore, the total number of valence electrons is: Total valence electrons = (2 x 1) + (1 x 4) + (1 x 6) = 2 + 4 + 6 = 12
02

Arrange the atoms and create the bonds

Now, we will arrange the atoms and create the bonds between them. Start with the central carbon atom and connect it to the two hydrogen atoms and the oxygen atom. Each bond represents a pair of shared electrons. H | H - C - O After creating the initial bonds, we have used 6 valence electrons (3 pairs).
03

Distribute the remaining valence electrons

We have 6 valence electrons left (12 - 6). Distribute them around the oxygen atom to satisfy the octet rule. Octet rule states that every atom in the molecule should have 8 electrons around it, including the shared electrons. H | H - C - O | O Now, the oxygen atom has an octet, and each hydrogen atom has 2 electrons around it. The carbon atom, however, has only 6 electrons around it (3 pairs).
04

Modify the structure to satisfy the octet rule for carbon

To satisfy the octet rule for carbon, we need to move one of the lone pairs from oxygen and form a double bond with carbon: H | H - C = O | The modified structure now depicts the bonding in formaldehyde. Carbon has 8 electrons around it (4 pairs), oxygen has 8 electrons (4 pairs), and each hydrogen atom has 2 electrons (1 pair). #Conclusion# In H₂CO, carbon is the central atom, double bonded to oxygen and single bonded to each hydrogen atom. The localized electron model helps illustrate the bonding in formaldehyde using the Lewis structure, depicting the distribution of valence electrons and satisfying the octet rule for all atoms involved.

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.

Start your free trial

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

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

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? $$ \text {a} \mathrm{Li}_{2} \quad \text { b. } \mathrm{C}_{2} \quad \text { c. } \mathrm{S}_{2} $$

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 SbF 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}\( SbFs (density \)=3.10 \mathrm{g} / \mathrm{mL}$ ) are allowed to react?

The molecules \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) are isoelectronic but their properties are quite different. Although as a first approximation we often use the same MO diagram for both, suggest how the \(\mathrm{MOs}\) in \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) might be different.

What modification to the molecular orbital model was made from the experimental evidence that \(\mathrm{B}_{2}\) is paramagnetic?

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

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.

Sign-up for free