Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 99

A toxic cloud covered Bhopal, India, in December 1984 when water leaked into a tank of methyl isocyanate, and the product escaped into the atmosphere. Methyl isocyanate is used in the production of many pesticides. Draw the Lewis structure for methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO}\), including resonance forms. The skeletal structure is

Short Answer

Expert verified
The Lewis structure for methyl isocyanate (\(\mathrm{CH}_{3} \mathrm{NCO}\)), including resonance forms, is as follows: Form 1: H | H-C-N-C=O | H Form 2: H | H-C=N-C-O | H

Step by step solution

01

Determine the total number of valence electrons

To draw the Lewis structure, the first step is to count the total number of valence electrons for all atoms in the molecule. Carbon (C) has 4 valence electrons Hydrogen (H) has 1 valence electron Nitrogen (N) has 5 valence electrons Oxygen (O) has 6 valence electrons In the \(\mathrm{CH}_{3} \mathrm{NCO}\) molecule, there is one carbon, three hydrogens, one nitrogen, and one oxygen. Total valence electrons = (1 x 4) + (3 x 1) + (1 x 5) + (1 x 6) = 4 + 3 + 5 + 6 = 18 valence electrons
02

Draw the skeletal structure of the molecule

The skeletal structure of methyl isocyanate is given as \(\mathrm{CH}_{3} \mathrm{NCO}\). Place the least electronegative atom (carbon) in the center and connect the other atoms around it by single bonds. Skeletal structure: H - C - N - C = O
03

Complete the octet rule for all atoms

Distribute the remaining valence electrons to complete the octet (8 electrons) for all atoms besides hydrogen (which can only have 2 electrons). H - C - N - C = O Each C atom requires 3 more electrons, N atom needs 2 more electrons (since it's connected to C with a single bond), and O atom needs 2 more electrons to fulfill the octet rule. Distribute the remaining 10 electrons (18 initial valence electrons - 8 electrons utilized in single bonds): Place an electron pair on nitrogen, complete the octet for both carbon atoms by adding three lone pairs to each, and place an electron pair on oxygen. H | H-C-N-C=O | H
04

Check for resonance structures

Evaluate if resonance structures are possible by redistributing the electrons. A resonance structure can be formed by moving the lone pair on nitrogen to form a double bond between nitrogen and the central carbon atom. Simultaneously, one of the pi bonds between central C and O can shift and create a lone pair on oxygen. Resonance structure: H | H-C=N-C-O | H
05

Present the final Lewis structures, including resonance forms

The Lewis structure for methyl isocyanate (\(\mathrm{CH}_{3} \mathrm{NCO}\)) has two possible resonance forms: Form 1: H | H-C-N-C=O | H Form 2: H | H-C=N-C-O | H Both structures must be considered, as resonance structures represent the real distribution of electrons in the molecule.

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

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide \(\quad\) c. calcium nitride b. strontium fluoride \(\quad\) d. aluminum bromide

Predict the molecular structure for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{BrFI}_{2} \quad\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2} \quad\) c. \(\mathrm{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Consider the following energy changes: $$\begin{array}{ll} \text {} & \quad { \Delta H} \\ \text {} & {(k J / m o l)} \\ \hline \\ {\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-}} & {735} \\ {\mathrm{Mg}^{+}(g) \rightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-}} & {1445} \\ {\mathrm{O}(g)+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(g)} & {-141} \\ {\mathrm{O}^{-}(g)+\mathrm{e}^{-} \rightarrow 0^{2-}(g)} & {878}\end{array}$$ Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) Explain.

Use bond energy values (Table 8.5\()\) to estimate \(\Delta H\) for each of the following reactions in the gas phase. a. \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}\) b. $\mathrm{N} \equiv \mathrm{N}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. PF_. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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.

Sign-up for free