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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 7

In the Lewis structure shown here, A, D, E, Q, X, and Z represent elements in the first two rows of the periodic table. Identify all six elements so that the formal charges of all atoms are zero.

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Hello everyone. So in this video we're given this skeleton of a Lewis structure and we're trying to figure out each and every single one of the possible items. It's labeled A. All the way through F. Let's go ahead and figure this out. So the possible choices that were given is all the elements from the 1st and 2nd periods of the PR table. So that does limit our options a little bit. And that's good because we have fewer items to kind of consider. So first one to consider. Let's see. I want to start off with our oxygen because that's something that we know. So oxygen likes to have So likes to have two bonds and in our skeletal structure here the one with two bonds is going to be Well, let's see here we have our F. So when we say that oxygen is equal to F. Now let's go ahead and consider our hydrogen. So hydrogen also lets indicated here. So we have an Oh, okay. And then for hydrogen, it likes to have or likes to be a terminal atom. And why is that? It's because of the duet rule, which means the hydrogen can only have two electrons for it to be happy. Instead of the octet rule where items like to have eight atoms or electrons. So looking for terminal atoms with a single bound, that's going to be our A and C. We don't need these extra electrons for hydrogen because of the duet role. So that's going to be again A and C. So we'll have an H and H right over here. Next thing to consider is of course going to be our carbons. So like our oxygen, we have a predicted number of bonds. Carbon likes to have four bonds. So you can see in our structure here, the one that's a central carbon would be R, B and R E. For E. We have a double bond. So that's 1234 bonds and that's still considered as the four. Again, that would be B and E. So put it in here, that's carbon and a carbon. Okay, And lastly, the last two, they're similar with one bond, a single bond and three lone pairs. So to follow the formal charge, we have one bond again and the three lone pairs and that gives us seven electrons. Flooring has seven valence electrons. So that would give a formal charge of zero. So then our last one, our flooring would be for Adams, D and G. This F and this is A F. So again, back to our carbon here, we know that it likes to have more bonds and it is usually the least electro negative, therefore it's going to be our central atom. So we see in chemistry usually that carbons will be our central item. So again, her answers then would be the oxygen would be for f, hydrogen would be a C. Carbon would be B and E. And our flooring would be D and G. And that's going to be my final answer for this problem. Thank you all so much for watching.
Related Practice
Textbook Question

The orbital diagram that follows shows the valence electrons for a 3+ ion of an element. (a) What is the element?

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Textbook Question

Which of the following charts shows the general periodic trends for the electronegativities of the representative elements? [Section 8.4]

Textbook Question

A molecule with formula C4H3NO has the connectivity shown in the figure. After the Lewis structure of the molecule is completed, how many of each of the following are there in the molecule:

a. single bonds,

b. double bonds,

c. triple bonds,

d. nonbonding pairs? [Sections 8.3 and 8.5]

Textbook Question

The molecule shown here is styrene, C8H8, a benzene derivative that is used to make a number of polymers, including polystyrene. The shorthand notation for the benzene ring (described in Section 8.6) is used. Three of the carbon–carbon bonds are numbered in the structure.

a. Which of the three bonds is the strongest?

b. Which of the three bonds is the longest?

c. Which of the three bonds is best described as halfway between a single and a double bond? [Sections 8.6 and 8.8]

Textbook Question

Consider the Lewis structure for the polyatomic oxyanion shown here, where X is an element from the third period (Na - Ar). By changing the overall charge, n, from 1- to 2- to 3- we get three different polyatomic ions. For each of these ions (b) determine the formal charge of the central atom, X;

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Textbook Question

(a) True or false: An element's number of valence electrons is the same as its atomic number.

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