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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 91b

The hypochlorite ion, ClO-, is the active ingredient in bleach. The perchlorate ion, ClO4-, is a main component of rocket propellants. Draw Lewis structures for both ions. (b) What is the formal charge of Cl in the perchlorate ion, assuming the Cl—O bonds are all single bonds?

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Hey everyone were asked to draw the lewis structure of the pariah date ion where all our iodine and oxygen bonds are all single bonds and were asked to determine the formal charge of our iodine first. Let's go ahead and calculate the total number of valence electrons will have. So for iodine since we have one of iodine and it's in our group seven A. This will give us seven valence electrons. For oxygen, oxygen is in our group six A. So this will give us six valence electrons and we're going to multiply this by four since we have four of oxygen, this will get us to a total of 24 valence electrons. When we add these two values up, we get 31 valence electrons. But since we have that -1 charge, this means that we'll have to add one valence electron. So in total we need to draw out 32 valence electrons, drawing this out. We know that our iodine and our oxygen will be connected by a single bond for all four oxygen's. And to complete oxygen, doctor will have to add three lone pairs onto each one Because we added those three lone pairs, oxygen will have a formal charge of -1. So we're going to denote this by adding a negative charge next to our oxygen And our ion as a whole we know has a -1 charge. So we'll add a superscript of -1. Now to calculate the formal charge of our iodine, This is going to be equal to our group number which is seven. And we're going to subtract our bonds plus our non bonding electrons. In this case we have four bonds and we're going to add four plus zero since we have zero non bonding electrons surrounding our iodine. When we calculate this out, we end up with a formal charge of positive three. So in our Lewis structure we can go ahead and add that three plus charge on our iodine. And this is going to be our final answers now. I hope that made sense and let us know if you have any questions.
Related Practice
Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion.

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Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (d) What is the formal charge on hydrogen in the polyatomic ion?

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Textbook Question

Although I3- is a known ion, F3- is not. (c) Another classmate says F3- does not exist because it would violate the octet rule. Is this classmate possibly correct?

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Textbook Question

The following three Lewis structures can be drawn for N2O:

(a) Using formal charges, which of these three resonance forms is likely to be the most important?

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Textbook Question

The following three Lewis structures can be drawn for N2O:

(b) The N—N bond length in N2O is 1.12 Å, slightly longer than a typical N ≡N bond; and the N— O bond length is 1.19 Å, slightly shorter than a typical N ═O bond (see Table 8.4). Based on these data, which resonance structure best represents N2O?

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Textbook Question

Ortho-Dichlorobenzene, C6H4Cl2, is obtained when two of the adjacent hydrogen atoms in benzene are replaced with Cl atoms. A skeleton of the molecule is shown here. (a) Complete a Lewis structure for the molecule using bonds and electron pairs as needed.

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