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Chapter 9: Problem 50

Draw three resonance structures for the chlorate ion, \(\mathrm{ClO}_{3}^{-}\). Show formal charges.

Short Answer

Expert verified
The three resonance structures for \(\mathrm{ClO}_{3}^{-}\) can be drawn, with Chlorine atom in the center, surrounded by three Oxygen atoms. For each structure, Chlorine makes a double bond with one Oxygen atom, and single bonds with the other two. The formal charges are 0 on the Oxygen with a double bond and -1 on all other Oxygen atoms.

Step by step solution

01

Determine the Total Number of Valence Electrons

Each Oxygen has 6 valence electrons and Chlorine has 7 as it is in the 7th group (or 17th group in 18 column format) in the periodic table. Further, there is an extra one electron due to the ion having -1 charge. Therefore, total valence electrons are \( (6 \times 3) + 7 + 1 = 26 \).
02

Draw the Mechanism

Place the Chlorine atom in the center, the Oxygen atoms surrounding it, and bond each Oxygen to Chlorine with a single bond. This will use 6 electrons. Now, place remaining 20 electrons on the 3 Oxygen atoms, providing each with 8 electrons (2 from the bond and 6 as lone pairs). So far, Chlorine has only 6 electrons and is not satisfying the octet rule.
03

Create Resonance Structures

To satisfy the octet rule for Chlorine, a double bond can be drawn with each Oxygen atom one after another. These give the three resonance structures. For each structure, there is a single Oxygen atom with a double bond sharing 2 extra electrons with Chlorine, and the other two Oxygen atoms have single bonds each. Each Oxygen atom will have a formal charge of -1 for those having single bonds and 0 for the one with a double bond.

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