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Chapter 8: Problem 91

The hypochlorite ion, \(\mathrm{ClO}^{-},\) is the active ingredient in bleach. The perchlorate ion, \(\mathrm{ClO}_{4}^{-},\) is a main component of rocket propellants. Draw Lewis structures for both ions.

Short Answer

Expert verified
The Lewis structure for the hypochlorite ion (ClO-) is: O || Cl- The Lewis structure for the perchlorate ion (ClO4-) is: O || O--Cl--O- || O

Step by step solution

01

Count the number of valence electrons

For both ClO- and ClO4-, we need to count the valence electrons of each atom and add an extra electron for the negative charge. Chlorine (Cl) has 7 valence electrons and Oxygen (O) has 6. For ClO-: Total valence electrons = 7 (from Cl) + 6 (from O) + 1 (negative charge) = 14 For ClO4-: Total valence electrons = 7 (from Cl) + 4*6 (from 4 O atoms) + 1 (negative charge) = 32
02

Arrange the atoms and distribute the electrons (ClO-)

First, we look at ClO-. Chlorine will be the central atom, and oxygen will be bonded to it. We will start by creating a single bond between chlorine and oxygen, which uses 2 electrons (1 pair). Then, distribute the remaining electrons to satisfy the octet rule for each atom: • Chlorine needs 6 more electrons or 3 lone pairs (total: 6 + 1*2 = 8). • Oxygen needs 6 more electrons or 3 lone pairs (total: 6 + 1*2 = 8). Now we have used all 14 valence electrons, and both atoms satisfy the octet rule.
03

Drawing the Lewis structure (ClO-)

The Lewis structure for the hypochlorite ion (ClO-) will look like this: O || Cl- Here, the central chlorine atom (Cl) has one single bond to the oxygen atom (O), and each atom has 3 lone pairs of electrons. The negative charge is assigned to the chlorine atom.
04

Arrange the atoms and distribute the electrons (ClO4-)

Now, we look at ClO4-. Chlorine will be the central atom, with the 4 oxygen atoms bonded around it. We will start by creating single bonds between chlorine and each of the 4 oxygen atoms, which uses 8 electrons (4 pairs). Then, distribute the remaining electrons to satisfy the octet rule for each atom: • Chlorine needs 4 more electrons or 2 lone pairs (total: 4 + 4*2 = 8). • Each oxygen needs 6 more electrons or 3 lone pairs (total: 6 + 1*2 = 8). Now we have used all 32 valence electrons, and all atoms satisfy the octet rule.
05

Drawing the Lewis structure (ClO4-)

The Lewis structure for the perchlorate ion (ClO4-) will look like this: O || O--Cl--O- || O Here, the central chlorine atom (Cl) has single bonds to each of the 4 oxygen atoms; each oxygen atom has 3 lone pairs of electrons. The negative charge is assigned to one of the oxygen atoms. These are the Lewis structures for hypochlorite ion (ClO-) and perchlorate ion (ClO4-).

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Most popular questions from this chapter

Consider the hypothetical molecule \(\mathrm{A}-\mathrm{A}=\mathrm{A}\) with a bent shape. Are the following statements true or false? (a) This molecule cannot exist. (b) If this molecule exists, it must possess an odd electron.

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: $(\mathbf{a}) \mathrm{OCS},(\mathbf{b}) \mathrm{SOCl}_{2}(\mathrm{~S}$ is the central atom), (c) \(\mathrm{BrO}_{3}^{-}\), (d) \(\mathrm{HClO}_{2}(\mathrm{H}\) is bonded to \(\mathrm{O})\).

We can define average bond enthalpies and bond lengths for ionic bonds, just like we have for covalent bonds. Which ionic bond is predicted to be stronger, \(\mathrm{Na}-\mathrm{Cl}\) or \(\mathrm{Ca}-\mathrm{O}\) ?

Draw Lewis structures for the following: $(\mathbf{a}) \mathrm{CH}_{2} \mathrm{Cl}_{2},(\mathbf{b}) \mathrm{ClCN},\( (c) \)\mathrm{AsF}_{5}$ (d) \(\mathrm{CH}_{2} \mathrm{O}\) (C the central atom), (e) \(\mathrm{OF}_{2}\) (f) \(\mathrm{NO}_{3}^{-}\).

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).
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