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Chapter 8: Problem 61

Draw the dominant Lewis structures for these chlorineoxygen molecules/ions: $\mathrm{ClO}, \mathrm{ClO}^{-}, \mathrm{ClO}_{2}^{-}, \mathrm{ClO}_{3}^{-}, \mathrm{ClO}_{4}^{-}$. Which of these do not obey the octet rule?

Short Answer

Expert verified
The dominant Lewis structures for the chlorine-oxygen molecules/ions are: 1. ClO: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 2 non-bonding electron pairs and one electron without a pair)] 2. ClO⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 3 non-bonding electron pairs)] with a negative charge on Cl 3. ClO₂⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 2 non-bonding electron pairs) - (single bond) - (O with 3 non-bonding electron pairs)] with a negative charge on Cl 4. ClO₃⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 1 non-bonding electron pair) - (single bond) - (O with 3 non-bonding electron pairs)] and a third O single bonded to Cl as well, with a negative charge on Cl 5. ClO₄⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl without non-bonding electron pairs) - (single bond) - (O with 3 non-bonding electron pairs)] with two other oxygens single bonded to Cl, and a negative charge on Cl Out of these, only the ClO molecule does not obey the octet rule, as its Cl atom has only 7 valence electrons.

Step by step solution

01

Determine the total number of valence electrons for each molecule/ion

First, let us find the number of valence electrons in each molecule or ion. Chlorine (Cl) has 7 valence electrons, while oxygen (O) has 6. And as we have negative charges on the ions, we'll need to add one electron for each negative charge. 1. ClO: 7 (from Cl) + 6 (from O) = 13 valence electrons 2. ClO⁻: 7 (from Cl) + 6 (from O) + 1 (for the negative charge) = 14 valence electrons 3. ClO₂⁻: 7 (from Cl) + 2×6 (from two O's) + 1 (for the negative charge) = 20 valence electrons 4. ClO₃⁻: 7 (from Cl) + 3×6 (from three O's) + 1 (for the negative charge) = 26 valence electrons 5. ClO₄⁻: 7 (from Cl) + 4×6 (from four O's) + 1 (for the negative charge) = 32 valence electrons
02

Draw the Lewis structures for each molecule/ion

Now, let's draw the dominant Lewis structures for each molecule/ion: 1. ClO: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 2 non-bonding electron pairs and one electron without a pair)] 2. ClO⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 3 non-bonding electron pairs)] with a negative charge on Cl 3. ClO₂⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 2 non-bonding electron pairs) - (single bond) - (O with 3 non-bonding electron pairs)] with a negative charge on Cl 4. ClO₃⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl with 1 non-bonding electron pair) - (single bond) - (O with 3 non-bonding electron pairs)] and a third O (with 3 non-bonding electron pairs) single bonded to Cl as well, with a negative charge on Cl 5. ClO₄⁻: [(O with 3 non-bonding electron pairs) - (single bond) - (Cl without non-bonding electron pairs) - (single bond) - (O with 3 non-bonding electron pairs)] with the remaining two oxygens also single bonded to Cl, each with 3 non-bonding electron pairs, and a negative charge on Cl
03

Identify which structures do not obey the octet rule

Now we'll identify which of these structures do not obey the octet rule: 1. ClO: The Cl atom has only 7 valence electrons, so it does not obey the octet rule. 2. ClO⁻: All atoms have 8 valence electrons, so it obeys the octet rule. 3. ClO₂⁻: All atoms have 8 valence electrons, so it obeys the octet rule. 4. ClO₃⁻: All atoms have 8 valence electrons, so it obeys the octet rule. 5. ClO₄⁻: All atoms have 8 valence electrons, so it obeys the octet rule. So, only the ClO molecule does not obey the octet rule.

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Most popular questions from this chapter

Ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a very soluble salt in water. (a) Draw the Lewis structures of the ammonium and chloride ions. (b) Is there an \(\mathrm{N}-\mathrm{Cl}\) bond in solid ammonium chloride? (c) If you dissolve \(14 \mathrm{~g}\) of ammonium chloride in \(500.0 \mathrm{~mL}\) of water, what is the molar concentration of the solution? (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \% \mathrm{~N}\). Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} / \mathrm{mol}\). The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(\mathrm{S}_{8}\) ring is \(205 \mathrm{pm} .\) ) \((\mathbf{d})\) The enthalpy of formation of the compound is estimated to be $480 \mathrm{~kJ} / \mathrm{mol}^{-1} . \Delta H_{f}^{\circ}\( of \)\mathrm{S}(g)\( is \)222.8 \mathrm{~kJ} / \mathrm{mol}$. Estimate the average bond enthalpy in the compound.

Which of the following statements about electronegativity is false? (a) Electronegativity is the ability of an atom in a molecule to attract electron density toward itself. (b) Electronegativity is the same thing as electron affinity. (c) The numerical values for electronegativity have no units. (d) Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{BF}_{3}\) from \(\mathrm{B}\) and \(\mathrm{F}\) atoms, showing valence- shell electrons. (a) How many valence electrons does B have initially? (b) How many bonds F has to make in order to achieve an octet? (c) How many valence electrons surround the \(\mathrm{B}\) in the \(\mathrm{BF}_{3}\) molecule? (d) How many valence electrons surround each \(\mathrm{F}\) in the \(\mathrm{BF}_{3}\) molecule? (e) Does \(\mathrm{BF}_{3}\) obey the octet rule?

List the individual steps used in constructing a Born-Haber cycle for the formation of BaI \(_{2}\) from the elements. Which of the steps would you expect to be exothermic?
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