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Which of the following chlorine oxides are odd-electron molecules? (a) \(\mathrm{Cl}_{2} \mathrm{O}_{7} ;\) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{6} ;\) (c) \(\mathrm{ClO}_{4} ;\) (d) \(\mathrm{ClO}_{3}\) (e) \(\mathrm{ClO}_{2}\)

Short Answer

Expert verified
Answer: The odd-electron chlorine oxides are ClO₄, ClO₃, and ClO₂.

Step by step solution

01

Calculate valence electrons for each molecule

For each chlorine oxide molecule, we have to calculate the number of valence electrons. For this, we need to know the valence electrons for chlorine (Cl) and oxygen (O). Cl has 7 valence electrons and O has 6 valence electrons. (a) \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) Total valence electrons = \((2 \times 7) + (7 \times 6) = 14 + 42 = 56\) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{6}\) Total valence electrons = \((2 \times 7) + (6 \times 6) = 14 + 36 = 50\) (c) \(\mathrm{ClO}_{4}\) Total valence electrons = \((1 \times 7) + (4 \times 6) = 7 + 24 = 31\) (d) \(\mathrm{ClO}_{3}\) Total valence electrons = \((1 \times 7) + (3 \times 6) = 7 + 18 = 25\) (e) \(\mathrm{ClO}_{2}\) Total valence electrons = \((1 \times 7) + (2 \times 6) = 7 + 12=19\)
02

Determine odd or even electron species

Now we will determine if the number of valence electrons for each molecule is odd or even. If odd, then the molecule is an odd-electron species. (a) \(\mathrm{Cl}_{2} \mathrm{O}_{7}\): 56 electrons (even) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{6}\): 50 electrons (even) (c) \(\mathrm{ClO}_{4}\): 31 electrons (odd) (d) \(\mathrm{ClO}_{3}\): 25 electrons (odd) (e) \(\mathrm{ClO}_{2}\): 19 electrons (odd)
03

Identify the odd-electron molecules

From our analysis in Step 2, we can conclude that the following chlorine oxides are odd-electron molecules: (c) \(\mathrm{ClO}_{4}\) (d) \(\mathrm{ClO}_{3}\) (e) \(\mathrm{ClO}_{2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in the formation of chemical bonds. Understanding them is essential for predicting how atoms will interact with each other. Each element has a distinct number of valence electrons that can be identified by its position in the periodic table. Chlorine, found in group 17, has 7 valence electrons, while oxygen in group 16 has 6 valence electrons.
  • These electrons are responsible for chemical reactivity and bonding.
  • Valence electrons determine the number of possible bonds an atom can form.
  • The total number of valence electrons in a molecule influences its stability and reactivity.
By calculating the number of valence electrons in molecules, we are able to determine their potential as odd-electron species, which often have unique chemical properties.
Odd-Electron Molecules
Odd-electron molecules, also known as radicals, are species with unpaired electrons. These molecules are often more reactive due to the presence of an odd number of total valence electrons. In our exercise, we calculated the total valence electrons of chlorine oxides to identify which among them are odd-electron molecules.
  • Odd-electron species are important in chemical reactions because they can easily gain or lose an electron to become more stable.
  • They usually have high reactivity and are often involved in combustion and oxidation processes.
  • Identifying these species is crucial in understanding reaction mechanisms in chemistry.
In this case, the molecules \(\mathrm{ClO}_{4}\), \(\mathrm{ClO}_{3}\), and \(\mathrm{ClO}_{2}\) were identified as having an odd number of valence electrons, making them odd-electron molecules.
Chemical Formulas
Chemical formulas represent the types and numbers of atoms in a molecule. They are an essential part of chemistry for understanding the composition of a compound. Each element in a formula is represented by its chemical symbol along with a subscript indicating the number of atoms present.
  • The formula provides insight into the ratio of atoms within a molecule.
  • It helps in determining properties such as molecular weight and reactivity.
  • Chemical formulas serve as the basis for understanding chemical structures and reactions.
For chlorine oxides, formulas like \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) and \(\mathrm{ClO}_{2}\) not only show the atoms' types but also suggest the molecules' geometric arrangements and bonding patterns. Analyzing the formulas is essential in predicting the properties and behavior of the compounds in various chemical contexts.

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Most popular questions from this chapter

How many valence electrons does each of the following species contain? (a) \(\mathrm{H}^{+} ;\) (b) \(\mathrm{H}_{3} \mathrm{O}^{+} ;\) (c) \(\mathrm{CO}_{2} ;\) (d) \(\mathrm{CH}_{4}.\)

In the typical Lewis structure of \(\mathrm{BF}_{3}\) there are only six valence electrons on the boron atom and each \(\mathrm{B}-\mathrm{F}\) bond is a single bond. However, the length and strength of these bonds indicate that they have a small measure of doublebond character-that is, their bond order is slightly greater than 1. a. Draw a Lewis structure of \(\mathrm{BF}_{3}\), including all resonance structures, in which there is one \(\mathrm{B}=\mathrm{F}\) double bond. b. What is the formal charge on the \(\mathrm{B}\) atom, and what is the average formal charge on each \(\mathrm{F}\) atom? c. Based on formal charges alone, what should be the bond order of each \(\mathrm{B}-\mathrm{F}\) bond in \(\mathrm{BF}_{3} ?\) d. What factor might support a bond order slightly greater than \(1 ?\)

Draw the Lewis structure for butane, \(\mathrm{C}_{4} \mathrm{H}_{10},\) given the structure contains four carbon atoms bonded in a row: \(\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}.\)

Draw Lewis structures for the following molecules and ions: (a) \(\mathrm{Br}_{2} ;\) (b) \(\mathrm{H}_{3} \mathrm{O}^{+} ;\) (c) \(\mathrm{N}_{2} ;\) (d) \(\mathrm{HF}\)

Ozone Depletion Methyl bromide \(\left(\mathrm{CH}_{3} \mathrm{Br}\right)\) is produced naturally by fungi. Methyl bromide has also been used in agriculture as a fumigant, but its use is being phased out because the compound has been linked to ozone depletion in the upper atmosphere. a. Draw the Lewis structure of \(\mathrm{CH}_{3} \mathrm{Br}\). b. Which bond in \(\mathrm{CH}_{3} \mathrm{Br}\) is more polar, carbon-hydrogen or carbon-bromine?

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