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Chapter 8: Problem 32

Total number of valence electrons present in \(6.4 \mathrm{~g}\) peroxides ion \(\left(\mathrm{O}_{2}^{2}\right)\) is (a) \(0.2 N_{A}\) (b) \(3.2 N_{A}\) (c) \(3.6 N_{A}\) (d) \(2.8 N_{A}\)

Short Answer

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Question: Determine the total number of valence electrons present in 6.4g of peroxides ion (O鈧偮测伝). Answer: The total number of valence electrons present in 6.4 grams of peroxide ions is 2.8 Avogadro's number (\(2.8 N_{A}\)).

Step by step solution

01

Calculate the molecular weight of oxygen

We have to determine the molecular weight of O鈧偮测伝, which comes from the sum of the atomic weights of the two Oxygen atoms: Molecular weight of O鈧偮测伝 = 2 * atomic weight of Oxygen = 2 * 16 g/mol = 32 g/mol.
02

Calculate the number of moles of peroxide ion in the given mass

Now, let's calculate the number of moles of peroxide ion in the given mass (6.4 g): Number of moles = (mass of peroxide ion) / (molecular weight of O鈧偮测伝) Number of moles = 6.4 g / 32 g/mol = 0.2 mol.
03

Calculate the number of peroxide ions in the given mass

To determine the number of O鈧偮测伝 ions in 0.2 mol, we use Avogadro's number (\(N_A = 6.022 \times 10^{23}\) particles/mol) as follows: Number of O鈧偮测伝 ions = (number of moles) * (\(N_A\)) Number of O鈧偮测伝 ions = (0.2 mol) * (\(6.022 \times 10^{23}\) particles/mol) = \(1.204 \times 10^{23}\) ions.
04

Determine the number of valence electrons in each peroxide ion

Let's calculate the number of valence electrons in the O鈧偮测伝 ion. Each oxygen atom has 6 valence electrons (from its electronic configuration), but the O鈧偮测伝 ion has 2 more electrons (hence the -2 charge). Thus, the ion has 6+6+2 = 14 valence electrons.
05

Calculate the total number of valence electrons in the given mass

Finally, we can calculate the total number of valence electrons in the 6.4 g sample of O鈧偮测伝: Total number of valence electrons = (number of O鈧偮测伝 ions) * (number of valence electrons per O鈧偮测伝 ion) Total number of valence electrons = (\(1.204 \times 10^{23}\) ions) * (14 electrons/ion) = \(1.686 \times 10^{24}\) electrons. To express the total number of valence electrons in terms of \(N_A\), we can divide by Avogadro's number: Total valence electrons / \(N_A\) = (\(1.686\times 10^{24}\) electrons) / (\(6.022 \times 10^{23}\) electrons/mol) = 2.8. So the total number of valence electrons present in 6.4 grams of peroxide ions is \(2.8 N_{A}\), which corresponds to option (d).

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Key Concepts

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

Understanding the Peroxide Ion
The peroxide ion ( O_2^{2-} ) is an interesting entity in chemistry. At its heart, it's composed of two oxygen atoms. Each oxygen atom usually has a 2- charge in the peroxide ion, giving the ion its 2- charge.
When discussing ions, it's important to remember that ions are essentially atoms that have gained or lost electrons. This means they can carry a charge.
  • In the case of the peroxide ion, the two extra electrons result in a net charge of 2- .
  • This charge affects the ion's behavior in chemical reactions.
Peroxide ions are common in chemistry, particularly in compounds like hydrogen peroxide. Understanding the fundamental structure of these ions helps in studying their reactions and properties.
Molecular Weight Calculation Made Simple
Calculating the molecular weight of a compound starts with knowing the atomic weights of its atoms. Molecular weight, also known as molar mass, is essentially the sum of the individual weights of the atoms in the molecule.
For the peroxide ion, O_2^{2-} , we calculate it by summing the atomic weights of the oxygen atoms.
  • Each oxygen atom has an atomic weight of about 16 ext{ g/mol} .
  • Therefore, the molecular weight of O_2^{2-} is 2 imes 16 ext{ g/mol} = 32 ext{ g/mol} .
This value is crucial in many calculations, such as determining how much substance is present in a given mass.
The Importance of Avogadro's Number
Avogadro's number, N_A = 6.022 imes 10^{23} , is a key figure in chemistry. It provides the number of atoms or molecules in one mole of a substance.
This means a mole of any substance contains 6.022 imes 10^{23} particles.
With this constant, chemists can scale up from individual atoms or molecules to quantities that can be seen and measured.
  • It acts as a bridge between the atomic and macroscopic worlds.
  • For example, in the peroxide ion, using Avogadro's number helps determine the number of molecules in a given mass.
Using Avogadro's number simplifies complex conversions from microscopic to macroscopic measurements.
A Peek into Electron Configuration
Electron configuration is the distribution of electrons in an atom's orbitals. In simpler terms, it's how electrons are arranged around the nucleus.
  • For an oxygen atom, this configuration is 1s^22s^22p^4 .
  • Each electron belongs to a specific shell or subshell, represented by numbers and letters such as 1s, 2s, 2p .
Understanding electron configuration is key to predicting and explaining chemical behaviors and properties.
In ions like O_2^{2-} , electron configuration reveals how added electrons affect reactions, stability, and even the structure of the ion.

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Most popular questions from this chapter

The fractional abundance of \(\mathrm{Cl}^{35}\) in a sample of chlorine containing only \(\mathrm{Cl}^{3 \mathrm{~s}}\) (atomic weight \(=34.9\) ) and \(\mathrm{Cl}^{37}\) (atomic weight \(=36.9\) ) isotopes, is \(0.6\). The average mass number of chlorine is (a) \(35.7\) (b) \(35.8\) (c) \(18.8\) (d) \(35.77\)

When \(12 \mathrm{~g}\) graphite is burnt in sufficient oxygen, \(\mathrm{CO}\) as well as \(\mathrm{CO}_{2}\) is formed. If the product contains \(40 \%\) CO and \(60 \%\) \(\mathrm{CO}_{2}\) by mass and none of the reactant is left, what is the mass of oxygen gas used in combustion? (a) \(24.0 \mathrm{~g}\) (b) \(21.33 \mathrm{~g}\) (c) \(23.8 \mathrm{~g}\) (d) \(15.6 \mathrm{~g}\)

The vapour density of a sample of \(\mathrm{SO}_{3}\) gas is 28 . Its degree of dissociation in to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is (a) \(1 / 7\) (b) \(1 / 6\) (c) \(6 / 7\) (d) \(2 / 5\)

When a mixture of aluminium powder and iron (III) oxide is ignited, it produces molten iron and aluminium oxide. In an experiment, \(5.4 \mathrm{~g}\) of aluminium was mixed with \(18.5 \mathrm{~g}\) of iron (III) oxide. At the end of the reaction, the mixture contained \(11.2 \mathrm{~g}\) of iron, \(10.2 \mathrm{~g}\) of aluminium oxide, and an undetermined amount of unreacted iron (III) oxide. No aluminium was left. What is the mass of the iron (III) oxide left? (a) \(2.5 \mathrm{~g}\) (b) \(7.3 \mathrm{~g}\) (c) \(8.3 \mathrm{~g}\) (d) \(2.9 \mathrm{~g}\)

Iron pyrites are converted to iron oxide by the process of
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