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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

Expert verified
Question: Determine the total number of valence electrons present in 6.4g of peroxides ion (O₂²⁻). 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 O₂²⁻, which comes from the sum of the atomic weights of the two Oxygen atoms: Molecular weight of O₂²⁻ = 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 O₂²⁻) 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 O₂²⁻ ions in 0.2 mol, we use Avogadro's number (\(N_A = 6.022 \times 10^{23}\) particles/mol) as follows: Number of O₂²⁻ ions = (number of moles) * (\(N_A\)) Number of O₂²⁻ 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 O₂²⁻ ion. Each oxygen atom has 6 valence electrons (from its electronic configuration), but the O₂²⁻ 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 O₂²⁻: Total number of valence electrons = (number of O₂²⁻ ions) * (number of valence electrons per O₂²⁻ 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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Most popular questions from this chapter

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