### Electron Transitions in a Hydrogen Atom**Consider these electron transitions in a hydrogen atom:**1. **n = 2 → n = 1**2. **n = 3 → n = 1**3. **n = 1 → n = 4****Question:**Which of the electron transitions would release the most energy?**Options:**- **A) I**- **B) II**- **C) III**- **D) I, II, and III release the same amount of energy**#### Explanation:In the context of electron transitions in a hydrogen atom, energy is released when an electron transitions from a higher energy level (higher n value) to a lower energy level (lower n value). The amount of energy released is proportional to the difference in the energy levels.- **Option I (n = 2 → n = 1):** This transition releases energy because the electron moves from a higher energy level (n = 2) to a lower energy level (n = 1).- **Option II (n = 3 → n = 1):** This transition also releases energy, and the energy released will be more than in Option I since the difference between the energy levels (n = 3 and n = 1) is greater.- **Option III (n = 1 → n = 4):** This transition would require the absorption of energy since the electron is moving from a lower energy level (n = 1) to a higher energy level (n = 4).Therefore, **Option II (n = 3 → n = 1)** would release the most energy. ### Determining the Heat of Formation for PbSUsing the provided table and the chemical equation below, we aim to determine the heat of formation (ΔH°f) of PbS in kJ/mol.#### Chemical Reaction\[ 2 \text{PbS (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + 2 \text{PbO (s)} \quad \Delta H^\circ = -828.4 \text{ kJ/mol} \]#### Given Data| Substance | $ \Delta H_f^\circ $ (kJ/mol) ||-----------|-------------------------------|| $ \text{O}_2 \text{(g)} $ | 0 || $ \text{SO}_2 \text{(g)} $ | -296.9 || $ \text{PbO (s)} $ | -217.3 |### ExplanationThis equation corresponds to the enthalpy change for the reaction given. To find the standard enthalpy change of formation (ΔH°f) for PbS, you can use Hess's Law, which involves the sum of the enthalpies of the products minus the reactants.#### Steps to Determine $ \Delta H_f^\circ $ for PbS:1. **Total Enthalpy Change for Products:** \[ \Delta H_{\text{products}} = 2 \times \Delta H_f^\circ (\text{SO}_2) + 2 \times \Delta H_f^\circ (\text{PbO}) \] Plug in the given values: \[ \Delta H_{\text{products}} = 2 \times (-296.9 \text{ kJ/mol}) + 2 \times (-217.3 \text{ kJ/mol}) \] \[ \Delta H_{\text{products}} = -593.8 \text{ kJ/mol} + (-434.6 \text{ kJ/mol}) = -1028.4 \text{ kJ/mol} \]2. **Using the Total Enthalpy Change for the Reaction:** \[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants

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Consider these electron transitions in a hydrogen atom: I. n = 2 n = 1 II. n = 3 n = 1 III. n= 1n = 4 Which of the electron transitions would release the most energy? A) I B) I| C) II D) I, II, and III release the same amount of energy

Consider these electron transitions in a hydrogen atom: I. n = 2 n = 1 II. n = 3 n = 1 III. n= 1n = 4 Which of the electron transitions would release the most energy? A) I B) I| C) II D) I, II, and III release the same amount of energy

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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