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The second electron affinity values for both oxygen and sulfur are unfavorable (endothermic). Explain.

Short Answer

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The second electron affinity values for both oxygen and sulfur are unfavorable (endothermic) because adding a second electron to their already negatively charged ions (O^- and S^-) is energetically unfavorable. This is due to the incoming electron facing repulsion from the other electrons already in the ion, and there is no significant stabilization by achieving a filled subshell or lower energy orbital. The negatively charged ions will repel the incoming electrons more strongly than neutral atoms, requiring additional energy to overcome the electrostatic repulsion, resulting in an endothermic process.

Step by step solution

01

Understand Electron Affinity

In general, when an electron is added to a neutral atom, energy is released, and the process is exothermic. However, the second electron affinity can be endothermic if it is energetically unfavorable for the atom to gain an additional electron. In the case of oxygen and sulfur, their second electron affinities are endothermic, which we need to explain.
02

Recall the Electronic Configurations of Oxygen and Sulfur

Recall the electronic configurations of oxygen and sulfur: Oxygen: \(1s^2 2s^2 2p^4\) (8 electrons) Sulfur: \(1s^2 2s^2 2p^6 3s^2 3p^4\) (16 electrons) Both oxygen and sulfur have six valence electrons.
03

Explain the Unfavorable Second Electron Affinity for Oxygen and Sulfur

When the first electron is added, both oxygen and sulfur form negatively charged ions that have stable configurations with filled p orbitals: Oxygen: \(1s^2 2s^2 2p^6\) (9 electrons; O^- ion) Sulfur: \(1s^2 2s^2 2p^6 3s^2 3p^6\) (17 electrons; S^- ion) However, the addition of a second electron to these ions becomes energetically unfavorable. This is because, the incoming electron faces repulsion from the other electrons already in the ion, and there is no significant stabilization by achieving a filled subshell or lower energy orbital. The negatively charged ions will repel the incoming electrons more strongly than neutral atoms, which require additional energy to overcome the electrostatic repulsion between the negatively charged ion and the incoming electron. That's why the second electron affinity value for both oxygen and sulfur is endothermic, as energy is absorbed, making it unfavorable.

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Most popular questions from this chapter

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl} \quad\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O} \quad\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Predict the molecular structure for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{BrFI}_{2} \quad\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2} \quad\) c. \(\mathrm{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Write Lewis structures that obey the octet rule for each of the following molecules and ions. (In each case the first atom listed is the central atom.) a. \(\mathrm{POCl}_{3}, \mathrm{SO}_{4}^{2-}, \mathrm{XeO}_{4}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}\) b. \(\mathrm{NF}_{3}, \mathrm{SO}_{3}^{2-}, \mathrm{PO}_{3}^{3-}, \mathrm{ClO}_{3}^{-}\) c. \(\mathrm{ClO}_{2}-, \mathrm{SCl}_{2}, \mathrm{PCl}_{2}^{-}\) d. Considering your answers to parts a, b, and c, what conclusions can you draw concerning the structures of species containing the same number of atoms and the same number of valence electrons?

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$: \mathrm{N}=\mathrm{N}=\dot{\mathrm{O}}\longleftrightarrow: \mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{Q}} : \longleftrightarrow : \dot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}$$ Given the following bond lengths, \(\mathrm{N}-\mathrm{N} \qquad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\) \(\mathrm{N}=\mathrm{N} \qquad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\) \(\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm}\) rationalize the observations that the N-N bond length in \(\mathrm{N}_{2} \mathrm{O}\) is 112 \(\mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is 119 \(\mathrm{pm}\) . Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\) . Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: \(\mathrm{N}(g), 472.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}(g),\) \(216.0 \mathrm{kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{kJ} / \mathrm{mol}\) . Compare your value to the one in Table \(8.5 .\)

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