Which statement best describes the amount of energy released by gas-phase oxygen and sulfur atoms upon accepting an electron (electron affinities)? (A) O releases less energy than S because the smaller size of O causes greater electron-electron repulsion. (B) O releases less energy than S because O has a lower mass than S. (C) O releases more energy than S because the electron enters a 2p orbital in O but a 3p orbital in S. (D) O releases more energy than S because O has a larger effective nuclear charge Zeff than S.

The correct answer is (C) O releases more energy than S because the electron enters a 2p orbital in O but a 3p orbital in S.

Here's why:

The energy released when an atom accepts an electron, also known as electron affinity, is influenced by the distance of the outermost shell from the nucleus and the effective nuclear charge.

In the case of oxygen (O) and sulfur (S), the electron enters a 2p orbital in O and a 3p orbital in S. The 2p orbital in O is closer to the nucleus than the 3p orbital in S. Therefore, the electron in O is more strongly attracted to the nucleus, resulting in a greater release of energy.

The effective nuclear charge (Zeff) is the net positive charge experienced by an electron in a multi-electron atom. The greater the Zeff, the more strongly the electron is attracted to the nucleus. However, both O and S are in the same group of the periodic table (Group 16), so they have the same Zeff. Therefore, the difference in electron affinity between O and S is not due to Zeff, ruling out option (D).

Option (A) is incorrect because electron-electron repulsion would make it harder, not easier, for an atom to accept an electron.

Option (B) is incorrect because the mass of an atom does not directly affect its electron affinity.

So, the best statement that describes the amount of energy released by gas-phase oxygen and sulfur atoms upon accepting an electron is (C) O releases more energy than S because the electron enters a 2p orbital in O but a 3p orbital in S.