Identify the elements correctly shown by decreasing radii size. Group of answer choices N > N3- K+ > K S- > S2- N3- > N Cu2+ > Cu+

Answer: \( \text{N}^{3-} > \text{N} \)

Explanation: When comparing ionic and atomic radii, the size of an ion or atom is influenced by the number of electrons and the effective nuclear charge.

  1. N vs. \( \text{N}^{3-} \): The nitride ion \( \text{N}^{3-} \) has gained three additional electrons compared to the neutral nitrogen atom. This increase in electron-electron repulsion causes the ion to be larger than the neutral atom.
  1. K vs. \( \text{K}^+ \): The potassium ion \( \text{K}^+ \) has lost an electron compared to the neutral potassium atom, resulting in a smaller ionic radius due to the reduced electron-electron repulsion and increased effective nuclear charge per electron.
  1. S vs. \( \text{S}^{2-} \): The sulfide ion \( \text{S}^{2-} \) has gained two additional electrons, making it larger than the neutral sulfur atom due to increased electron-electron repulsion.
  1. Cu vs. \( \text{Cu}^+ \) and \( \text{Cu}^{2+} \): The copper ions \( \text{Cu}^+ \) and \( \text{Cu}^{2+} \) have lost electrons compared to the neutral copper atom, with \( \text{Cu}^{2+} \) being smaller than \( \text{Cu}^+ \) due to the higher positive charge and greater effective nuclear charge per electron.

Therefore, the correct order by decreasing radii size is \( \text{N}^{3-} > \text{N} \).

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