To solve this problem, we need to determine which element forms the largest ion to achieve the nearest noble gas configuration.$\\$The elements given are:$\\$• Oxygen (O)$\\$• Fluorine (F)$\\$• Nitrogen (N)$\\$• Sodium (Na)$\\$Let's analyze each element:$\\$1. Oxygen (O):$\\$• Electronic configuration: $1s^2 2s^2 2p^4 \\$• To achieve the nearest noble gas configuration (Neon), it gains 2 electrons to form $O^{2-}. \\$2. Fluorine (F):$\\$• Electronic configuration: $1s^2 2s^2 2p^5 \\$• To achieve the nearest noble gas configuration (Neon), it gains 1 electron to form $F^{-}. \\$3. Nitrogen (N):$\\$• Electronic configuration: $1s^2 2s^2 2p^3 \\$• To achieve the nearest noble gas configuration (Neon), it gains 3 electrons to form $N^{3-}. \\$4. Sodium (Na):$\\$• Electronic configuration: $1s^2 2s^2 2p^6 3s^1 \\$• To achieve the nearest noble gas configuration (Neon), it loses 1 electron to form $Na^{+}. \\$The size of an ion is influenced by the number of electrons and the effective nuclear charge.$\\$• When an atom gains electrons, the size of the ion increases due to increased electron-electron repulsion.$\\$• Among the ions formed, $N^{3-}$ has the largest number of electrons added, resulting in the largest increase in size.$\\$Therefore, the element expected to form the largest ion to achieve the nearest noble gas configuration is Nitrogen (N).$\\$This corresponds to Option 3.