Question

7 multiple choice 1 point o²⁻, f⁻, na⁺, and mg²⁺ are isoelectronic with ne meaning they have the same electron configurations even though they have a different number of protons. what is true based on this information? o o²⁻ is the largest because it has the most electrons. o na⁺ is the largest because of the atomic radius periodic trend. o all four ions are the same size as ne. o mg²⁺ is the smallest because it has the most protons.

Explanation:

1. Analyze the first option: Isoelectronic species have the same number of electrons. \(O^{2 -}\), \(F^-\), \(Na^+\), \(Mg^{2+}\) and \(Ne\) all have 10 electrons. So the claim that \(O^{2 -}\) is largest because it has most electrons is wrong (they have same number of electrons).
2. Analyze the second option: For isoelectronic species, the one with more protons has a smaller radius (since more protons pull electrons more tightly). \(Na^+\) has 11 protons, but \(Mg^{2+}\) has 12 protons and \(O^{2 -}\) has 8 protons. The atomic radius trend for isoelectronic ions is based on proton number, not the general atomic radius trend of elements. So \(Na^+\) is not the largest.
3. Analyze the third option: Ions and atoms (Ne is an atom) of isoelectronic species have different sizes because the number of protons differs, affecting the electron - proton attraction. So they are not the same size as Ne.
4. Analyze the fourth option: For isoelectronic species (\(O^{2 -}\) (8 protons), \(F^-\) (9 protons), \(Na^+\) (11 protons), \(Mg^{2+}\) (12 protons)), as the number of protons increases, the effective nuclear charge on the electrons increases, pulling the electrons closer to the nucleus, thus decreasing the ionic radius. \(Mg^{2+}\) has the most protons among them, so it has the smallest ionic radius.

Answer:

D. \(Mg^{2+}\) is the smallest because it has the most protons.