Question

Among the ions \(\mathrm{O^{2-}},\ \mathrm{Na^+},\ \mathrm{F^-},\ \mathrm{N^{3-}},\ \mathrm{Mg^{2+}}\), the ions with smallest and largest radii are respectively

• A. \(\mathrm{F^-},\ \mathrm{N^{3-}}\)
• B. \(\mathrm{Mg^{2+}},\ \mathrm{N^{3-}}\)
• C. \(\mathrm{Na^+},\ \mathrm{F^-}\)
• D. \(\mathrm{F^-},\ \mathrm{Na^+}\)

Hint:

In an isoelectronic series, all species have the same number of electrons. The ion with the highest nuclear charge has the smallest radius, while the ion with the lowest nuclear charge has the largest radius.

Answer 1

The Correct Option is B

Step 1: Identify the isoelectronic series.
All five ions -- N\(^{3-}\), O\(^{2-}\), F\(^-\), Na\(^+\), Mg\(^{2+}\) -- have 10 electrons. They form an isoelectronic series.

Step 2: Apply the isoelectronic trend.
In an isoelectronic series, higher nuclear charge means smaller radius. Mg\(^{2+}\) (Z=12) has the highest Z, so smallest radius; N\(^{3-}\) (Z=7) has the lowest Z, so largest radius. \[ oxed{Mg^{2+},\ N^{3-}} \]