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
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
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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.
Updated On: Jun 26, 2026
- \(\mathrm{F^-},\ \mathrm{N^{3-}}\)
- \(\mathrm{Mg^{2+}},\ \mathrm{N^{3-}}\)
- \(\mathrm{Na^+},\ \mathrm{F^-}\)
- \(\mathrm{F^-},\ \mathrm{Na^+}\)
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The Correct Option is B
Solution and Explanation
Step 1: Identify the electronic configurations.
The ions \[ \mathrm{N^{3-}},\ \mathrm{O^{2-}},\ \mathrm{F^-},\ \mathrm{Na^+},\ \mathrm{Mg^{2+}} \] all contain \[ 10 \text{ electrons} \] Therefore, they form an isoelectronic series.
Their atomic numbers are \[ N=7,\quad O=8,\quad F=9,\quad Na=11,\quad Mg=12 \]
Step 2: Recall the trend in an isoelectronic series.
For an isoelectronic series, the ionic radius decreases as the nuclear charge increases.
This is because a greater number of protons attracts the same number of electrons more strongly.
Thus, \[ \text{Radius} \propto \frac{1}{\text{Nuclear Charge}} \]
Step 3: Arrange the ions according to nuclear charge.
The nuclear charges are \[ \mathrm{N^{3-}}(7) \lt \mathrm{O^{2-}}(8) \lt \mathrm{F^-}(9) \lt \mathrm{Na^+}(11) \lt \mathrm{Mg^{2+}}(12) \] Therefore, the order of ionic radii is \[ \mathrm{N^{3-}} \gt \mathrm{O^{2-}} \gt \mathrm{F^-} \gt \mathrm{Na^+} \gt \mathrm{Mg^{2+}} \]
Step 4: Identify the smallest and largest ions.
From the above order, \[ \text{Largest radius}=\mathrm{N^{3-}} \] and \[ \text{Smallest radius}=\mathrm{Mg^{2+}} \]
Step 5: Final conclusion.
Hence, the ions with the smallest and largest radii respectively are \[ \boxed{\mathrm{Mg^{2+}},\ \mathrm{N^{3-}}} \]
The ions \[ \mathrm{N^{3-}},\ \mathrm{O^{2-}},\ \mathrm{F^-},\ \mathrm{Na^+},\ \mathrm{Mg^{2+}} \] all contain \[ 10 \text{ electrons} \] Therefore, they form an isoelectronic series.
Their atomic numbers are \[ N=7,\quad O=8,\quad F=9,\quad Na=11,\quad Mg=12 \]
Step 2: Recall the trend in an isoelectronic series.
For an isoelectronic series, the ionic radius decreases as the nuclear charge increases.
This is because a greater number of protons attracts the same number of electrons more strongly.
Thus, \[ \text{Radius} \propto \frac{1}{\text{Nuclear Charge}} \]
Step 3: Arrange the ions according to nuclear charge.
The nuclear charges are \[ \mathrm{N^{3-}}(7) \lt \mathrm{O^{2-}}(8) \lt \mathrm{F^-}(9) \lt \mathrm{Na^+}(11) \lt \mathrm{Mg^{2+}}(12) \] Therefore, the order of ionic radii is \[ \mathrm{N^{3-}} \gt \mathrm{O^{2-}} \gt \mathrm{F^-} \gt \mathrm{Na^+} \gt \mathrm{Mg^{2+}} \]
Step 4: Identify the smallest and largest ions.
From the above order, \[ \text{Largest radius}=\mathrm{N^{3-}} \] and \[ \text{Smallest radius}=\mathrm{Mg^{2+}} \]
Step 5: Final conclusion.
Hence, the ions with the smallest and largest radii respectively are \[ \boxed{\mathrm{Mg^{2+}},\ \mathrm{N^{3-}}} \]
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