Ionic Radius
For an isoelectronic series of ions such as sodium ion, magnesium ion, fluoride ion, and oxide ion, how does ionic radius depend on nuclear charge?
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About This Question
- Subject
- chemistry
- Chapter
- classification of elements and periodicity in properties
- Topic
- ionic radius
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Radius decreases as nuclear charge increases
Isoelectronic species have the same number of electrons but different nuclear charges, so the size of each ion is determined by how strongly its nucleus pulls the fixed electron cloud. In the series O^2-, F^-, Na^+, Mg^2+, all have ten electrons, but the number of protons rises from 8 in oxygen to 12 in magnesium. As nuclear charge increases, the same ten electrons are pulled inward more strongly, so the ionic radius decreases in the order O^2- > F^- > Na^+ > Mg^2+. Therefore radius decreases as nuclear charge increases across an isoelectronic series. The option that radius increases with nuclear charge reverses the trend. The option that radius is independent of charge ignores the differing proton counts. The option that all radii are equal is false despite equal electron numbers. This isoelectronic trend is a standard NCERT periodicity application. It is worth emphasising that this is not a special case but a representative example of how ionic radius operates throughout classification of elements and periodicity in properties. Plausibility check: the anion O^2- with the fewest protons holds its electrons most loosely and is largest, while Mg^2+ with the most protons is smallest, confirming the decrease.
This medium difficulty chemistry question is from the chapter classification of elements and periodicity in properties, covering the topic of ionic radius. It appeared in the 2025 exam.
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