Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. n 3−, o 2−, f −, na+, mg2+ 2. na+, mg2+, o 2−, n 3−, f − 3. na+, mg2+, n 3−, o 2−, f − 4. mg2+, na+, f −, o 2−, n 3− 5. f −, o 2−, n 3−, na+, mg2+

Option 1:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

Explanation:

Ionic radius is the radius of an atom´s ion in ionic crystal structure. In an ion that lose an electron, to form a cation, the radius of the ion gets smaller, because the repulsion between electrons decrease because fewer electrons are present. Conversely, adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.                  

Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers.                        

In a isolelectronic series (same number of electrons), the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

We will use this principle to solve our problem.  

In our case, the given ions are:  

where Z= number of protons, and e⁻ = number of electrons.

Hence the decreasing order of ionic radius is:

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

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there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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