Animated Solution for Chemistry - Periodicity in Properties: The ionic radii of $F^{-}$ and $O^{2-}$ respectively are $1.33\text{ A˚}$ and $1.4\text{ A˚}$ while the covalent radius of N is $0.74\text{ A˚}$. The correct statement for the ionic radius of $N^{3-}$ from the following is

When we look at the ions $F^{-}$, $O^{2-}$, and $N^{3-}$, they might seem completely different at first glance. However, if we write down their electronic configurations, a beautiful pattern emerges. Fluorine (atomic number 9) gains one electron to become $F^{-}$, giving it a total of 10 electrons. Oxygen (atomic number 8) gains two electrons to become $O^{2-}$, also reaching 10 electrons. Nitrogen (atomic number 7) gains three electrons to become $N^{3-}$, again hitting that magic number of 10 electrons.

Because they all possess exactly 10 electrons, they are known as isoelectronic species. They all share the same noble gas configuration of Neon: $1s^{2}2s^{2}2p^6$.

Protons vs. Electrons

If they all have the same number of electrons, why are their sizes different? The secret lies in the nucleus. The size of an ion is determined by a microscopic tug-of-war between the positively charged protons in the nucleus and the negatively charged electrons in the outer shells.

For isoelectronic species, the number of electrons is constant, so the deciding factor is the Effective Nuclear Charge ($Z_{eff}$). This is directly proportional to the number of protons ($Z$).

Let's compare them: - $N^{3-}$ has 7 protons pulling on 10 electrons. - $O^{2-}$ has 8 protons pulling on 10 electrons. - $F^{-}$ has 9 protons pulling on 10 electrons.

More protons mean a stronger inward pull on the electron cloud. Therefore, Fluoride, with the highest number of protons (9), pulls its 10 electrons the tightest, making it the smallest. Nitride, with only 7 protons, has the weakest grip on its 10 electrons, allowing the electron cloud to expand the most.

Based on our logic, the order of ionic radii must be:

The problem states that the radius of $F^{-}$ is $1.33\text{ A˚}$ and $O^{2-}$ is $1.40\text{ A˚}$. Since $N^{3-}$ has the weakest effective nuclear charge among the three, its radius must be strictly greater than $1.40\text{ A˚}$.

Therefore, the ionic radius of $N^{3-}$ is bigger than both $O^{2-}$ and $F^{-}$. The mention of the covalent radius of Nitrogen ($0.74\text{ A˚}$) is simply a distractor, though it beautifully illustrates how much an atom expands when it gains electrons to become an anion!