The ionic radii of O2-, F-, Na+ and Mg2+ are in the order
(a) O2- F- Mg2+ Na+
(b) Mg2+ Na+

Q: Why does a negative charge make an ion larger?

When an atom gains electrons to become a negative ion (anion), the number of electrons increases while the number of protons remains the same. This increases the electron-electron repulsion in the outer shell, causing the electron cloud to expand and making the ion larger than its parent atom.

Q: Why does a positive charge make an ion smaller?

When an atom loses electrons to become a positive ion (cation), the electron-electron repulsion decreases. Additionally, the same number of protons is now pulling on fewer electrons, so the effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus.

Q: What exactly does 'isoelectronic' mean?

Isoelectronic species are atoms, ions, or molecules that have the exact same number of total electrons and the same electron configuration. For example, $O^{2-}$, $F^{-}$, $Ne$, $Na^{+}$, and $Mg^{2+}$ all have exactly 10 electrons.

Q: Does the number of neutrons affect the ionic radius?

No, neutrons are electrically neutral. The size of the electron cloud is determined by the electrostatic attraction between the positively charged protons in the nucleus and the negatively charged electrons. Neutrons only affect the mass of the atom, not its radius.

Q: Can we compare the sizes of ions that are NOT isoelectronic using this exact same method?

No. If the ions have different numbers of electrons, you cannot simply look at the number of protons. You must also consider the number of electron shells (principal quantum number, $n$) and the shielding effect caused by inner electrons. The $1/Z$ rule strictly applies only when the electron count is identical.

Question

The ionic radii of $O^{2-}, F^{-}, Na^{+}$ and $Mg^{2+}$ are in the order

Options:

(A) $O^{2-} > F^{-} > Mg^{2+} > Na^{+}$

(B) $Mg^{2+} > Na^{+} > F^{-} > O^{2-}$

(C) $O^{2-} > F^{-} > Na^{+} > Mg^{2+}$

(D) $F^{-} > O^{2-} > Na^{+} > Mg^{2+}$

Sigma Percentile · Accepted Answer

### The Setup: A Battle of Ions

Imagine you are looking at a microscopic lineup of four different ions: the Oxide ion ($O^{2-}$), the Fluoride ion ($F^{-}$), the Sodium ion ($Na^{+}$), and the Magnesium ion ($Mg^{2+}$). Our mission is to arrange them in the correct decreasing order of their ionic radii. At first glance, they look completely different—some are negatively charged anions, while others are positively charged cations. But there is a hidden secret that connects them all.

### The Secret of Isoelectronic Species

To uncover this secret, let's count the number of electrons in each ion.

Oxygen normally has an atomic number of 8, meaning it has 8 electrons. By gaining 2 extra electrons to become $O^{2-}$, it now has $8 + 2 = 10$ electrons. Fluorine has an atomic number of 9, and gaining 1 electron gives it $9 + 1 = 10$ electrons. Sodium, with an atomic number of 11, loses 1 electron to become $Na^{+}$, leaving it with $11 - 1 = 10$ electrons. Finally, Magnesium has an atomic number of 12, and losing 2 electrons leaves it with $12 - 2 = 10$ electrons.

Notice the pattern? Every single one of these ions has exactly **10 electrons**. In chemistry, we call these **isoelectronic species**—a group of atoms or ions that share the exact same electron configuration.

### The Nuclear Tug-of-War

Since all these ions have the same number of electrons, what determines their size? The answer lies in the nucleus. The nucleus contains positively charged protons that pull on the negatively charged electrons. This is a classic electrostatic tug-of-war.

For isoelectronic species, the rule is simple: **The greater the nuclear charge (number of protons), the stronger the attractive pull on the electrons.** A stronger pull draws the electron cloud closer to the nucleus, resulting in a smaller ionic radius. Mathematically, we can say that for isoelectronic species, the ionic radius is inversely proportional to the nuclear charge ($Z$):

$$ 	ext{Ionic Radius} \propto \frac{1}{Z} $$

### The Final Verdict

Let's line up our ions based on their atomic numbers ($Z$):
- $O^{2-}$ has $Z = 8$
- $F^{-}$ has $Z = 9$
- $Na^{+}$ has $Z = 11$
- $Mg^{2+}$ has $Z = 12$

Magnesium has the highest number of protons (12). It acts like a super-strong magnet, pulling its 10 electrons very tightly, making $Mg^{2+}$ the smallest ion in the group. On the other end of the spectrum, Oxygen has the fewest protons (8). Its pull is the weakest, allowing the electron cloud to expand, making $O^{2-}$ the largest ion.

Therefore, the correct decreasing order of their ionic radii is:

$$ \mathbf{O^{2-} > F^{-} > Na^{+} > Mg^{2+}} $$

This perfectly matches option (c). Always remember this golden rule for isoelectronic species: more protons mean a smaller size. It is a high-yield concept that frequently appears in competitive exams!

Animated Solution for Chemistry - Periodicity in Properties: The ionic radii of $O^{2 -}, F^{-}, N a^{+}$ and $M g^{2 +}$ are in the order

Select Answer:

Visualized Solution

$G i v e n I o n s$

$E l ec t r o n C o u n t$

$I soe l ec t r o ni c S p ec i es R u l e$

$C o m p a r in g N u c l e a r C ha r g e$

$F ina l O r d er o f R a d ii$

$T h e W a y F or w a r d$

The Sigma Insight: Periodic Table and Periodic Properties

The Setup

The Secret of Isoelectronic Species

The Nuclear Tug-of-War

The Final Verdict

Similar Questions

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The increasing order of the ionic radii of the given isoelectronic species is

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Animated Solution for Chemistry - Periodicity in Properties: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ionic radii of O2−,F−,Na+ and Mg2+ are in the order