The increasing order of the ionic radii of the given isoelectronic species is
(a) Cl-, Ca2+, K+, S2-

Q: Why do we only look at protons here?

Because the number of electrons is identical across all the given species (they are isoelectronic). Since the electron-electron repulsion is roughly the same, the only varying factor that affects the size is the nuclear pull, which depends on the number of protons.

Q: Does the sign of the charge matter?

The sign of the charge helps you calculate the number of electrons. Once you establish that they are isoelectronic, you can ignore the charge and just look at the atomic number (number of protons) to determine the size.

Q: What if the number of electrons were different?

If the species are not isoelectronic, you have to consider both the number of shells (principal quantum number) and the effective nuclear charge. Generally, more shells mean a larger size, and within the same period, a higher effective nuclear charge means a smaller size.

Q: Why is a cation smaller than its parent atom?

When an atom loses an electron to become a cation, the electron-electron repulsion decreases, but the number of protons remains the same. This means the effective nuclear pull per electron increases, drawing the remaining electrons closer and reducing the size.

Q: Can an anion be smaller than a cation?

Yes, but usually only if they belong to different periods. For example, a Fluoride ion ($\text{F}^-$) is smaller than a Cesium ion ($\text{Cs}^+$) because Cesium has many more electron shells, even though it's a cation. However, for isoelectronic species, the anion is always larger than the cation.

Question

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

Options:

(A) $	ext{Cl}^-, 	ext{Ca}^{2+}, 	ext{K}^+, 	ext{S}^{2-}$

(B) $	ext{S}^{2-}, 	ext{Cl}^-, 	ext{Ca}^{2+}, 	ext{K}^+$

(C) $	ext{Ca}^{2+}, 	ext{K}^+, 	ext{Cl}^-, 	ext{S}^{2-}$

(D) $	ext{K}^+, 	ext{S}^{2-}, 	ext{Ca}^{2+}, 	ext{Cl}^-$

Sigma Percentile · Accepted Answer

### The Mystery of Isoelectronic Species

Imagine a tug-of-war where one team always has exactly 18 players. No matter what, there are always 18 people pulling on one side of the rope. This is exactly what happens with **isoelectronic species**. The prefix "iso-" means "same," so isoelectronic species are atoms or ions that have the exact same number of electrons.

In our problem, we are given four ions: $	ext{Cl}^-$, $	ext{Ca}^{2+}$, $	ext{K}^+$, and $	ext{S}^{2-}$. Let's do a quick headcount of their electrons. Chlorine normally has 17 electrons, but the negative charge means it gained one, bringing the total to 18. Calcium starts with 20, but the $2+$ charge means it lost two, leaving it with 18. Potassium (19) loses one to become 18, and Sulfur (16) gains two to reach 18.

They all have exactly 18 electrons! So, if the electron cloud is the same size in terms of the number of particles, what makes one ion larger or smaller than another?

### The Power of the Nucleus

This is where the other team in our tug-of-war comes in: the protons in the nucleus. The nucleus is positively charged and pulls the negatively charged electrons inward.

Since the number of electrons is constant (18) for all these ions, the size of the ion depends entirely on how strong the nucleus is. The strength of the nucleus is determined by the number of protons, which is the atomic number ($Z$).

Let's look at the atomic numbers:
- Calcium ($Z = 20$) has 20 protons.
- Potassium ($Z = 19$) has 19 protons.
- Chlorine ($Z = 17$) has 17 protons.
- Sulfur ($Z = 16$) has 16 protons.

### The Final Verdict

Think about it: 20 protons pulling on 18 electrons will have a much stronger grip than 16 protons pulling on the same 18 electrons. The stronger the pull, the closer the electrons are drawn to the nucleus, and the smaller the ion becomes.

Therefore, for isoelectronic species, the ionic radius is inversely proportional to the atomic number ($Z$). Mathematically, we can write this as:
$$r \propto \frac{1}{Z}$$

Applying this logic, Calcium ($	ext{Ca}^{2+}$) has the most protons, so it pulls its electrons the tightest, making it the smallest. Sulfur ($	ext{S}^{2-}$) has the fewest protons, so its grip is the weakest, allowing the electron cloud to expand and making it the largest.

Arranging them in increasing order of size, we get:
**$	ext{Ca}^{2+} < 	ext{K}^+ < 	ext{Cl}^- < 	ext{S}^{2-}$**

This perfectly matches option (c). It's a beautiful demonstration of how subatomic forces dictate the physical properties of elements!

Animated Solution for Chemistry - Periodicity in Properties: The increasing order of the ionic radii of the given isoelectronic species is

Select Answer:

Visualized Solution

$Isoelectronic Species$

$Electron Count$

$Atomic Numbers (Z)$

$Effective Nuclear Charge$

$Final Order$

$The Way Forward$

The Sigma Insight: Periodic Table and Periodic Properties

The Mystery of Isoelectronic Species

The Power of the Nucleus

The Final Verdict

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