It is not possible to measure the atomic radius precisely since the electron cloud surrounding the atom does not have a sharp boundary. One practical approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and then dividing by two. For metals we define the term "metallic radius" which is taken as half the inter nuclear distance separating the metal cores in the metallic crystal. Then van der waal's radius represents the over all size of the atoms which includes its valence shell in a non bonded situation. It is the half of the distance between two similar atoms in separate molecules in a solid. The atomic radius decreases across a period and increases down the group. Same trends are observed in case of ionic radius. Ionic radius of the species having same number of electrons depends on the number of protons in their nuclei. Sometimes, atomic and ionic radii give unexpected trends due to poor shielding of nuclear charge by `d`-and `f-`orbital electrons.
Now answer the following three questions:
`K^(+), CI^(-), Ca^(2+), S^(2-)` ions are isoelectronic. The decreasing order of their size is:

Options

A. `Ca^(2+) gt K^(+) gt CI^(-) gt S^(2-)`
B. `S^(2-) gt CI^(-) gt K^(+) gt Ca^(2+)`
C. `K^(+) gt CI^(-) gt Ca^(2+) gt S^(2-)`
D. `S^(2-) gt CI^(-) gt Ca^(2+) gt K^(+)`

Explanation

To determine the decreasing order of size for the isoelectronic ions \( K^+, Cl^-, Ca^{2+}, S^{2-} \), we need to analyze the charges and their effects on ionic size. Here’s a step-by-step solution:

### Step 1: Identify the Isoelectronic Species
All given ions \( K^+, Cl^-, Ca^{2+}, S^{2-} \) are isoelectronic, meaning they all have the same number of electrons. In this case, they all have 18 electrons, corresponding to the electron configuration of Argon (Ar).

### Step 2: Understand the Effect of Charge on Ionic Size
In isoelectronic species, the size of the ions is influenced by their charge:
- Anions (negatively charged ions) are larger than cations (positively charged ions).
- Among anions, the one with the highest negative charge will be the largest.
- Among cations, the one with the highest positive charge will be the smallest.

### Step 3: Compare the Charges of the Ions
- \( K^+ \) has a charge of +1.
- \( Cl^- \) has a charge of -1.
- \( Ca^{2+} \) has a charge of +2.
- \( S^{2-} \) has a charge of -2.

### Step 4: Determine Relative Sizes Based on Charge
1. **Cations**:
- \( Ca^{2+} \) (smallest due to highest positive charge)
- \( K^+ \) (larger than \( Ca^{2+} \) due to lower positive charge)

2. **Anions**:
- \( S^{2-} \) (largest due to highest negative charge)
- \( Cl^- \) (larger than \( K^+ \) and \( Ca^{2+} \) but smaller than \( S^{2-} \))

### Step 5: Arrange the Ions in Decreasing Order of Size
Based on the analysis:
- Largest: \( S^{2-} \)
- Next: \( Cl^- \)
- Next: \( K^+ \)
- Smallest: \( Ca^{2+} \)

Thus, the decreasing order of their size is:
\[ S^{2-} > Cl^- > K^+ > Ca^{2+} \]

### Final Answer
The decreasing order of size of the ions is:
\[ S^{2-} > Cl^- > K^+ > Ca^{2+} \]

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