The correct order of ionic radii for the ions, P\(^{3-}\), S\(^{2-}\), Ca\(^{2+}\), K\(^{+}\), Cl\(^{-}\) is:
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For isoelectronic species, remember this simple rule: More protons, smaller size. Anions will always be larger than cations in an isoelectronic series because they have fewer protons pulling the same number of electrons.
Step 1: Understanding the Question:
We need to arrange a given set of ions in decreasing order of their ionic radii. Step 2: Key Formula or Approach:
The key is to first check the number of electrons in each ion. If they have the same number of electrons, they are called isoelectronic species. For isoelectronic species, the ionic radius decreases as the nuclear charge (number of protons or atomic number, Z) increases. A higher nuclear charge pulls the same number of electrons more strongly, resulting in a smaller radius. Step 3: Detailed Explanation:
Let's find the number of electrons and protons (Z) for each ion:
- P\(^{3-}\): Phosphorus (Z=15) gains 3 electrons. Electrons = 15 + 3 = 18.
- S\(^{2-}\): Sulfur (Z=16) gains 2 electrons. Electrons = 16 + 2 = 18.
- Cl\(^{-}\): Chlorine (Z=17) gains 1 electron. Electrons = 17 + 1 = 18.
- K\(^{+}\): Potassium (Z=19) loses 1 electron. Electrons = 19 - 1 = 18.
- Ca\(^{2+}\): Calcium (Z=20) loses 2 electrons. Electrons = 20 - 2 = 18.
All the given ions are isoelectronic, as they all have 18 electrons.
Now, we arrange them according to their nuclear charge (Z):
P (Z=15) \(<\) S (Z=16) \(<\) Cl (Z=17) \(<\) K (Z=19) \(<\) Ca (Z=20)
Since the ionic radius decreases with increasing nuclear charge for isoelectronic species, the order of decreasing radii will be the reverse of the order of nuclear charge:
P\(^{3-}\) \(>\) S\(^{2-}\) \(>\) Cl\(^{-}\) \(>\) K\(^{+}\) \(>\) Ca\(^{2+}\) Step 4: Final Answer:
The correct order of decreasing ionic radii is P\(^{3-}\) \(>\) S\(^{2-}\) \(>\) Cl\(^{-}\) \(>\) K\(^{+}\) \(>\) Ca\(^{2+}\).
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