Problem 169 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following sets of ions represents a collection of isoelectronic species? [2006] (a) \(\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}\) (b) \(\mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}\) (c) \(\mathrm{N}^{3}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{S}^{2-}\) (d) \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+} \mathrm{Ca}^{2+}\)

Step-by-Step Solution

Answer

Option (a) represents isoelectronic species (all ions have 18 electrons).

Step 1: Understand Isoelectronic Species

Isoelectronic species are atoms, ions, or molecules that have the same number of electrons. To identify which options contain isoelectronic species, count the electrons in each ion present in the options.

Step 2: Determine Electrons in Option (a)

Calculate the number of electrons for each ion in option (a):- \( \mathrm{K}^{+} \): Potassium has 19 protons and loses 1 electron to form \( \mathrm{K}^{+} \), so it has 18 electrons.- \( \mathrm{Cl}^{-} \): Chlorine has 17 protons and gains 1 electron to form \( \mathrm{Cl}^{-} \), so it has 18 electrons.- \( \mathrm{Ca}^{2+} \): Calcium has 20 protons and loses 2 electrons to form \( \mathrm{Ca}^{2+} \), so it has 18 electrons.- \( \mathrm{Sc}^{3+} \): Scandium has 21 protons and loses 3 electrons to form \( \mathrm{Sc}^{3+} \), so it has 18 electrons. All species in option (a) have 18 electrons.

Step 3: Determine Electrons in Option (b)

Calculate the number of electrons for each ion in option (b):- \( \mathrm{Ba}^{2+} \): Barium has 56 protons and loses 2 electrons to form \( \mathrm{Ba}^{2+} \), so it has 54 electrons.- \( \mathrm{Sr}^{2+} \): Strontium has 38 protons and loses 2 electrons to form \( \mathrm{Sr}^{2+} \), so it has 36 electrons.- \( \mathrm{K}^{+} \): Same as in option (a), it has 18 electrons.- \( \mathrm{Ca}^{2+} \): Same as in option (a), it has 18 electrons. The numbers of electrons are not the same, so these ions are not isoelectronic.

Step 4: Determine Electrons in Option (c)

Calculate the number of electrons for each ion in option (c):- \( \mathrm{N}^{3-} \): Nitrogen has 7 protons and gains 3 electrons to form \( \mathrm{N}^{3-} \), so it has 10 electrons.- \( \mathrm{O}^{2-} \): Oxygen has 8 protons and gains 2 electrons to form \( \mathrm{O}^{2-} \), so it has 10 electrons.- \( \mathrm{F}^{-} \): Fluorine has 9 protons and gains 1 electron to form \( \mathrm{F}^{-} \), so it has 10 electrons.- \( \mathrm{S}^{2-} \): Sulfur has 16 protons and gains 2 electrons to form \( \mathrm{S}^{2-} \), so it has 18 electrons.Three ions have 10 electrons, but one does not, so they are not all isoelectronic.

Step 5: Determine Electrons in Option (d)

Calculate the number of electrons for each ion in option (d):- \( \mathrm{Li}^{+} \): Lithium has 3 protons and loses 1 electron to form \( \mathrm{Li}^{+} \), so it has 2 electrons.- \( \mathrm{Na}^{+} \): Sodium has 11 protons and loses 1 electron to form \( \mathrm{Na}^{+} \), so it has 10 electrons.- \( \mathrm{Mg}^{2+} \): Magnesium has 12 protons and loses 2 electrons to form \( \mathrm{Mg}^{2+} \), so it has 10 electrons.- \( \mathrm{Ca}^{2+} \): Same as in option (a), it has 18 electrons.The numbers of electrons are not the same, so these ions are not isoelectronic.

Key Concepts

Ions

In chemistry, ions are atoms or molecules with a net electric charge due to the loss or gain of electrons. This charge alteration occurs because atoms seek to achieve a stable electronic configuration, often resembling that of the noble gases.
When an atom loses electrons, it forms a positively charged ion, known as a cation. On the other hand, gaining electrons results in a negatively charged ion called an anion.
These charges play a crucial role in determining the properties and reactivity of a substance.
To better understand this, consider these examples:

• Potassium (K) loses one electron to become \(\mathrm{K}^{+}\), a cation.
• Chlorine (Cl) gains one electron to become \(\mathrm{Cl}^{-}\), an anion.

The science of ions is fundamental to areas like electrochemistry and is essential when discussing isoelectronic species, where the electron count is key.

Electronic Configuration

Electronic configuration describes how electrons are distributed in an atom's orbitals around the nucleus.
It is structured based on energy levels and follows principles such as the Pauli exclusion principle and Hund's rule.
Each element has a unique electronic configuration that reflects its position on the periodic table. This configuration greatly influences the chemical and physical properties of elements. Consider the electronic configuration of neutral and ionized states of potassium:

• Neutral Potassium (K): \[1s^2 \,2s^2 \,2p^6 \,3s^2 \,3p^6 \,4s^1 \]
• Ionic Potassium (\(\mathrm{K}^{+}\)): \[1s^2 \,2s^2 \,2p^6 \,3s^2 \,3p^6 \]

Losing an electron to form \(\mathrm{K}^{+}\) results in a stable electronic configuration similar to argon. This stability is central to understanding why atoms form ions and why some configurations are more prevalent.

Electron Counting

Electron counting involves determining the number of electrons in atoms or ions.
This process is essential for identifying their electronic configuration and understanding chemical bonding.
Counting electrons is straightforward for neutral atoms; the atomic number equals the electron count.
However, for ions, the process requires adjusting for gained or lost electrons:

• For cations, subtract the number of lost electrons from the atomic number.
• For anions, add the number of gained electrons to the atomic number.

Taking potassium, chloride, calcium, and scandium as examples:

• \(\mathrm{K}^{+}\) starts with 19 electrons but loses 1, remaining with 18 electrons.
• \(\mathrm{Cl}^{-}\) starts with 17 electrons and gains 1, also reaching a total of 18 electrons.

Understanding electron counting aids in identifying isoelectronic species, which are crucial in this chemistry challenge.

Chemistry Problem Solving

Chemistry problem-solving often involves various techniques, but a fundamental approach is crucial for success.
Solving specific problems requires a strong foundation in concepts, accurate analytical skills, and effective strategies.
Here are some helpful steps for tackling chemistry problems:

• Understand the problem: Identify the main concepts and what the problem is asking.
• Gather relevant data: Collect information about the elements or compounds involved.
• Apply appropriate methods: Use concepts like electron counting and known principles to address the problem.
• Check your work: Verify calculations and whether the solution makes sense in context.

For the isoelectronic species problem, students counted electrons for each ion to confirm their suitability as isoelectronic groups. Option (a) proved to demonstrate isoelectronic behavior as all ions had 18 electrons.
By honing these skills, students can approach chemistry questions with confidence and clarity, leading to successful problem resolution.