Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Aqueous solution of which salt will not contain ions with the electronic configuration \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} ?\) [Main Online April 10, 2016] (a) \(\mathrm{NaF}\) (b) \(\mathrm{KBr}\) (c) \(\mathrm{NaCl}\) (d) \(\mathrm{CaI}_{2}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(\mathrm{NaF}\).

Step 1: Understanding the Problem

The question asks us to identify which salt, when dissolved in water, will not produce any ions with the electronic configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\). This configuration corresponds to the noble gas configuration of \(\text{Ar}\) (Argon).

Step 2: Identifying Ion Configurations

Each salt when dissolved in water dissociates into its constituent ions. We need to examine the electronic configuration of the resulting ions from each salt. 1. \(\mathrm{NaF}\) gives \(\mathrm{Na}^+\) and \(\mathrm{F}^-\).2. \(\mathrm{KBr}\) gives \(\mathrm{K}^+\) and \(\mathrm{Br}^-\).3. \(\mathrm{NaCl}\) gives \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\).4. \(\mathrm{CaI}_{2}\) gives \(\mathrm{Ca}^{2+}\) and \(\mathrm{I}^-\).

Step 3: Determining Each Ion's Configuration

We must determine the electronic configuration of each ion:- \(\mathrm{Na}^+\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}\) (Neon).- \(\mathrm{F}^-\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}\) (Neon).- \(\mathrm{K}^+\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\) (Argon).- \(\mathrm{Br}^-\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{d}^{10} 4 \mathrm{s}^{2} 4 \mathrm{p}^{6}\).- \(\mathrm{Cl}^-\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\) (Argon).- \(\mathrm{Ca}^{2+}\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\) (Argon).- \(\mathrm{I}^-\) has a configuration of \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{d}^{10} 4 \mathrm{s}^{2} 4 \mathrm{p}^{6} 4 \mathrm{d}^{10} 5 \mathrm{s}^{2} 5 \mathrm{p}^{6}\).

Step 4: Choosing the Correct Answer

By analyzing the configurations, we see that \(\mathrm{K}^+\), \(\mathrm{Cl}^-\), \(\mathrm{Ca}^{2+}\) all have the desired configuration of Argon (\(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6}\)).Only ions from \(\mathrm{NaF}\) do not reach this configuration. Hence, the correct answer is \(\mathrm{NaF}\).

Key Concepts

Ionic Dissociation

When substances like salts dissolve in water, they undergo a process known as ionic dissociation. This means that the salt breaks down into its respective positive and negative ions. For instance, when you place sodium fluoride (\(\text{NaF}\)) in water, it separates into sodium ions (\(\text{Na}^+\)) and fluoride ions (\(\text{F}^-\)).
This process is crucial in chemistry because it dictates many reactions that involve ions. Dissociated ions are often the active participants in chemical reactions.

Sodium Fluoride (NaF): Separates into \(\text{Na}^+\) and \(\text{F}^-\).

Potassium Bromide (KBr): Splits into \(\text{K}^+\) and \(\text{Br}^-\).

Sodium Chloride (NaCl): Dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\).

Calcium Iodide (CaI}_2): Breaks down into \(\text{Ca}^{2+}\) and \(\text{I}^-\).

By understanding how ionic dissociation works, we can predict behaviors of solutions, such as conductivity and reactivity, based on the ions present after dissociation.

Noble Gas Configuration

The concept of noble gas configuration refers to an atom or ion having an electron arrangement similar to that of a noble gas. Noble gases are known for their stability due to having full valence shells, which makes them chemically inert.
In chemistry, atoms often gain or lose electrons to achieve this type of stable configuration. For example, when an atom achieves a full set of electrons in its outermost shell, it mirrors the electron arrangement of the nearest noble gas.

• Argon (Ar): With an electron configuration of \(1 \text{s}^2 2 \text{s}^2 2 \text{p}^6 3 \text{s}^2 3 \text{p}^6\), it's a goal for many elements in the third period to emulate this structure.
• Example: The chloride ion (\(\text{Cl}^-\)) and calcium ion (\(\text{Ca}^{2+}\)) both adjust their electron counts to achieve the noble gas configuration of argon.

Achieving a noble gas configuration is often a driving force for chemical reactions and stability in compounds.

Electronic Structure of Ions

The electronic structure or configuration of ions is a key concept in chemistry that describes how electrons are arranged in an ion's orbitals. These configurations differ from those of neutral atoms because ions have lost or gained electrons to become charged.
When determining the electronic configuration for any ion, it’s crucial to account for this electron change. This involves subtracting electrons for positive ions (cations) and adding electrons for negative ions (anions).

• Sodium Ion (Na⁺): Has lost one electron, resulting in a configuration similar to neon, \(1 \text{s}^2 2 \text{s}^2 2 \text{p}^6\).
• Fluoride Ion (F⁻): Has gained an electron, also reaching a configuration like neon.
• Calcium Ion (Ca²⁺): By losing two electrons, it has the same configuration as argon.
• Chloride Ion (Cl⁻): Gains an electron to mimic argon's stable structure.

Understanding the electronic structure of ions helps in predicting their chemical behavior, especially in reactions where electron transfer is key.