Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) \(\mathrm{A}\) is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: Third ionization energy of phosphorus is larger than sulphur. Reason: There is a larger amount of stability associated with filled s- and p-sub-shells (a noble gas electron configuration) which corresponds to having eight electrons in the valence shell of an atom or iron.

Step-by-Step Solution

Answer

Option (b): Both A and R are correct, but R is not the correct explanation of A.

Step 1: Understand the Assertion

The assertion says that the third ionization energy of phosphorus is larger than that of sulfur. This means removing the third electron from phosphorus requires more energy than removing it from sulfur.

Step 2: Assess the Third Ionization Energies

To understand why the third ionization energy of phosphorus is larger than sulfur, consider the electronic configuration. After removing the first two electrons both phosphorus and sulfur have similar states; however, the third ionization would complete a half-filled 'p' sub-shell for phosphorus, which has extra stability compared to sulfur.

Step 3: Examine the Reason Statement

The reason mentioned is concerned with the extra stability associated with having eight electrons in the valence shell, which gives the electron configuration of a noble gas.

Step 4: Evaluate the Correction of the Reason

Phosphorus does not achieve a noble gas configuration by losing three electrons, as it only loses electrons from the 'p' orbitals, resulting in a stable half-full 'p' sub-shell, not a noble gas configuration. Therefore, although stability is relevant, the specific reason cited in R is incorrect.

Step 5: Determine the Relationship between Assertion and Reason

We conclude both the assertion and reason are correct: the assertion is true about third ionization energies, but the reason provided is not the correct explanation of the assertion as it inaccurately describes a noble gas configuration.

Key Concepts

Electronic Configuration

The electronic configuration of an atom describes how electrons are distributed in its atomic orbitals. Each electron's position is defined by four quantum numbers. Electrons fill orbitals in a specific order based on the principles of quantum mechanics, notably the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
The Aufbau principle states that electrons occupy the lowest energy orbital available. Hund's rule states that electrons will fill each degenerate (equal energy) orbital singly before pairing up. The Pauli exclusion principle asserts that no two electrons can have the same set of four quantum numbers.
For example, phosphorus has an atomic number of 15, with an electronic configuration of \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3 \]. Sulfur, with an atomic number of 16, has a configuration of \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4 \].

• Phosphorus has three electrons in its 3p orbital, resulting in a half-filled subshell.
• Sulfur has four electrons, with one paired electron in the 3p subshell.

These configurations play a crucial role in an atom's chemical behavior and energetic stability, especially during ionization.

Periodic Table Trends

The periodic table organizes elements in a way that highlights periodic trends, including ionization energy. Ionization energy is the energy required to remove an electron from an atom in the gaseous state. Typically, ionization energy increases across a period and decreases down a group.
Across a period, increasing proton number leads to stronger nuclear attraction, making electrons harder to remove. However, specific electron configurations can affect these trends. For example, removing an electron from a stable, half-filled orbital, such as phosphorus's 3p subshell, requires more energy due to its additional stability.

• In a comparison between phosphorus and sulfur, even though sulfur is to the right of phosphorus, phosphorus's stable half-filled configuration makes its third ionization energy higher.

The arrangement and energy state of electrons indirectly affect how trends manifest, creating exceptions based on molecular orbital stability.

Stability of Half-Filled Subshells

The stability of half-filled subshells is a crucial concept in understanding ionization energy anomalies. Subshells that are either exactly half-filled or fully filled possess extra stability compared to other configurations.
For instance, a 3p subshell, such as that in phosphorus, is half-filled with three electrons. This half-filled state is energetically favorable due to electron-electron interactions and exchanges providing a symmetric distribution.
This stability arises from two main factors:

• Minimized electron repulsion due to spread electrons.
• Exchange energy, where electrons with parallel spins in different orbitals contribute to a lower energy state.

This isn't the same as achieving a noble gas configuration, which involves a fully filled outer shell. However, such configurations are key in determining the energy needed to remove additional electrons, such as during the third ionization of phosphorus, leading to greater energy requirements compared to sulfur, despite general periodic table trends.