Problem 130 - 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: The first ionization energy of Be is greater than that of \(\mathrm{B}\). Reason: \(2 \mathrm{p}\) orbital is lower in energy than \(2 \mathrm{~s}\). 131 . Assertion: Third ionization energy of phosphorus is larger than sulphur.

Step-by-Step Solution

Answer

(c) A is true but R is false. (a) Both assertions are true and valid.

Step 1: Understanding the Assertion about Ionization Energies

The assertion states that the first ionization energy of Beryllium (Be) is greater than that of Boron (B). Generally, ionization energy increases across a period, but Be has a stable electron configuration (2s²), making it more difficult to remove an electron compared to B, which has an electron in a higher energy 2p orbital.

Step 2: Evaluating the Reason Given

The reason claims that the 2p orbital is lower in energy than the 2s orbital. This is incorrect as in any given principal energy level, the s orbital is lower in energy than the p orbital.

Step 3: Analyzing the Assertion Correctness

The assertion that the first ionization energy of Be is greater than B is true because the electron in Be is in a 2s orbital which is fully filled and stable, while in B, the electron is being removed from a higher energy 2p orbital.

Step 4: Determining the Correct Explanation

Since the reason states the opposite of the actual energy levels (where 2s is indeed lower in energy than 2p), the reason is false and cannot be the correct explanation for the assertion.

Step 5: Understanding the Assertion about Third Ionization Energies

The assertion states that the third ionization energy of phosphorus is larger than that of sulfur. Phosphorus has half-filled 3p orbitals after the removal of the first two electrons, giving it extra stability compared to sulfur. Therefore, it's easier to remove an additional electron from sulfur, making the third ionization energy of phosphorus higher.

Step 6: Conclusion on the Second Assertion

Since each electron removed from phosphorus destabilizes it less than sulfur, the assertion about phosphorus having a higher third ionization energy than sulfur is true.

Key Concepts

beryllium (Be) ionization

Beryllium (\( \text{Be} \)) sits in the second period of the periodic table. Its first ionization energy is higher than that of boron (\( \text{B} \)). Ionization energy refers to the energy required to remove an electron from a neutral atom in its gaseous state. \( \text{Be} \) has an electron configuration of \( 1s^2 2s^2 \). This configuration is particularly stable because the \( 2s \) orbital is completely filled with two electrons.

A filled orbital provides stability and makes it energetically less favorable to remove an electron. This high stability results in a higher ionization energy, making electron removal more difficult. \( \text{Be} \) contrasts with \( \text{B} \), which has an electron in the higher energy \( 2p \) orbital. Therefore, the assertion that the ionization energy of \( \text{Be} \) is larger than \( \text{B} \) is accurate.

boron (B) ionization

Boron (\( \text{B} \)) is located to the right of beryllium on the periodic table, indicating that, generally, it should have a higher ionization energy. However, its first ionization energy is actually lower than that of beryllium. This anomaly can be explained by examining its electron configuration.\( \text{B} \) has the configuration \( 1s^2 2s^2 2p^1 \). The presence of a single electron in the \( 2p \) orbital leads to less stability compared to the fully-filled \( 2s \) orbital in \( \text{Be} \). \( 2p \) orbitals are at higher energy levels compared to \( 2s \), making the electron easier to remove.

Thus, although \( \text{B} \) has more total electrons, the presence of a single \( 2p \) electron actually decreases the overall ionization energy when compared to the more stable configuration of \( \text{Be} \).

electron configuration stability

Electron configuration stability is a key concept in understanding periodic trends. Atoms strive for the lowest energy and most stable arrangement of electrons. Beryllium's \( 1s^2 2s^2 \) configuration is more stable due to a fully filled \( 2s \) orbital. This contrasts with boron, which transitions to \( 1s^2 2s^2 2p^1 \), introducing a new subshell with less stability.

A stable electron configuration minimizes repulsion between electrons and maximizes attraction to the nucleus. Shells and subshells prefer to be either completely filled or half-filled for maximum stability:

• Filled orbitals provide stability by completing electron pairs.
• Half-filled orbitals stabilize due to electron distribution symmetry.

These principles are integral to explaining ionization energies and electron affinities across the periodic table.

energy levels of orbitals

Energy levels of atomic orbitals are foundationally important in chemistry. They determine how tightly an electron is bound to an atom and affect its ionization energy. In any principal energy level, the \( s \) orbitals are lower in energy than \( p \) orbitals. This explains why removing an electron from the \( 2s \) in beryllium is harder than from the \( 2p \) in boron.

• Lower energy means electrons are held more tightly by the nucleus.
• Higher energy orbitals have electrons that are easier to remove.

A clear understanding of these energy levels helps explain anomalies in expected ionization trends, such as the case with beryllium and boron.

third ionization energy comparison

When comparing the third ionization energy of elements such as phosphorus and sulfur, electron configuration plays a crucial role. Phosphorus, upon losing two electrons, maintains a half-filled \( 3p \) configuration, which is exceptionally stable.

For phosphorus, \( ext{P}^+ ightarrow ext{P}^{3+} \), the loss of a third electron is more significant energetically due to this stability. Conversely, sulfur has a different electron arrangement and, after two electron removals, is less stable than phosphorus. Thus, stripping a third electron requires less energy compared to phosphorus:

• Third ionization involves removing electrons from the \( 3p \) orbitals.
• A half-filled \( 3p \) in phosphorus makes the third ionization energy larger and more challenging compared to sulfur.

This comparison underscores the importance of electron configuration stability in ionization energy discussions.