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) 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

(b) A is true; R is true but not the correct explanation of A.

Step 1: Evaluate the Assertion

The assertion states that the third ionization energy of phosphorus is larger than that of sulfur. Ionization energy generally increases across a period. For phosphorus and sulfur, removing the third electron from phosphorus is more challenging because it disrupts the half-filled stable configuration of the 3p orbital. Hence, the assertion is true.

Step 2: Evaluate the Reason

The reason provided mentions the stability of atoms with filled s- and p-sub-shells, referring to noble gas configurations. While stable configurations typically elevate ionization energies, phosphorus does not have a noble gas configuration during its third ionization; rather, it has a half-filled p sub-shell which is stable, but not a filled s- and p-sub-shell configuration. Therefore, the reason is unrelated.

Step 3: Relate Assertion and Reason

Since the assertion is true, but the reason does not correctly explain why the third ionization energy of phosphorus is larger (it's actually due to the stability of the half-filled 3p sub-shell), the reason, while referring to ionization energy trends, does not apply here.

Key Concepts

Periodic Trends

When we talk about ionization energy, the concept of periodic trends becomes crucial. Ionization energy is the amount of energy required to remove an electron from an atom in its gaseous state. This energy tends to show predictable patterns across the periodic table.

In general, ionization energy increases as you move across a period from left to right. The primary reason is that the number of protons increases, leading to a greater effective nuclear charge. This stronger pull from the nucleus makes it more challenging to remove an electron.

• Higher nuclear charge attracts the valence electrons more strongly.
• Electrons added to the same shell experience similar shielding, reinforcing the nuclear attraction.

Conversely, ionization energy generally decreases down a group in the periodic table. The added electron shells result in increased distance from the nucleus, which weakens the nuclear attraction. Thus, it becomes easier to remove the outermost electron.

Understanding these trends helps predict and explain the ionization energies of elements, such as why phosphorus has a higher third ionization energy than sulfur due to the stability associated with its particular electron configuration.

Electron Configuration

Electron configuration describes the arrangement of electrons in an atom's orbitals. This is vital when discussing ionization energy as it can greatly influence the difficulty of removing electrons.

Every element has a unique electron configuration that corresponds to its position in the periodic table. These configurations are built up according to the Pauli exclusion principle, which states each orbital can hold two electrons with opposite spins, and Hund's rule, which states that electrons will occupy degenerate orbitals singly before pairing.

• Shells are filled in the order of increasing energy levels, influenced by both principal quantum numbers and subshell energies.
• The approximate order of filling is given by the Aufbau principle: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.

Considering phosphorus and sulfur:

• Phosphorus has a half-filled 3p sub-shell, making it stable.
• Sulfur has one more electron than phosphorus, breaking this half-filled stability.

These configurations play a direct role in the ionization energies observed, as removing an electron from a stable configuration requires more energy.

Stability of Half-Filled and Fully Filled Orbitals

The stability of electron orbitals can significantly affect ionization energy. Half-filled and fully filled orbitals have a special status due to their symmetry and energy configurations.

Half-filled orbitals, as seen in phosphorus with a 3p configuration, offer stability because of optimized electron repulsion and exchange energy. These electrons are unpaired across the orbitals, reducing repulsion.

• Paired electrons in the same orbital according to Pauli's exclusion principle experience more repulsion.
• Exchange energy is maximized in half-filled orbitals due to symmetrical distribution, offering extra stability.

Fully filled orbitals resemble noble gas configurations, renowned for their extraordinary stability, but not pertinent in phosphorus's case for its third ionization.

While sulfur almost reaches a fully filled subshell, it's this very advantage that contributes to its lower third ionization energy compared to phosphorus. The disruption of half-filled orbitals in phosphorus upon removal of an electron further explains the high energy needed for its third ionization.