Problem 42 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following, the element with the highest ionization energy is (a) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{1}\) (b) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\) (c) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}\) (d) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p} 4\)

Step-by-Step Solution

Answer

The element with the highest ionization energy is (b) Phosphorus \([ ext{Ne}] 3s^2 3p^3\).

Step 1: Understand the Ionization Energy Concept

Ionization energy is the energy required to remove an electron from an atom. Elements with a more stable electronic configuration have higher ionization energies.

Step 2: Identify Electronic Configurations

List the given electronic configurations:(a) \([ ext{Ne}] 3s^2 3p^1\) is Al (Aluminum),(b) \([ ext{Ne}] 3s^2 3p^3\) is P (Phosphorus),(c) \([ ext{Ne}] 3s^2 3p^2\) is Si (Silicon),(d) \([ ext{Ne}] 3s^2 3p^4\) is S (Sulfur).

Step 3: Compare Ionization Energies of Elements

Generally, the ionization energy increases across a period in the periodic table. Review the elements: Al, Si, P, and S. Noting that the trend for first ionization energy is Al < Si < P < S because P has a half-filled 3p subshell, which is relatively stable.

Step 4: Determine the Element with Highest Ionization Energy

In this sequence, Phosphorus \([ ext{Ne}] 3s^2 3p^3\), with its half-filled 3p subshell, has the electron configuration with the highest ionization energy.

Key Concepts

Electronic Configuration

Electronic configuration describes how electrons are arranged in an atom. Each element is made up of a nucleus surrounded by electrons. These electrons reside in energy levels known as shells. Each shell can contain different subshells labeled as s, p, d, or f. Each of these subshells can hold a certain number of electrons. For example:

• The 's' subshell can hold up to 2 electrons
• The 'p' subshell can hold up to 6 electrons
• The 'd' subshell can hold up to 10 electrons
• The 'f' subshell can hold up to 14 electrons

In the electronic configurations given in the exercise, electrons are arranged following the order of increasing energy levels. The number and arrangement of electrons in these configurations play a crucial role in determining the chemical behavior of elements.

Periodic Trends

The periodic table is a powerful tool for predicting trends in element properties, including ionization energy. `
` Ionization energy is one of the key properties that shows periodic trend patterns. Across a period (from left to right), the ionization energy generally increases due to increased nuclear charge, which pulls electrons closer to the nucleus. `
` As we move down a group (from top to bottom), the ionization energy generally decreases because the added shells increase the distance between the nucleus and the outer electrons, making them easier to remove. `
` For example, in the context of ionization energy, the exercise reveals that phosphorus

• Has a configuration of \(\mathrm{[Ne] 3s^2 3p^3}\)
• exhibits a higher ionization energy compared to its neighbors Aluminum (Al), Silicon (Si), and Sulfur (S) because of the stability of its electron arrangement.

Understanding these trends helps in predicting and comparing the ionization energies of different elements.

Stability of Electron Configurations

The stability of an element's electron configuration greatly affects its ionization energy. Atoms tend to be more stable when their electron configurations are filled or half-filled. `
` For instance, phosphorus (

• Configuration: \(\mathrm{[Ne] 3s^2 3p^3}\)
• has a half-filled 3p subshell, leading to an increased stability.

This increased stability is reflected in its higher ionization energy compared to other nearby elements with less stable configurations. `
` Why does a half-filled subshell provide stability? It’s because each orbital in the subshell is partially filled, ensuring a symmetrical distribution of electrons, reducing repulsion and making the atom more stable. `
`When studying ionization energies,

• always consider how close or far the configuration is to a full or half-full subshell.
• More stable configurations require significantly more energy to remove an electron, reflecting higher ionization energies.

This concept is key to understanding the differences in ionization energies among different elements.