Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

The electronic configuration of elements A, B and C are \([\mathrm{He}] 2 \mathrm{~s}^{1},[\mathrm{Ne}] 3 \mathrm{~s}^{1}\) and \([\mathrm{Ar}] 4 \mathrm{~s}^{1}\) respectively. Which one of the following order is correct for the first ionization potentials (in \(\mathrm{kJ} \mathrm{mol}^{-}\)) of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) ? (a) \(\mathrm{A}>\mathrm{B}>\mathrm{C}\) (b) \(\mathrm{O}>\mathrm{B}>\mathrm{A}\) (c) \(\mathrm{B}>\mathrm{O}>\mathrm{A}\) (d) \(C>A>B\)

Step-by-Step Solution

Answer

The correct order is (a) \(\text{A} > \text{B} > \text{C}\).

Step 1: Identify Elements from Electronic Configuration

The electronic configurations provided are \([\mathrm{He}] 2\mathrm{~s}^{1}\), \([\mathrm{Ne}] 3\mathrm{~s}^{1}\), and \([\mathrm{Ar}] 4\mathrm{~s}^{1}\), which correspond to the elements lithium (Li), sodium (Na), and potassium (K) respectively. This is because each configuration ends with a single electron in the s-orbital after a noble gas core: Li is \([\mathrm{He}] 2\mathrm{~s}^{1}\), Na is \([\mathrm{Ne}] 3\mathrm{~s}^{1}\), and K is \([\mathrm{Ar}] 4\mathrm{~s}^{1}\).

Step 2: Recall Ionization Energy Trends

First ionization energy generally decreases as you move down a group in the periodic table due to increasing atomic size and shielding effect. Lithium, being in the second period, should have the highest ionization energy, followed by sodium and then potassium, because as we move from Li to K down Group 1 (the alkali metals), the energy required to remove the outermost electron decreases.

Step 3: Compare Ionization Energies for A, B, and C

Given the trend identified for the Group 1 elements (Li, Na, K), the expected order based on decreasing ionization potential would be that Li has the highest ionization potential, followed by Na, and then K: \(\text{Li (A)} > \text{Na (B)} > \text{K (C)}\).

Step 4: Match the Order with Given Options

From the calculated order \(\text{A} > \text{B} > \text{C}\), we can see that this order matches with option (a), which states \(\mathrm{A} > \mathrm{B} > \mathrm{C}\). Therefore, this is the correct option.

Key Concepts

Electronic Configuration

Electronic configuration is the arrangement of electrons within an atom. Each electron shell has subshells (s, p, d, f) that fill according to precise rules.
It follows the Aufbau principle, where electrons occupy orbitals in order of increasing energy levels.

• In the configurations provided, \( [\mathrm{He}] 2\mathrm{~s}^{1} \), \( [\mathrm{Ne}] 3\mathrm{~s}^{1} \), and \( [\mathrm{Ar}] 4\mathrm{~s}^{1} \), these elements have one electron in the s-orbital outside a noble gas core.
• This setup makes them members of the alkali metal family.
• Alchemy arrangement in this way is pivotal to their chemical behavior, including interaction and ionization energy.

Recognizing these configurations is crucial for determining chemical properties and predicting behaviors.

Periodic Trends

Periodic trends refer to patterns in the periodic table concerning elements' properties. One of these trends is ionization energy.
Ionization energy is the energy needed to remove an electron from an atom.

• It generally increases across a period as the effective nuclear charge attracts electrons stronger.
• Moving down a group usually leads to a decrease in ionization energy, as the atomic size increases.

These trends help predict and compare behaviors like how easily an atom loses or gains electrons, a crucial concept to grasp in chemistry.

Alkali Metals

Alkali metals are found in Group 1 of the periodic table and include lithium (Li), sodium (Na), and potassium (K).
These metals have several distinct characteristics:

• They have one electron in their outermost s-orbital, making them very reactive.
• They readily lose this electron to form positive ions or cations (Li\(^+\), Na\(^+\), K\(^+\)).
• They are soft and exhibit a metallic luster in their pure form.
• React aggressively with water, releasing hydrogen and forming hydroxides.

Alkali metals exhibit a straightforward trend in ionization energies, decreasing down the group due to increasing atomic radii and nuclear size.

Periodic Table Group 1

Group 1 of the periodic table is synonymous with alkali metals. These elements are characterized by the same electronic configuration of having one valence electron in the s-orbital.
Be aware of some of the following points:

• The similarities among these elements lead to common chemical behaviors.
• They exhibit increasing reactivity as you move down from lithium to cesium.
• Ionization energy decreases down the group, making each successive element more reactive.

Understanding Group 1 behavior is essential in grasping larger periodic trends and their implications on chemistry at large.

Shielding Effect

The shielding effect describes how inner shell electrons can block outer electrons from the full effect of the nucleus' positive charge. This occurs because electrons repel each other.
Here is how it works:

• As more inner electrons are added, the outer electron feels less attraction to the nucleus.
• This decreased pull results in a lower ionization energy.
• Consequently, outer electrons are more easily lost compared to those closer to the nucleus.

This effect grows down a group, playing a significant role in explaining the trends observed in ionization energies and reactivity, particularly in alkali metals like Li, Na, and K.