Periodic Table Revision series

Periodic Table Revision for JEE

The periodic table is one of the most fundamental concepts in chemistry, especially for students preparing for competitive exams like JEE. It provides a structured way to understand the properties, trends, and behaviors of elements. This article will help you revise the periodic table for JEE with a focus on key trends, important groups, and exam-centric insights.

1. Understanding the Periodic Table

The periodic table is arranged in periods and groups, where elements are placed based on their atomic number, electronic configuration, and recurring chemical properties. Each element has a unique position that reflects its properties. Knowing these properties is crucial for solving JEE questions effectively.

Quick Fact: The periodic table consists of 18 groups (vertical columns) and 7 periods (horizontal rows). Elements are categorized into metals, non-metals, metalloids, and noble gases.

2. Key Trends to Remember

• Atomic Radius: Decreases across a period from left to right and increases down a group.
• Ionization Energy: Increases across a period and decreases down a group.
• Electronegativity: Increases across a period and decreases down a group.
• Electron Affinity: Generally increases across a period, with exceptions in some groups.
• Metallic and Non-metallic Character: Metallic character decreases across a period and increases down a group, whereas non-metallic character shows the opposite trend.

3. Important Groups and Their Properties

Group	Group Name	Characteristics
1	Alkali Metals	Highly reactive, form strong bases (alkalis) when combined with water.
2	Alkaline Earth Metals	Less reactive than Group 1, form oxides that are basic in nature.
17	Halogens	Highly reactive non-metals, react with metals to form salts.
18	Noble Gases	Inert, very low reactivity due to their stable electronic configuration.

4. Tips for JEE Aspirants

• Focus on understanding periodic trends and exceptions, as they are often tested in JEE.
• Practice questions related to the properties of elements and their compounds.
• Memorize key group properties, especially for Alkali Metals, Alkaline Earth Metals, Halogens, and Noble Gases.
• Use mnemonic devices to remember the order of elements in specific groups.

5. Example Questions from JEE Advanced with Solutions

To help you better understand the types of questions asked in JEE Advanced, here are a few examples along with their detailed solutions:

• Example 1: Calculate the first ionization energy of Sodium (Na) given that it is an s-block element and trends in ionization energy across periods. Discuss how this value compares with Magnesium (Mg).
Solution: The first ionization energy of Sodium (Na) is lower than that of Magnesium (Mg) because Na has only one electron in its outermost shell, which is easier to remove. The ionization energy of Na is around 496 kJ/mol, while that of Mg is about 738 kJ/mol. This difference is due to the increased nuclear charge in Mg, which holds the electrons more tightly.
• Example 2: Given the electron affinity of Fluorine (F) is higher than that of Chlorine (Cl), explain this anomaly based on their electronic configurations and atomic sizes.
Solution: Although Fluorine (F) is expected to have a higher electron affinity due to its smaller size and greater electronegativity, the actual electron affinity of Chlorine (Cl) is slightly higher. This anomaly is due to the very small atomic radius of F, causing electron-electron repulsion in the already compact 2p orbital, making it less favorable to add an extra electron compared to Cl.
• Example 3: Identify the correct sequence of metallic character in the following elements: Li, Be, Na, Mg. Justify your answer based on periodic trends.
Solution: The correct sequence of metallic character is Na > Mg > Li > Be. This is because metallic character increases down a group and decreases across a period. Na is more metallic than Li, and Mg is more metallic than Be.
• Example 4: Explain why the ionization energy of Boron (B) is less than that of Beryllium (Be), even though B is to the right of Be in the periodic table.
Solution: Boron (B) has a lower ionization energy than Beryllium (Be) due to its electronic configuration. Be has a completely filled 2s subshell (1s² 2s²), which is more stable. Boron has an additional 2p electron (1s² 2s² 2p¹), which is easier to remove due to its higher energy and less stable configuration, resulting in lower ionization energy.
• Example 5: Determine which element among the given choices has the highest electronegativity and provide a reasoning based on periodic trends and electronic configurations.
Solution: Among most elements, Fluorine (F) has the highest electronegativity (3.98 on the Pauling scale). This is because electronegativity increases across a period and decreases down a group. Fluorine, being in the top right of the periodic table (excluding noble gases), has the greatest tendency to attract electrons.

6. Conclusion

Thorough revision of the periodic table and its trends is essential for JEE preparation. Understanding these concepts not only helps in solving direct questions but also in grasp

Thorough revision of the periodic table and its trends is essential for JEE preparation. Understanding these concepts not only helps in solving direct questions but also in grasping the fundamentals of inorganic chemistry, which is crucial for both JEE Main and Advanced. Keep revising, practicing, and testing yourself to build a strong foundation in this topic.

We hope these examples and their detailed solutions have provided you with a clearer understanding of how to approach periodic table questions in JEE Advanced. Regular practice and conceptual clarity are key to mastering this topic. Happy studying and best of luck with your preparation!