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) 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 .

Step-by-Step Solution

Answer

(c) A is true but R is false.

Step 1: Understanding the Assertion

The assertion is about the first ionization energy of Beryllium (Be) and Boron (B). The ionization energy is the energy required to remove the most loosely bound electron. The assertion claims that Be has a higher ionization energy compared to B.

Step 2: Analyzing Ionization Energies

Beryllium has a filled 2s subshell, making it more stable and requiring more energy to remove an electron from it. Boron, on the other hand, has its outermost electron in a higher energy 2p subshell, making it easier to remove. Therefore, Be indeed has greater first ionization energy than B.

Step 3: Understanding the Reason

The reason states that the 2p orbital is lower in energy than the 2s orbital. This statement is incorrect because the 2s orbital is actually lower in energy than the 2p orbital, not vice versa.

Step 4: Evaluating the Connection

Although the assertion is correct, the reason given does not explain the assertion correctly, as the energy comparison given between 2p and 2s is false.

Step 5: Conclusion

Since the assertion is true, but the reason provided is false, the correct choice is (c) A is true but R is false.

Key Concepts

Beryllium

Beryllium (Be) is an element with the atomic number 4, located in Group 2 of the periodic table, known as the alkaline earth metals. One of its key characteristics is its high ionization energy compared to elements with similar atomic sizes. This is due to its electronic configuration of

• 1s² 2s², which indicates a full 2s orbital.

This stable electronic structure leads to a strong attraction between the nucleus and the 2s electrons, making it challenging to remove these electrons. Therefore, Beryllium requires more energy to ionize, giving it a higher ionization energy.
Understanding Beryllium's ionization helps explain why it behaves differently when compared to other elements such as Boron, that are nearby on the periodic table.

Boron

Boron (B), with the atomic number 5, sits in Group 13 of the periodic table. Its configuration is

• 1s² 2s² 2p¹.

The presence of an electron in a 2p orbital, which is higher in energy compared to the 2s, gives Boron unique chemical properties. This means that its outer electron can be removed more easily than those in Beryllium, as it experiences a lower effective nuclear charge. As a result, Boron's first ionization energy is lower than that of Beryllium.

Recognizing the position of the outermost electron in Boron helps to understand the differences in their chemical behavior and the energies required to ionize these elements.

2s and 2p orbitals

The 2s and 2p orbitals are part of the second shell of electrons in atoms. They each have unique energy levels and shapes.

• The 2s orbital is spherical and lower in energy.
• The 2p orbitals (three in total, oriented orthogonally) have a dumbbell shape and are higher in energy.

When electrons fill these orbitals, they generally fill the 2s before the 2p. This energy difference in the orbitals is critical in understanding why elements have different ionization energies.
In the case of the exercise about Beryllium and Boron, the 2s electrons are more tightly bound to the nucleus due to their slightly lower energy level and spherical symmetry, leading to higher ionization energy in Beryllium compared to Boron. In contrast, Boron's 2p electron is less firmly held, resulting in lower ionization energy. This understanding is essential for analyzing and predicting the chemical properties of elements.