Question

Assertion (A): First ionization energy of beryllium is greater than that of boron. Reason (R): Boron has larger size than beryllium. (a) Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is not the correct explanation of \(A\). (c) \(\mathrm{A}\) is true, \(\mathrm{R}\) is false (d) \(\mathrm{A}\) is false, \(\mathrm{R}\) is true

Short answer

Answer: (b) Both A and R are true, and R is not the correct explanation of A. Explanation: Although both Assertion (A) and Reason (R) are true statements, the fact that boron has a larger size than beryllium does not account for the greater ionization energy in beryllium compared to boron. The correct explanation for Assertion (A) is related to the electron shielding by the filled 2s orbital in B and the effective nuclear charge experienced by the 2p electron in B.

Step-by-step solution

Understand ionization energy

Ionization energy is the energy required to remove the most loosely bound electron from a neutral atom in its gaseous state. The lesser the energy required to remove an electron, the lower will be the ionization energy. There are several factors that affect ionization energy, including nuclear charge, atomic size, and electron shielding.

Analyze the ionization energy trend

Across a period in the periodic table, the ionization energy generally increases from left to right. This is due to the increase in the effective nuclear charge experienced by the valence electrons, which pulls them more tightly towards the nucleus and makes it harder to remove these electrons. However, there are some exceptions to this trend, and one of them is the case of beryllium and boron.

Compare the first ionization energies of beryllium and boron

Beryllium (Be) has an atomic number of 4 and its electronic configuration is \(1s^2\ 2s^2\). Boron (B) has an atomic number of 5 and its electronic configuration is \(1s^2\ 2s^2\ 2p^1\). To remove an electron from Be, we need to remove one of the 2s electrons. On the other hand, removing an electron from B would involve removing the single 2p electron. Since the 2p electron in B is located farther from the nucleus than the 2s electron in Be, it might be expected that B will have a lower ionization energy than Be. However, electron shielding by the filled 2s orbital in B is less effective than the shielding in Be. As a result, the effective nuclear charge experienced by the 2p electron in B is greater than that experienced by the 2s electron in Be, making it harder to remove the 2p electron in B. Therefore, the Assertion (A) is true.

Determine the validity of Reason (R)

The Reason (R) given is that "Boron has a larger size than beryllium". This statement is true as atomic size increases while going down a group and decreases while going from left to right in a period. However, this fact alone does not account for the greater ionization energy in beryllium compared to boron, as already discussed in Step 3.

Relate Assertion (A) and Reason (R) and choose the correct option

From the analysis above, we can conclude that both Assertion (A) and Reason (R) are true statements. However, the Reason (R) is not the correct explanation for Assertion (A), as the discussion in Step 3 demonstrates. Therefore, the correct answer is (b) Both A and R are true, and R is not the correct explanation of A.

Key concepts

Periodic Table Trends

When looking at the periodic table, trends help us understand how elements behave. One crucial trend is ionization energy, which typically increases as you move across a period from left to right. As elements across a period gain protons, their nuclei have greater nuclear charges. This causes stronger attractions between the nucleus and the valence electrons, which makes electrons harder to remove.
However, there are exceptions to this trend. For instance, despite appearing later in the period, boron has a lower first ionization energy than beryllium. This is because electron configurations and subshell energies also play significant roles in ionization energy.
Remember, trends are general rules but should be considered alongside specific atomic structure details.

Atomic Structure

The atomic structure of an element provides essential insights into its chemical properties and reactivity. Every atom consists of a nucleus containing protons and neutrons, with electrons orbiting this nucleus in various energy levels and subshells.
For example, beryllium, with an atomic number of 4, has the electronic configuration of \(1s^2\ 2s^2\). Meanwhile, boron, with an atomic number of 5, carries the configuration \(1s^2\ 2s^2\ 2p^1\). Understanding the atomic structure allows us to predict reactivity and properties like ionization energy.
In this case, removing an electron from beryllium involves its full 2s subshell, while boron's ionization involves its single 2p electron. Knowing these structures helps explain why boron, contrary to periodic trends, has lower ionization energy than beryllium.

Electron Configuration

Electron configuration describes the distribution of electrons in an atom's energy levels and orbitals. It determines many properties of an atom, including its ionization energy.
Beryllium's electron configuration consists of fully filled \(1s^2\ 2s^2\), making it quite stable, as filled subshells provide more stability. Boron, with its \(1s^2\ 2s^2\ 2p^1\) configuration must remove an electron from a higher energy and less stable 2p subshell.
This difference in electron configuration is essential for understanding the ionization energies of these elements. Removing an electron from a less stable subshell, such as 2p in boron, requires less energy compared to removing one from a more stable subshell like 2s in beryllium.

Effective Nuclear Charge

Effective nuclear charge ( Z_eff) is a concept used to describe the net positive charge experienced by an electron in a multi-electron atom. It affects how tightly an electron is held by the nucleus.

• As you move across a period, the effective nuclear charge generally increases as more protons are added to the nucleus, enhancing the attraction felt by electrons and increasing ionization energy.
• In our comparison, boron appears to have a more positive effective nuclear charge than beryllium when considering the outer electron. This is unexpected, as boron’s ionization energy is lower.

This confusion is clarified by considering electron shielding, particularly by filled subshells. The 2p subshell in boron is less shielded and less bonded compared to the 2s subshell in beryllium, despite boron's higher effective nuclear charge. This makes boron's outer electron easier to remove.