Problem 65 - A Complete Resource Book in Chemistry for JEE Main

Question

Sum of first three ionization energies of \(\mathrm{Al}\) is \(53.0 \mathrm{eV}\) atom \(^{-1}\) and the sum of first two ionization energies of Na is \(52.2 \mathrm{eV}\) atom \(^{-1}\). Out of Al (III) and Na (II) (a) \(\mathrm{Al}\) (III) is more stable than \(\mathrm{Na}\) (II) (b) \(\mathrm{Na}\) (II) is more stable than Al (III) (c) Both are equally unstable (d) Both are equally stable

Step-by-Step Solution

Answer

(a) Al (III) is more stable than Na (II).

Step 1: Understand Ionization Energy

The ionization energy is the amount of energy required to remove an electron from a neutral atom in its gaseous phase. Higher ionization energy indicates a more stable cation because it requires more energy to remove an electron.

Step 2: Analyze Given Information

We have the sum of the first three ionization energies for Aluminum (Al) as 53.0 eV and the sum of the first two for Sodium (Na) is 52.2 eV. Aluminum's three ionization energies mean it transitions to a +3 charge state, and Sodium's two ionization energies mean it transitions to a +2 charge state.

Step 3: Compare Ionization Energies for Stability

A higher total ionization energy typically implies greater stability because more energy must be supplied to further ionize, thus indicating the charge state is more stable. Aluminum (III) with a total of 53.0 eV and Na (II) with 52.2 eV are to be compared.

Step 4: Conclusion from Ionization Energies

Since the total ionization energy for Aluminum (III) is higher than that of Sodium (II), Aluminum (III) requires more energy to lose more electrons compared to Sodium (II). Therefore, Aluminum (III) is more stable than Sodium (II).

Key Concepts

Aluminum Ionization

Aluminum (Al) is a metallic element with the atomic number 13. Ionization of an atom refers to the process of removing electrons to form cations or positively charged ions. For aluminum, the ionization process involves the removal of electrons to eventually form an Al³⁺ ion. In the case of aluminum, to remove electrons and achieve this

• First Ionization: An electron is removed from the outer 3p orbital.
• Second Ionization: An electron is removed from the now outer 3s orbital.
• Third Ionization: Another electron is removed from the 3s orbital.

The high amount of energy - 53.0 eV - needed for all three ionizations indicates that aluminum becomes increasingly stable as electrons are removed. This stability is due to the increasing electropositive charge as it moves from Al to Al³⁺, signaling a strong attraction between the cation and electrons that remain.

Sodium Ionization

Sodium (Na) is another metallic element with a lower atomic number of 11. When sodium undergoes ionization, it is generally focused on achieving a +1 charge state by removing a single electron. However, in this context, we are considering the energy needed to form an Na²⁺ ion. Here's how sodium ionization occurs:

• First Ionization: An electron is removed from the outer 3s orbital, forming Na⁺.
• Second Ionization: Another electron from the 2p orbital is removed, forming Na²⁺.

The sum of energies required for these steps is 52.2 eV. Although sodium starts off with less ionization energy needed for initial electron removal, additional electrons are increasingly more difficult to remove. This causes the energy to cumulatively approach aluminum's. Yet, Na⁺ remains more common and stable than Na²⁺ because of its typical role in chemical contexts.

Cation Stability

Cation stability is a key factor in interpreting ionization energies. It reflects how favorably a cation can exist without losing further electrons. Typically, a more positive charge on a cation equates to greater electron attraction. This reduces the likelihood of additional ionization.
For Aluminum (III), its +3 state contributes to its stability. The strong electropositive charge allows effective electron density retention in the surrounding orbitals.

• Very stable because of the efficient energy distribution around the Al³⁺ ion.
• Completes sublevels more efficiently, reducing potential energy.

For Sodium (II), although it has a relatively high ionization energy, a +2 state interrupts the more energetically favorable configurations.

• Sodium tends to form Na⁺, and deviation from this often lowers stability.
• Less electron shielding compared to Al³⁺, making Na²⁺ less stable.

Ionization Potential Comparison

Comparing ionization potentials directly indicates which element is more stable in its ionized form. More ionization energy implies the electrons are more tightly bound to the nucleus, generally signifying increased stability.
For aluminum and sodium:

• Aluminum’s total ionization energy (53.0 eV for Al³⁺) is slightly higher than sodium’s (52.2 eV for Na²⁺).
• This small difference evidences that Al³⁺ is more stable than Na²⁺.
• Aluminium achieves a noble gas configuration, achieving a minimized energy state.

Therefore, based on ionization energy, Al³⁺ is more energetically stable. Sodium, while less tightly bound at this ionization level, is typically found stable in other ionic forms. Summarily, Al requires more energy to lose another electron than Na does, confirming aluminum's relative cationic stability.