Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

The correct statement among the following is (a) the first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\) (b) the second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of \(\mathrm{Na}\) (c) the first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\) (d) the third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of Al

Step-by-Step Solution

Answer

The correct statement is (a).

Step 1: Understand Ionization Potential

Ionization potential (IP) is the energy required to remove an electron from an atom in the gaseous state. The first ionization potential refers to removing the first electron, the second to the second electron, and so on. Factors affecting IP include atomic size and nuclear charge.

Step 2: Review First Ionization Potential

Compare the first ionization potentials of Al and Mg, and Na and Mg. Generally, ionization potential decreases down a group and increases across a period. Mg, being to the left of Al and Na in period 3, will generally have a higher IP than both Al and Na.

Step 3: Examine Second Ionization Potential

Consider the second ionization potentials of Mg and Na. The second IP of any element is higher than the first because it involves removing an electron from a positively charged ion. Compare the second IPs by considering the stability after removal of the first electron.

Step 4: Review Third Ionization Potential

Compare the third ionization potentials of Mg and Al. For Mg, removing the third electron means removing from a complete filled energy level, which costs more energy. For Al, the third electron removal leads to a stable filled-state, requiring less energy.

Step 5: Evaluate Statements

(a) True, as Al has increased electron shielding. (b) True, since Mg has a full outer electron shell that provides more stability and requires more energy to ionize. (c) True, because Na loses its single valence electron more easily than the more stable Mg configuration. (d) False, as Mg having a filled shell makes it difficult to remove the third electron compared to Al.

Key Concepts

First Ionization Potential

The first ionization potential refers to the energy required to remove the outermost electron from a neutral atom in its gaseous state. This process creates a positive ion. Several factors influence the magnitude of the first ionization potential:

• Atomic Size: Atoms with larger radii have outer electrons located further from the nucleus and are therefore easier to remove. This results in a lower ionization potential.
• Nuclear Charge: A higher positive charge in the nucleus attracts electrons more strongly, increasing the ionization potential.

For instance, when comparing aluminum (Al) with magnesium (Mg), we notice that Mg has a higher first ionization potential than Al due to its smaller size and greater effective nuclear charge. Similarly, between sodium (Na) and Mg, magnesium again has a higher first ionization potential for similar reasons. This is because ionization potential generally increases across a period in the periodic table as the nuclear charge increases and the atomic radius decreases.

Second Ionization Potential

The second ionization potential is the energy needed to remove a second electron after the first one has already been removed, creating a doubly positive ion. This potential is always higher than the first ionization, because the electron is being removed from a positively charged ion, which has a stronger electrostatic attraction to the electrons left. When comparing the second ionization potential of magnesium (Mg) with that of sodium (Na), Mg indeed has a higher potential. After losing one electron, Na achieves a stable noble gas configuration with a full outer shell, making the removal of a second electron quite energy-intensive. On the other hand, Mg loses its valence electron and attempts to reach the same stable state, thereby requiring more energy than Na to remove its second electron.

Third Ionization Potential

The third ionization potential involves the removal of yet another electron, making it even higher in magnitude because it involves disrupting an atom that is already doubly ionized. When comparing magnesium (Mg) and aluminum (Al), the situation becomes more nuanced. For Mg, removing a third electron means taking an electron from a filled inner shell, which is energetically costly. Conversely, with Al, the removal of the third electron results in a stable arrangement, having removed its outer p-electron initially. Thus, the third ionization potential of Mg is lower than that of Al, because Mg's configuration after the third electron removal is energetically unfavorable compared to Al’s resulting stability.