Problem 91
Question
The first ionization potential of \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}\) and \(\mathrm{Si}\) are in the order: (a) \(\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}\) (b) \(\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}\) (c) \(\mathrm{Na}<\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}\) (d) \(\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}\)
Step-by-Step Solution
Verified Answer
The correct order is option (d): \(\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}\).
1Step 1: Understanding Ionization Potential
The ionization potential (or ionization energy) is the energy required to remove an electron from a neutral atom in the gaseous phase. Generally, ionization energy increases across a period from left to right due to increasing nuclear charge, making it more difficult to remove an electron.
2Step 2: Analyze Periodic Trends
Sodium (Na), Magnesium (Mg), Aluminum (Al), and Silicon (Si) all belong to the same period (Period 3) of the periodic table. According to periodic trends, we expect the ionization energy to increase from Na to Si. Therefore, typical order without exceptions would be Na < Mg < Al < Si.
3Step 3: Consider Electron Configuration Exceptions
The electron configuration of each element can cause variations in expected trends. Mg's electron configuration is [Ne]3s², which is a full subshell and more stable than Al's [Ne]3s²3p¹, making Mg's ionization energy higher than Al's despite Al being further to the right.
4Step 4: Examine Options
Based on stability due to filled sublevels (Mg) being higher than partially filled sublevels (Al), the order should reflect Mg > Al. The potential sequence considering ionic energies and the exceptions should be: Na < Mg > Al < Si. This aligns with option (d).
Key Concepts
Periodic Trends in Ionization PotentialElectron Configuration and Its ImpactNavigating Periodic Table Trends
Periodic Trends in Ionization Potential
Ionization potential, also known as ionization energy, is significantly influenced by periodic trends observed in the periodic table. This energy refers to the amount needed to remove an electron from an atom in its gaseous state. As a general trend, ionization energy increases as we move from left to right across a period. This is because the nuclear charge (or the positive charge of the nucleus) increases, leading to a stronger attraction between the nucleus and the electrons.
As this attraction strengthens, it becomes increasingly harder to remove an electron, thus elevating the ionization energy. For elements within the same period, such as sodium (Na), magnesium (Mg), aluminum (Al), and silicon (Si), this trend can be noted: ionization potential typically increases from Na to Si. However, exceptions due to electron configuration can occur, requiring a closer inspection to understand the nuances.
As this attraction strengthens, it becomes increasingly harder to remove an electron, thus elevating the ionization energy. For elements within the same period, such as sodium (Na), magnesium (Mg), aluminum (Al), and silicon (Si), this trend can be noted: ionization potential typically increases from Na to Si. However, exceptions due to electron configuration can occur, requiring a closer inspection to understand the nuances.
Electron Configuration and Its Impact
The arrangement of electrons within an atom, known as electron configuration, plays a crucial role in determining ionization potential. For instance, magnesium (Mg) has the electron configuration of [Ne]3s², which is a filled subshell. This configuration imparts stability, making it require more energy to remove an electron compared to an atom with a less stable configuration.
On the other hand, aluminum (Al) has an electron configuration of [Ne]3s²3p¹. The presence of an extra electron in the 3p sublevel makes Al less stable compared to Mg, resulting in a lower energy requirement to remove this electron. This is why Mg, despite being to the left of Al on the periodic table, has a higher ionization energy than Al. Understanding these configurations helps clarify why ionization trends might not strictly follow the expected trend across a period.
On the other hand, aluminum (Al) has an electron configuration of [Ne]3s²3p¹. The presence of an extra electron in the 3p sublevel makes Al less stable compared to Mg, resulting in a lower energy requirement to remove this electron. This is why Mg, despite being to the left of Al on the periodic table, has a higher ionization energy than Al. Understanding these configurations helps clarify why ionization trends might not strictly follow the expected trend across a period.
Navigating Periodic Table Trends
The periodic table is more than just a collection of elements; it offers crucial insights into elemental properties, such as ionization potential. Elements in the same period often show a gradual increase in ionization energy as one moves rightward due to increased nuclear charge. However, subtleties within this trend, like the stability associated with filled or half-filled subshells (as seen with Mg versus Al), can cause deviations.
Therefore, effectively navigating the periodic table involves recognizing these trends, but also understanding when and why exceptions occur. For elements like Na, Mg, Al, and Si, knowing their placement and considering electron configurations provides a full picture of why their ionization potentials are ordered as Na < Mg > Al < Si, rather than strictly increasing. This comprehensive approach allows students to grasp how periodic trends and electron configuration intricately interact.
Therefore, effectively navigating the periodic table involves recognizing these trends, but also understanding when and why exceptions occur. For elements like Na, Mg, Al, and Si, knowing their placement and considering electron configurations provides a full picture of why their ionization potentials are ordered as Na < Mg > Al < Si, rather than strictly increasing. This comprehensive approach allows students to grasp how periodic trends and electron configuration intricately interact.
Other exercises in this chapter
Problem 89
Which of the following has the maximum number of unpaired electrons? (a) \(\mathrm{Mg}^{2+}\) (b) \(\mathrm{Ti}^{3+}\) (c) \(\mathrm{V}^{3+}\) (d) \(\mathrm{Fe}
View solution Problem 90
The statement that is not correct for periodic classification of elements is (a) the properties of elements are a periodic function of their atomic numbers. (b)
View solution Problem 92
The electronegativity of the following elements increases in the order: (a) Si, P, C, N (b) \(\mathrm{N}, \mathrm{Si}, \mathrm{C}, \mathrm{P}\) (c) \(\mathrm{P}
View solution Problem 93
Atomic radii of fluorine and neon in Angstorm units are given respectively by (a) \(1.60,1.60\) (b) \(0.72,0.72\) (c) \(0.72,1.60\) (d) none of these
View solution