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 (d) \(\text{Na}<\text{Mg}>\text{Al}<\text{Si}\).
1Step 1: Understanding Ionization Potential
The ionization potential, also known as ionization energy, is the energy required to remove the outermost electron from a gaseous atom. Generally, ionization energy increases across a period from left to right on the periodic table due to increasing nuclear charge, which attracts the electrons more strongly.
2Step 2: Identify Element Positions and Trends
Sodium (Na), Magnesium (Mg), Aluminum (Al), and Silicon (Si) are elements in the same period (Period 3) of the periodic table. Ionization potential typically increases across the period from Na to Si. However, there are exceptions when comparing Mg and Al, due to electronic configuration differences.
3Step 3: Analyze Magnesium and Aluminum Exception
Magnesium has a complete 3s subshell ( ext{[Ne]}3s^2), whereas Aluminum starts filling the 3p subshell ( ext{[Ne]}3s^2 3p^1). The ionization energy of Mg is higher than Al because it is more stable with a filled 3s subshell, while Al has only one electron in the 3p orbital.
4Step 4: Apply Knowledge to Given Orders
Considering the trend and the Mg-Al exception: 1. \( ext{Na} < ext{Mg}\) because of the usual increase in ionization energy across the period.2. \( ext{Mg} > ext{Al}\) due to the electronic configuration exception.3. \( ext{Al} < ext{Si}\) as the trend of increasing ionization energy resumes.
5Step 5: Determine the Correct Order
Based on the analysis, the correct order based on known ionization energy values is \( ext{Na} < ext{Mg} > ext{Al} < ext{Si}\). This matches option (d): \( ext{Na}< ext{Mg}> ext{Al}> ext{Si}\).
Key Concepts
Periodic TrendsElectronic ConfigurationIonization Energy ExceptionsPeriodic Table
Periodic Trends
Periodic trends refer to the patterns observed in the periodic table as we move across periods or down groups. One of the important trends is the ionization energy which typically increases across a period from left to right. This happens because as the atomic number increases, the number of protons in the nucleus also increases. This results in a greater nuclear charge, which pulls the electrons closer to the nucleus and makes it more difficult to remove an electron, thus increasing the ionization energy.
Across a period, the added electrons fill the same valence shell and don't shield each other very effectively. Therefore, the effective nuclear charge felt by the outermost electrons increases, causing them to be held more tightly. As a result, the ionization energy tends to rise from the left side of a period to the right.
Understanding these trends is crucial for predicting the behavior of elements and their reactions.
Across a period, the added electrons fill the same valence shell and don't shield each other very effectively. Therefore, the effective nuclear charge felt by the outermost electrons increases, causing them to be held more tightly. As a result, the ionization energy tends to rise from the left side of a period to the right.
Understanding these trends is crucial for predicting the behavior of elements and their reactions.
Electronic Configuration
Electronic configuration describes the distribution of electrons in an atom's orbitals and is vital in explaining trends in ionization energy. Elements in the same period will experience an increase in ionization energy due to their increasing effective nuclear charge.
For instance, consider sodium ([ext{Na]}) with an electronic configuration of [ext{[Ne]}3s^1]. As you move to magnesium ([ext{Mg]}) with configuration [ext{[Ne]}3s^2], the increase in nuclear charge results in a higher ionization energy. However, although aluminum ([ext{Al]}) with configuration [ext{[Ne]}3s^2 3p^1] follows magnesium, its ionization energy is lower. This is due to the additional 3p electron, which is slightly shielded by the filled 3s subshell, making it easier to remove.
The electronic configuration provides a deeper insight into the exceptions observed in periodic trends and helps in predicting the chemical behavior of various elements.
For instance, consider sodium ([ext{Na]}) with an electronic configuration of [ext{[Ne]}3s^1]. As you move to magnesium ([ext{Mg]}) with configuration [ext{[Ne]}3s^2], the increase in nuclear charge results in a higher ionization energy. However, although aluminum ([ext{Al]}) with configuration [ext{[Ne]}3s^2 3p^1] follows magnesium, its ionization energy is lower. This is due to the additional 3p electron, which is slightly shielded by the filled 3s subshell, making it easier to remove.
The electronic configuration provides a deeper insight into the exceptions observed in periodic trends and helps in predicting the chemical behavior of various elements.
Ionization Energy Exceptions
While periodic trends give a general idea of how ionization energy should behave, exceptions exist due to subtle differences in electronic configuration. One prominent exception involves magnesium ([ext{Mg]}) and aluminum ([ext{Al]}).
Magnesium, with its electronic configuration, ends in a filled 3s subshell, [ext{[Ne]}3s^2], providing extra stability. Aluminum's configuration ends in [ext{[Ne]}3s^2 3p^1], where the single 3p electron is less securely held due to the absence of a completely filled subshell.
This difference in electronic structure leads to higher ionization energy for magnesium even though aluminum is further across the period. It highlights how the electron configuration at certain positions can dramatically affect ionization energies, deviating from expected trends.
Magnesium, with its electronic configuration, ends in a filled 3s subshell, [ext{[Ne]}3s^2], providing extra stability. Aluminum's configuration ends in [ext{[Ne]}3s^2 3p^1], where the single 3p electron is less securely held due to the absence of a completely filled subshell.
This difference in electronic structure leads to higher ionization energy for magnesium even though aluminum is further across the period. It highlights how the electron configuration at certain positions can dramatically affect ionization energies, deviating from expected trends.
Periodic Table
The periodic table is an organizational chart that helps us understand and predict the chemical properties of elements, including ionization energy. Elements are arranged in order of increasing atomic number and are placed into periods and groups based on similarities in their electronic configurations.
Within the periodic table, elements progressively acquire additional protons and electrons as you move from left to right across a period. This addition leads to greater nuclear charge and usually an increase in ionization energy.
However, recognizing exceptions like those between magnesium and aluminum, provides valuable insights into the nature of chemical bonding and the factors which contribute to an element's reactivity. Thus, the periodic table is not just a simple list, but a tool for understanding the subtle workings of atoms and their interactions.
Within the periodic table, elements progressively acquire additional protons and electrons as you move from left to right across a period. This addition leads to greater nuclear charge and usually an increase in ionization energy.
However, recognizing exceptions like those between magnesium and aluminum, provides valuable insights into the nature of chemical bonding and the factors which contribute to an element's reactivity. Thus, the periodic table is not just a simple list, but a tool for understanding the subtle workings of atoms and their interactions.
Other exercises in this chapter
Problem 89
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}\
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