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Chapter 3: Problem 13

The first ionization energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) of \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}\) and \(\mathrm{Si}\) respectively, are: (a) \(496,737,577,786\) (b) \(496,577,737,786\) (c) \(786,737,577,496\) (d) \(496,577,786,737\)

Short Answer

Expert verified
Option (b) \(496, 577, 737, 786\) is the correct order of increasing ionization energy for Na, Mg, Al, and Si.

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove the outermost electron from a neutral atom in its gaseous state. Generally, ionization energy increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table.
02

Locate Elements in the Periodic Table

The elements given are \(\text{Na (Sodium)}, \text{Mg (Magnesium)}, \text{Al (Aluminum)},\) and \(\text{Si (Silicon)}\). All of them belong to the same period (Period 3) of the periodic table. The order from left to right is: \(\text{Na}, \text{Mg}, \text{Al}, \text{Si}\).
03

Determine General Ionization Energy Trend

Within a given period, ionization energy increases across the period from left to right. Thus we expect the order of increasing ionization energy to be: \(\text{Na} < \text{Mg} < \text{Al} < \text{Si}\).
04

Match Data with Trends

According to the trend, the increasing order should be \(496, 577, 737,\) and \(786\). Comparing this to the given options: (a) \(496, 737, 577, 786\) (b) \(496, 577, 737, 786\) (c) \(786, 737, 577, 496\) (d) \(496, 577, 786, 737\), option (b) matches the expected trend.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Table Trends
Understanding the periodic table trends is essential for predicting various properties of elements, one of which is ionization energy. Ionization energy generally increases as we move across a period from left to right. This is because elements on the right side of the periodic table have more protons, leading to a greater effective nuclear charge. This increase in positive charge means that electrons are held more tightly, which requires more energy to remove one.
On the other hand, ionization energy tends to decrease as we move down a group. As you go down a group, the outermost electron is further from the nucleus due to added energy levels or shells. This increased distance reduces the nuclear attraction compared to elements higher in the group, causing less energy to be needed to remove an electron.
These trends are vital for predicting how an element will react, its bonding tendencies, and its general chemical behavior.
Periodic Table Groups
Groups in the periodic table are columns that share similar chemical properties. Each group contains elements that have the same number of electrons in their outermost shell, also known as valence electrons. This similarity in electron configuration means that elements in the same group typically show similar reactivity and other chemical behaviors.
For example, Group 1 contains alkali metals like sodium (\(\text{Na}\)) which are very reactive, especially with water, due to their single valence electron. Groups can help predict the behavior of elements; elements within the same group often form similar compounds.
It's also crucial to note that as you move down a group, the elements increase in atomic size, which influences properties such as ionization energy, because the outer electrons are held less tightly by the nucleus.
Periodic Table Periods
Periods are rows on the periodic table and represent elements organized by increasing atomic number. As you move across a period from left to right, the characteristics of elements often change significantly. One major trend is the increasing ionization energy. This occurs because as protons are added to the nucleus, electrons are held more tightly.
The periodic table is structured such that each period corresponds to the filling of a different electron shell. For instance, in Period 3, where elements like sodium (\(\text{Na}\)), magnesium (\(\text{Mg}\)), aluminum (\(\text{Al}\)), and silicon (\(\text{Si}\)) are located, electrons are added to the third shell. This electron addition results in various changes in atomic properties across the period.
Understanding periods can help one predict how elements react with one another and how their properties change systematically across the table.

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Most popular questions from this chapter

The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are :(a) 16,5 and 2 (b) 15,5 and 3 (c) 16,6 and 3 (d) 15,6 and 2

B has a smaller first ionization enthalpy than Be. Consider the following statements: (I) it is easier to remove \(2 p\) electron than \(2 s\) electron (II) \(2 p\) electron of \(\mathrm{B}\) is more shielded from the nucleus by the inner core of electrons than the \(2 s\) electrons of \(\mathrm{Be}\) (III) \(2 s\) electron has more penetration power than \(2 p\) electron (IV) atomic radius of \(\mathrm{B}\) is more than \(\mathrm{Be}\) (atomic number \(\mathrm{B}=5, \mathrm{Be}=4\) ) The correct statements are: (a) (I), (II) and (IV) (b) (II), (III) and (IV) (c) (I), (II) and (III) (d) (I), (III) and (IV)

The \(1^{\text {st }}, 2^{\text {nd }}\), and the \(3^{\text {rd }}\) ionization enthalpies, \(I_{1}, I_{2}\), and \(I_{3}\), of four atoms with atomic numbers \(n, n+1, n+2\), and \(n+3\), where \(n<10\), are tabulated below. What is the value of \(n ?\) $$ \begin{array}{|c|c|c|c|} \hline {\begin{array}{c} \text { Atomic } \\ \text { number } \end{array}} & & {\text { Ionization Enthalpy }(\mathrm{kJ} / \mathrm{mol})} \\\ { 2 - 4 } & I_{1} & I_{,} & I_{3} \\ \hline n & 1681 & 3374 & 6050 \\ \hline n+1 & 2081 & 3952 & 6122 \\ \hline n+2 & 496 & 4562 & 6910 \\ \hline n+3 & 738 & 1451 & 7733 \\ \hline \end{array} $$

The following statements concern elements in the periodic table. Which of the following is true? [Main Online April 10, 2016] (a) For Group 15 elements, the stability of \(+5\) oxidation state increases down the group (b) Elements of Group 16 have lower ionization enthalpy values compared to those of Group 15 in the corresponding periods. (c) The Group 13 elements are all metals (d) All the elements in Group 17 are gases.

Amongst \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) and \(\mathrm{H}_{2} \mathrm{Te}\), the one with the highest boiling point is [2000S] (a) \(\mathrm{H}_{2} \mathrm{O}\) because of hydrogen bonding (b) \(\mathrm{H}_{2}\) Te because of higher molecular weight (c) \(\mathrm{H}_{2} \mathrm{~S}\) because of hydrogen bonding (d) \(\mathrm{H}_{2} \mathrm{Se}\) because of lower molecular weight
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