/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Based on their positions in the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Based on their positions in the periodic table, predict which atom of the following pairs will have the smaller first ionization energy\(:(\mathbf{a}) \mathrm{Cl}, \mathrm{Ar} ;(\mathbf{b}) \mathrm{Be}, \mathrm{Ca} ;(\mathbf{c}) \mathrm{K}, \mathrm{Co} ;(\mathbf{d}) \mathrm{S}, \mathrm{Ge} ;(\mathbf{e}) \mathrm{Sn}, \mathrm{Te}.\)

Short Answer

Expert verified
In summary, based on their positions in the periodic table, the atoms with smaller first ionization energies for each given pair are: (a) Cl (chlorine) (b) Ca (calcium) (c) K (potassium) (d) Ge (germanium) (e) Sn (tin)

Step by step solution

01

Pair (a): Cl and Ar

In the periodic table, Cl (chlorine) is in Group 17 and Period 3, while Ar (argon) is in Group 18 and Period 3. Since they are in the same period, we should focus on the group difference. As we move from left to right across a period, the atomic number increases, so the ionization energy increases. Thus, Cl will have a smaller first ionization energy compared to Ar.
02

Pair (b): Be and Ca

In the periodic table, Be (beryllium) is in Group 2 and Period 2, while Ca (calcium) is in Group 2 and Period 4. Since they are in the same group, we should focus on the period difference. As we move down a group, the atomic radius increases due to the addition of electron shells, so the ionization energy decreases. Thus, Ca will have a smaller first ionization energy compared to Be.
03

Pair (c): K and Co

In the periodic table, K (potassium) is in Group 1 and Period 4, while Co (cobalt) is in Group 9 and Period 4. Since they are in the same period, we should focus on the group difference. As we move from left to right across a period, the atomic number increases, so the ionization energy increases. Thus, K will have a smaller first ionization energy compared to Co.
04

Pair (d): S and Ge

In the periodic table, S (sulfur) is in Group 16 and Period 3, while Ge (germanium) is in Group 14 and Period 4. In this case, there are differences in both group and period. Generally, the ionization energy trend is more dominating in the period direction. Therefore, as we move down a group, the atomic radius increases due to the addition of electron shells, so the ionization energy decreases. Thus, Ge will have a smaller first ionization energy compared to S.
05

Pair (e): Sn and Te

In the periodic table, Sn (tin) is in Group 14 and Period 5, while Te (tellurium) is in Group 16 and Period 5. Since they are in the same period, we should focus on the group difference. As we move from left to right across a period, the atomic number increases, so the ionization energy increases. Thus, Sn will have a smaller first ionization energy compared to Te.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Periodic Table Trends
Understanding periodic table trends is essential for predicting the behavior of elements, such as their ionization energy. The periodic table is arranged in such a way that it reflects the periodic recurrence of chemical properties. Ionization energy tends to increase as you move from left to right across a period, due to a greater nuclear charge attracting the electrons more strongly. This makes electrons more difficult to remove. Conversely, as you move down a group, ionization energy generally decreases. This is because electrons are further from the nucleus due to the larger atomic radius and more electron shells, which makes them easier to remove.

Another trend is atomic radius, which decreases from left to right across a period and increases down a group. This trend is important because it affects ionization energy. A smaller atomic radius means electrons are closer to the attractive force of the nucleus, resulting in higher ionization energy. These periodic trends explain why ionization energies vary among elements and are key when comparing first ionization energies across the periodic table.
Atomic Radius
The atomic radius is a term used to describe the size of an atom. It's not a fixed value but rather an average distance from the center of the nucleus to the outer boundary of the electron cloud. Atomic radius decreases from left to right within a period because, although the number of electron shells remains the same, each added proton in the nucleus creates a stronger pull on the electron cloud, drawing it closer. This increased nuclear charge without an increase in electron shielding results in a smaller radius.

Down a group, the atomic radius increases due to the addition of more electron shells, which outweighs the increase in nuclear charge. As these outer electrons are further from the nucleus, they are less tightly bound and hence, increase the size of the atom. The atomic radius is a fundamental property that influences other properties such as ionization energy and electronegativity.
Electron Shells
Electron shells, also known as energy levels, are the orbits surrounding an atom's nucleus that contain the electrons. Each shell can only hold a certain number of electrons, and they fill up in a specific order: closer shells fill before farther ones. The importance of electron shells in ionization energy becomes clear when you consider that adding shells increases the distance between the outermost electrons and the nucleus.

When an atom has more shells, the electrons feel less pull from the nucleus due to the shielding effect from inner-shell electrons. This makes it easier to remove an outer electron—hence, the atom has a lower first ionization energy. Understanding how electron shells fill is crucial for predicting the chemical reactivity and bonding behavior of an element, as well as its ionization energy.
Ionization Energy Comparison
Comparing ionization energies of different elements is vital in understanding their reactivity. First ionization energy refers to the energy required to remove the outermost electron from a neutral atom. When comparing elements, we often see that those closer to the top right of the periodic table (excluding noble gases) have higher ionization energies because they have a smaller radius and greater nuclear charge, thus holding onto their electrons more tightly.

For elements in the same period, ionization energy generally increases as you move from left to right. For elements in the same group, ionization energy decreases as you move down the group. By applying these trends, students can predict which element in a given pair will have a lower or higher first ionization energy, as seen in the step-by-step exercise provided. This comparison of ionization energies helps us to infer the strength of an atom's bond with its electrons and has implications in its chemical behavior and bonding ability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the relationship between the ionization energy of an anion with a \(1-\) charge such as \(F^{-}\) and the electron affinity of the neutral atom, \(\mathrm{F}\) ?

[7.113]When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of MgO and magnesium nitride after burning is 0.470 g. Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is 0.486 g of MgO. What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a \(6.3-\mathrm{g}\) Mg ribbon reacts with 2.57 \(\mathrm{g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{kJ} / \mathrm{mol} .\) Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

Which neutral atom is isoelectronic with each of the following ions? \(\mathrm{Ga} ^{3+}, \mathrm{Zr}^{4+}, \mathrm{Mn}^{7+}, \mathrm{I}^{-}, \mathrm{Pb}^{2+}.\)

Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury’s 4\(f\) orbitals at 105 eV, from an X-ray source that provided 1253.6 \(\mathrm{eV}\) of energy \(\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{J}\right)\) The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV},\) corresponding to the 1 \(\mathrm{s}\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\) (a) Calculate the wavelength of the X rays used in this experiment. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-} ;\) which electrons are the valence electrons in each case?

(a) Why is calcium generally more reactive than magnesium? (b) Why is calcium generally less reactive than potassium?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.