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Identify each statement as true or false: (a) Cations are larger than their corresponding neutral atoms. (b) \(\mathrm{Li}^{+}\) is smaller than Li. (c) \(\mathrm{Cl}^{-}\) is bigger than I .

Short Answer

Expert verified
(a) False. Cations are smaller than their corresponding neutral atoms, as the loss of electrons results in a stronger attraction towards the nucleus. (b) True. \(\mathrm{Li}^{+}\) is smaller than Li because the ion has lost one electron, resulting in a smaller size. (c) False. The neutral iodine atom has a larger atomic size compared to the \(\mathrm{Cl}^{-}\) anion due to their positions in the Periodic Table.

Step by step solution

01

(Statement a: Cations are larger than their corresponding neutral atoms)

False. Cations are formed when neutral atoms lose one or more electrons, resulting in a positively charged ion. The loss of electrons causes the remaining electrons to experience a stronger attraction towards the nucleus, shrinking the atomic size. Therefore, cations are typically smaller than their corresponding neutral atoms.
02

(Statement b: \(\mathrm{Li}^{+}\) is smaller than Li)

True. Losing an electron in the process of forming a cation generally results in the ion being smaller than the neutral atom. In the case of \(\mathrm{Li}^{+}\), this Li ion has lost one electron, resulting in a smaller size compared to its corresponding neutral state.
03

(Statement c: \(\mathrm{Cl}^{-}\) is bigger than I)

False. This statement is concerning two different elements: chlorine (Cl) and iodine (I). The statement refers to the size of the anion \(\mathrm{Cl}^{-}\) and compares it to the neutral atom of iodine (I). As we move down a group in the Periodic Table, the atomic size increases due to the increased number of electron shells. Since iodine is located below the chlorine in Group 17 (halogens), the neutral iodine atom would have a larger atomic size compared to the neutral chlorine atom. As for the \(\mathrm{Cl}^{-}\) anion, it has gained an extra electron, making it larger than the neutral Cl atom. However, even with the increased size of the anion, the neutral iodine atom would still be larger than the \(\mathrm{Cl}^{-}\) anion, making this statement false.

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Key Concepts

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

Cation and Anion Size
When atoms form ions, their size changes as they gain or lose electrons. This change in size is particularly noticeable when an atom turns into a cation or anion. Cations are formed by the loss of one or more electrons, which results in a positive charge. Since electrons are negatively charged, removing them reduces the electron-electron repulsion in the atom. This allows the remaining electrons to be pulled closer to the nucleus by the positive charge of the protons. As a result, cations tend to be smaller than their corresponding neutral atoms.

For example, when a lithium atom (Li) loses one electron to become a \(Li^{+}\) cation, its size decreases. This is because the electron that was in the outermost shell is removed, reducing the atomic radius.

On the other hand, anions are formed by gaining additional electrons. This induces more electron-electron repulsion, pushing electrons further from the nucleus. Thus, anions tend to be larger than their corresponding neutral atoms. For instance, when chlorine gains an electron, it becomes \(Cl^{-}\), and its size increases compared to its neutral state.
Periodic Trends
The Periodic Table organizes elements in a way that reveals predictable patterns known as periodic trends. One of these trends is the change in atomic size as you move across periods (rows) and groups (columns).

• **Across a period**: As you move from left to right across a period, atoms gain more protons and electrons. However, the additional electrons enter the same energy level. The increased positive charge in the nucleus draws these electrons closer, causing the atomic radius to decrease across a period.

• **Down a group**: When moving down a group, additional electron shells are added. These extra shells increase the distance between the outermost electrons and the nucleus, resulting in a larger atomic radius. Consequently, the size of the atoms increases as you move down the group.

For example, iodine (I) appears below chlorine (Cl) in the Periodic Table. This means iodine has a larger atomic size than chlorine because iodine has more electron shells.
Ionization Process
The ionization process involves the removal or addition of electrons to create ions. This is crucial for understanding how ions are formed and why they exhibit certain properties.

• **Formation of cations**: Typically involves the loss of electrons. When an electron is removed, the atom requires energy, known as the ionization energy. This energy input results in a positively charged ion (cation).

• **Formation of anions**: Occurs when atoms gain electrons, which generally does not require as much energy compared to forming a cation. In fact, atoms release energy when gaining electrons to form negatively charged ions (anions).

The effectiveness of an atom to lose or gain electrons and form cations or anions is influenced by its position on the Periodic Table. Elements in Group 1, for example, easily lose one electron to form cations due to their low ionization energies. Conversely, elements like Group 17 readily gain electrons to form anions because they have high electronegativity and readily accept electrons to complete their outer shell.

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Most popular questions from this chapter

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Based on their positions in the periodic table, predict which atom of the following pairs will have the smaller first ionization energy: (a) \(\mathrm{Br}, \mathrm{Kr} ; \mathbf{( b )} \mathrm{C}, \mathrm{Ca} ;(\mathbf{c}) \mathrm{Li}, \mathrm{Rb} ;\); (d) \(\mathrm{Pb}, \mathrm{Si} ;\) (e) \(\mathrm{Al}, \mathrm{B}\).

(a) One of the alkali metals reacts with oxygen to form a solid white substance. When this substance is dissolved in water, the solution gives a positive test for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\). When the solution is tested in a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal? (b) Write a balanced chemical equation for the reaction of the white substance with water.

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group 11 and group 12 metals are as follows: (a) Why are the electron affinities of the group 12 elements greater than zero? (b) Why do the electron affinities of the group 11 elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}{ }^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

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