Using only the periodic table, predict the most stable ion for Na, Mg, Al, S, Cl, K, Ca, and Ga. Arrange these from largest to smallest radius, and explain why the radius varies as it does. Compare your predictions with Fig. \(8.8\).

Short Answer

Expert verified
The most stable ions for the given elements are: Na\(^+\), Mg\(^{2+}\), Al\(^{3+}\), S\(^{2-}\), Cl\(^-\), K\(^+\), Ca\(^{2+}\), and Ga\(^{3+}\). Arranging these ions from largest to smallest radius, we have K\(^+\), Na\(^+\), Ca\(^{2+}\), Mg\(^{2+}\), Ga\(^{3+}\), Al\(^{3+}\), Cl\(^-\), and S\(^{2-}\). The ionic radius varies mainly due to atomic number and charge of the ion; it increases down a group and decreases across a period. Positive ions have smaller radii due to electron loss, and negative ions have larger radii due to electron gain. Compare this prediction with Fig. 8.8 for verification.

Step by step solution

01

Determine the most stable ions for each element

Using the periodic table, we can determine the most stable ion by identifying the group to which the element belongs. Elements in groups 1, 2, and 13 tend to lose electrons to form positive ions, while elements in groups 15, 16, and 17 tend to gain electrons to form negative ions. Here are the most stable ions for the given elements: 1. Na (Group 1) - Na\(^+\) Ion 2. Mg (Group 2) - Mg\(^{2+}\) Ion 3. Al (Group 13) - Al\(^{3+}\) Ion 4. S (Group 16) - S\(^{2-}\) Ion 5. Cl (Group 17) - Cl\(^-\) Ion 6. K (Group 1) - K\(^+\) Ion 7. Ca (Group 2) - Ca\(^{2+}\) Ion 8. Ga (Group 13) - Ga\(^{3+}\) Ion
02

Analyze the trends in ionic radius and arrange ions in order

Ionic radius usually increases down a group (vertical column) and decreases across a period (horizontal row) in the periodic table. When elements form positive ions, they lose electrons from their outermost shell, and when elements form negative ions, they gain electrons in their outermost shell. Considering the trends, we can now arrange the ions from largest to smallest: 1. K\(^+\) 2. Na\(^+\) 3. Ca\(^{2+}\) 4. Mg\(^{2+}\) 5. Ga\(^{3+}\) 6. Al\(^{3+}\) 7. Cl\(^-\) 8. S\(^{2-}\)
03

Explain the variation in ionic radius

The ionic radius varies mainly due to two main factors: atomic number and charge of the ion. As we move across a period, the atomic number increases, which leads to an increase in the nuclear charge and a decrease in the ionic radius. Similarly, as we move down the group, the ionic radius increases due to the addition of electron shells. Also, the varying charges of the ions can influence their size. Positive ions are smaller than their parent atom because they lose electrons, which decreases their electron cloud size. Meanwhile, negative ions are larger than their parent atom due to the gain of electrons in their outermost shell, resulting in increased repulsion between electrons and a larger electron cloud. Now, you can compare this prediction with Fig. 8.8 to see their resemblance.

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 Vaia!

Key Concepts

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

Periodic Table Trends
Understanding the periodic table is fundamental in predicting ionic radii, as it is organized to showcase patterns in elements' properties. The table is arranged in rows called periods and columns known as groups. One crucial trend regarding ionic radius involves its variation across periods and down groups.

As we journey across a period from left to right, the ionic radius typically decreases. This shrinkage occurs because the atomic number rises, leading to a stronger positive charge in the nucleus that pulls electrons closer. However, as we travel down a group, ionic radii increase. This increment is because, with every step downwards, we introduce a new electron shell, further from the nucleus, offering more room and sometimes, shielded from the nucleus' pull by inner electrons.

Ion Size Comparisons

In a particular group, a cation will always be smaller than its parent atom, and an anion will be larger than its parent atom. The logic behind this is grounded in electron loss or gain — the former shrinks the electronic cloud surrounding the nucleus, while the latter increases electron-electron repulsion, leading to a more swollen cloud. These trends help explain why, for example, a potassium ion (K+) is larger than a magnesium ion (Mg2+), despite both belonging to the same period.
Cation and Anion Formation
The formation of cations and anions is a response to an atom's urge to achieve a more stable electron configuration, often resembling the nearest noble gas. To reach this desired state, atoms will either lose or gain electrons.

In the process, ions with a positive charge, or cations, emerge when an atom surrenders electrons. Commonly, this occurs with metals like sodium (Na) and aluminum (Al), which will shed one or more electrons. This results in a smaller ionic radius due to the reduced repulsion between the remaining electrons, and the increased attraction from the positively charged nucleus.

Conversely, anions are born when an atom acquires extra electrons, becoming negatively charged. Nonmetals like sulfur (S) and chlorine (Cl) tend to follow this path, accepting electrons to fulfill their outer shells. Subsequently, the ionic radius enlarges since the added electrons increase repulsion among one another and expand the electron cloud.
Atomic Number Influence
The effect of an element's atomic number on ionic radius is a key concept in understanding ionic sizes. The atomic number, which is the count of protons in the nucleus, not only defines the element but also influences the size of an ion that the atom can form.

Every step up the atomic number scale represents an additional proton in the nucleus and typically an added electron orbiting the nucleus. This extra proton exerts a stronger pull on the surrounding electrons, a phenomenon known as an increased effective nuclear charge. As a result, the electrons are drawn in tighter, leading to a smaller atomic or ionic radius for ions across a period.

However, when examining the effect down a group, an increase in atomic number means the introduction of new electron shells, which outweighs the heightened nuclear pull. This causes the ionic radius to grow, as electrons are further removed from the nucleus, with each new shell. In this context, the atomic number serves as a guiding parameter in predicting ionic sizes and their trends within the periodic table.

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

Use Coulomb's law, $$ V=\frac{Q_{1} Q_{2}}{4 \pi \epsilon_{0} r}=2.31 \times 10^{-19} \mathrm{~J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right) $$ to calculate the energy of interaction for the following two arrangements of charges, each having a magnitude equal to the electron charge. a. \(\stackrel{1 \times 10^{-10} \mathrm{~m}}{\longrightarrow(-1) \longleftrightarrow \infty \longrightarrow(+1) \longleftrightarrow 10^{-10} \mathrm{~m}}{\longleftrightarrow} \stackrel{\leftarrow}{\longleftrightarrow}\) b.

Consider the following energy changes: Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\). Explain.

Draw a Lewis structure for the \(N, N\) -dimethylformamide molecule. The skeletal structure is Various types of evidence lead to the conclusion that there is some double bond character to the \(\mathrm{C}-\mathrm{N}\) bond. Draw one of more resonance structures that support this observation.

Predict the molecular structure, bond angles, and polarity (has a net dipole moment or has no net dipole moment) for each of the following compounds. a. \(\mathrm{SeCl}_{4}\) b. \(\mathrm{SO}_{2}\) c. \(\mathrm{KrF}_{4}\) d. \(\mathrm{CBr}_{4}\) e. \(\mathrm{IF}_{3}\) f. \(\mathrm{ClF}_{5}\)

Which of the following incorrectly shows the bond polarity? Show the correct bond polarity for those that are incorrect. a. \(\delta^{+} \mathrm{H}-\mathrm{F}^{\delta-}\) d. \(^{\delta+} \mathrm{Br}-\mathrm{Br}^{\delta-}\) b. \(^{\delta+} \mathrm{Cl}-\mathrm{I}^{\delta-}\) e. \(^{\delta+} \mathrm{O}-\mathrm{P}^{\delta-}\) c. \(\delta+\mathrm{Si}-\mathrm{S}^{\delta-}\)

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.

Sign-up for free