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Chapter 8: Problem 40

By referring only to the periodic table, select (a) the most electronegative element in group \(6 A ;(b)\) the least electronegative element in the group \(\mathrm{Al}, \mathrm{Si}, \mathrm{P}\); (c) the most electronegative element in the group Ga, \(\mathrm{P}, \mathrm{Cl}, \mathrm{Na}\) (d) the element in the group \(\mathrm{K}, \mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with Ba.

Short Answer

Expert verified
(a) The most electronegative element in group 6A is Oxygen (O). (b) The least electronegative element in the group Al, Si, P is Aluminum (Al). (c) The most electronegative element in the group Ga, P, Cl, Na is Chlorine (Cl). (d) The element most likely to form an ionic compound with Ba among the group K, C, Zn, F is Fluorine (F).

Step by step solution

01

(a) Finding most electronegative element in group 6A

Refer to the periodic table and locate group 6A. This group includes the elements Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). Using the electronegativity trend, we know that it increases from left to right and bottom to top. Since Oxygen (O) is the first element in this group (at the top-left corner), it is the most electronegative element in group 6A.
02

(b) Finding least electronegative element in the group Al, Si, P

In this question, we need to find the least electronegative element in the group of Aluminum (Al), Silicon (Si), and Phosphorus (P). By looking at their positions in the periodic table: - Aluminum (Al) is in Group 13, Period 3 - Silicon (Si) is in Group 14, Period 3 - Phosphorus (P) is in Group 15, Period 3 Since electronegativity increases from left to right across a period, Aluminum (Al) has the lowest electronegativity among these elements.
03

(c) Finding most electronegative element in the group Ga, P, Cl, Na

Here, we need to find the most electronegative element among Gallium (Ga), Phosphorus (P), Chlorine (Cl), and Sodium (Na). Their positions in the periodic table are: - Gallium (Ga) is in Group 13, Period 4 - Phosphorus (P) is in Group 15, Period 3 - Chlorine (Cl) is in Group 17, Period 3 - Sodium (Na) is in Group 1, Period 3 As electronegativity increases from left to right across a period, Chlorine (Cl) is the most electronegative among these elements.
04

(d) Finding the element most likely to form an ionic compound with Ba

In this case, we are given a group of elements - Potassium (K), Carbon (C), Zinc (Zn), and Fluorine (F) - and need to find which of them is most likely to form an ionic compound with Barium (Ba). When an ionic compound is formed, one of the elements loses electrons (usually a metal) while the other gains electrons (usually a non-metal). Barium (Ba) is an alkaline earth metal, meaning it tends to lose two electrons. By looking at the given elements: - Potassium (K) is a metal in Group 1, Period 4 and thus, prone to losing one electron - Carbon (C) is a non-metal in Group 14, Period 2 and tends to gain four electrons - Zinc (Zn) is a metal in Group 12, Period 4 and thus, prone to losing two electrons - Fluorine (F) is a non-metal in Group 17, Period 2 and tends to gain one electron The most likely option to form an ionic compound with Barium (Ba) would be a non-metal that can gain two electrons, which in this case is not present among the given elements. However, Fluorine (F) is the element that comes closest to that requirement, so it would be the most likely to form an ionic compound with Barium (Ba). Results: (a) Oxygen (O) (b) Aluminum (Al) (c) Chlorine (Cl) (d) Fluorine (F)

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Key Concepts

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

Electronegativity
Electronegativity refers to the ability of an atom to attract electrons towards itself in a chemical bond. It is a crucial concept in chemistry because it helps us understand how atoms interact with each other. The higher the electronegativity, the stronger an atom attracts electrons.

- **Trends:** Electronegativity trends on the periodic table are essential when predicting how atoms will bond. Generally, electronegativity increases across a period from left to right and decreases down a group. This is because atoms have more protons (positive charge) in their nuclei as you move across a period, enhancing their ability to attract electrons. Moving down a group, the outer electrons are further from the nucleus, reducing attraction due to increased electron shielding.

- **Examples:** In Group 6A, Oxygen (O) is the most electronegative due to its position at the top-right corner. Elements such as Fluorine (F) and Chlorine (Cl) also showcase high electronegativity. Comparing elements like Aluminum (Al), Silicon (Si), and Phosphorus (P), Aluminum is the least electronegative as it's further left in the periodic table.
Group Trends
Group trends in the periodic table describe the patterns or behaviors of elements within the same group. These trends help predict an element's properties, aiding in understanding and predicting chemical reactions.

- **Properties:** Elements in the same group generally have similar chemical behaviors because they have the same number of valence electrons. However, some properties like atomic size, ionization energy, and electronegativity show clear trends within groups. For instance, as you move down a group, atomic size increases because more electron shells are added. Conversely, electronegativity and ionization energy typically decrease because the electron attraction weakens with increased distance from the nucleus.

- **Chemical Reactivity:** Group trends also define how reactive an element might be. In metals (such as those in Group 1, alkali metals), reactivity increases down the group. In nonmetals (like halogens in Group 17), reactivity decreases down the group. This understanding is fundamental in predicting how elements will form compounds, for example, how Fluorine (F) easily forms compounds due to its high electronegativity.
Ionic Compounds
Ionic compounds are formed when metals and nonmetals chemically combine through ionic bonds. Ionic bonds occur due to the electrostatic attraction between positively charged ions (cations) and negatively charged ions (anions).

- **Formation Process:** Typically, metals lose electrons and become positively charged cations, while nonmetals gain electrons to form negatively charged anions. For instance, Sodium (Na) can lose one electron to become Na鈦�, and Chlorine (Cl) can gain an electron to become Cl鈦�, resulting in the formation of the ionic compound sodium chloride (NaCl).

- **Properties:** Ionic compounds typically feature high melting and boiling points due to strong ionic bonds. They also tend to be hard and brittle, and they conduct electricity when molten or dissolved in water, as the ions are free to move. An example is Barium (Ba) potentially forming ionic compounds with elements like Fluorine (F), where Ba虏鈦� and F鈦� ions interact.
Chemical Bonding
Chemical bonding describes how atoms combine to form compounds, providing insight into how molecules are structured and interact.

- **Types of Bonds:** Atoms can form several types of bonds, with the most common being ionic, covalent, and metallic bonds. Ionic bonds involve the transfer of electrons, typically between a metal and a nonmetal, as seen in compounds like sodium chloride (NaCl). Covalent bonds involve sharing electrons between atoms, important in organic compounds. Metallic bonds, seen in elements like copper, involve a 'sea' of free electrons shared among a lattice of metal ions.

- **Determining Factors:** Factors such as electronegativity differences between atoms influence the type of chemical bond formed. A large difference typically results in ionic bonds, while smaller differences favor covalent bonding. Recognizing these factors allows chemists to predict compounds' properties and reactivity.

Understanding these basics of chemical bonding is key, especially when considering compound stability and reactivity, like how Fluorine (F) can influence bonding due to its high electronegativity.

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Most popular questions from this chapter

(a) Do the \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene alternate shortlong-short-long around the ring? Why or why not? (b) Are \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene shorter than \(\mathrm{C}-\mathrm{C}\) single bonds? (c) Are \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene shorter than \(\mathrm{C}=\mathrm{C}\) double bonds?

Illustrated are four ions - \(A, B, X\), and \(Y-\) showing their relative ionic radii. The ions shown in red carry positive charges: a \(2+\) charge for A and a \(1+\) charge for B. Ions shown in blue carry negative charges: a \(1-\) charge for \(X\) and a \(2-\) charge for \(Y\). (a) Which combinations of these ions produce ionic compounds where there is a \(1: 1\) ratio of cations and anions? (b) Among the combinations in part (a), which leads to the ionic compound having the largest lattice energy? [Section 8.2]

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s)\), and \(\mathrm{H}_{2}(g) . \mathrm{NaAlH}_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\). (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion. (d) What is the formal charge on hydrogen in the polyatomic ion?

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) \(\mathrm{OCS}\), (b) \(\mathrm{SOCl}_{2}\) ( \(\mathrm{S}\) is the central atom), (c) \(\mathrm{BrO}_{3}^{-}\), (d) \(\mathrm{HClO}_{2}\) (H is bonded to \(\left.\mathrm{O}\right)\).

Which one of these statements about formal charge is true? (a) Formal charge is the same as oxidation number. (b) To draw the best Lewis structure, you should minimize formal charge. (c) Formal charge takes into account the different electronegativities of the atoms in a molecule. (d) Formal charge is most useful for ionic compounds. (e) Formal charge is used in calculating the dipole moment of a diatomic molecule.
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