Problem 40
Question
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.
Step-by-Step Solution
Verified Answer
(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).
1Step 1: (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.
2Step 2: (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.
3Step 3: (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.
4Step 4: (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)
Key Concepts
ElectronegativityGroup TrendsIonic CompoundsChemical Bonding
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.
- **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.
- **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⁺, and Chlorine (Cl) can gain an electron to become Cl⁻, 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 Ba²⁺ and F⁻ ions interact.
- **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⁺, and Chlorine (Cl) can gain an electron to become Cl⁻, 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 Ba²⁺ and F⁻ 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.
- **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.
Other exercises in this chapter
Problem 38
(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going dow
View solution Problem 39
Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) \(\mathrm{Na}, \mathrm{Mg}\), K, Ca; (b) P,
View solution Problem 41
Which of the following bonds are polar? (a) B-F, (b) \(\mathrm{Cl}-\mathrm{Cl}\), (c) Se-O, (d) H-I. Which is the more electronegative atom in each polar bond?
View solution Problem 42
Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\); (
View solution