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Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c.\(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Short Answer

Expert verified
In summary, the most polar bonds for each group are: a. \(C-H\) b. \(Al-Br\) c. \(Si-O\) d. \(O-F\)

Step by step solution

01

List the given bonds and their electronegativity values.

First, we need to write down the list of given bonds and their electronegativity values. Electronegativity values can be looked up from the periodic table or other resources. Here are the given bonds and their electronegativity values: a. C-H: (C: 2.55, H: 2.20) Si-H: (Si: 1.90, H: 2.20) Sn-H: (Sn: 1.96, H: 2.20) b. Al-Br: (Al: 1.61, Br: 2.96) Ga-Br: (Ga: 1.81, Br: 2.96) In-Br: (In: 1.78, Br: 2.96) Tl-Br: (Tl: 1.62, Br: 2.96) c. C-O: (C: 2.55, O: 3.44) Si-O: (Si: 1.90, O: 3.44) d. O-F: (O: 3.44, F: 3.98) O-Cl: (O: 3.44, Cl: 3.16)
02

Calculate the electronegativity difference for each bond

Now, we need to calculate the electronegativity difference between the bonded atoms for each bond: a. C-H: |2.55 - 2.20| = 0.35 Si-H: |1.90 - 2.20| = 0.30 Sn-H: |1.96 - 2.20| = 0.24 b. Al-Br: |1.61 - 2.96| = 1.35 Ga-Br: |1.81 - 2.96| = 1.15 In-Br: |1.78 - 2.96| = 1.18 Tl-Br: |1.62 - 2.96| = 1.34 c. C-O: |2.55 - 3.44| = 0.89 Si-O: |1.90 - 3.44| = 1.54 d. O-F: |3.44 - 3.98| = 0.54 O-Cl: |3.44 - 3.16| = 0.28
03

Determine the most polar bonds in each group

Now we can compare the electronegativity differences within each group to identify the most polar bond: a. C-H is the most polar bond, as it has the highest electronegativity difference (0.35). b. Al-Br is the most polar bond, as it has the highest electronegativity difference (1.35). c. Si-O is the most polar bond, as it has the highest electronegativity difference (1.54). d. O-F is the most polar bond, as it has the highest electronegativity difference (0.54). Thus, the most polar bonds for each group are: a. C-H b. Al-Br c. Si-O d. O-F

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Key Concepts

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

Polar Bonds
A polar bond is a type of covalent bond where the electrons shared between the atoms are unequally distributed. This occurs because the atoms involved have different electronegativities, with one atom attracting the shared electrons more strongly. As a result, the bond has a slight charge difference across its ends, creating a dipole moment. This dipole moment leads to one end being slightly positive and the other slightly negative.

Polar bonds are important because they can influence properties like solubility, reactivity, and boiling point of the compounds. Understanding polar bonds is key when predicting how molecules will interact with each other and with different solvents. If a bond is polar, it might contribute to the molecule being overall polar, depending on the molecule's geometry.

When you compare bonds within the same group, the bond with the highest electronegativity difference will usually be the most polar, as seen in the exercise.
Electronegativity Difference
Electronegativity difference is the key factor in determining the polarity of a bond. Electronegativity itself is a measure of an atom's ability to attract and hold onto electrons. The greater the difference in electronegativity between two bonded atoms, the more polar the bond will be.

For example, in the exercise provided, we calculate the electronegativity difference by taking the absolute difference \[| ext{Electronegativity of Atom A} - ext{Electronegativity of Atom B} |\] This calculation helps to determine which atom possesses a stronger pull on the shared electrons. When the difference is zero, the bond is considered nonpolar covalent. When the difference is small, the bond is often considered polar covalent. If the difference is very large, the bond might even be considered ionic.

Understanding these differences gives insight into the nature of chemical bonds and helps to predict molecular interactions and properties.
Periodic Table Trends
The periodic table is an essential tool for predicting electronegativity and understanding why atoms behave the way they do in bonds. General trends are observed when moving across periods or down groups in the table.

  • As you move from left to right across a period, electronegativity tends to increase. This is because the nuclear charge increases, pulling electrons closer to the nucleus.
  • Conversely, as you move down a group, electronegativity tends to decrease. This decrease happens because the atomic size increases, making it difficult for the nucleus to attract bonding electrons effectively.

These trends allow us to predict the relative electronegativities of different elements without looking up specific values. For instance, knowing that oxygen is more electronegative than silicon helps to predict that the Si-O bond is more polar than the C-O bond, as seen in the exercise.

By leveraging these periodic trends, you can effectively predict bond polarities in a variety of chemical compounds without needing to memorize extensive data.

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Most popular questions from this chapter

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)

Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{C}, \mathrm{N}, \mathrm{O} \quad\) c. \(\mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}\) b. \(\mathrm{S}, \mathrm{Se}, \mathrm{Cl} \quad\) d. \(\mathrm{TI}, \mathrm{S}, \mathrm{Ge}\)

Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being 鈥減ulled鈥 from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}\)

Carbon and sulfur form compounds with each other with the formulas \(\mathrm{CS}_{2}\) and \(\mathrm{C}_{3} \mathrm{S}_{2}\) . Draw a Lewis structure for each compound that has a formal charge of zero for all atoms in the structure.

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