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Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\) , \(\mathrm{O}-\mathrm{H}, \mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H} .\)

Short Answer

Expert verified
The bonds are ranked in order of decreasing polarity as follows: \(\mathrm{F-H > O-H > N-H > C-H > P-H}\)

Step by step solution

01

Remember the electronegativity values and difference for each bond

To rank the bonds based on polarity, we need to know the electronegativity values for each of the elements involved. Here are the electronegativity values for the elements given: - Fluorine (F): 3.98 - Oxygen (O): 3.44 - Nitrogen (N): 3.04 - Carbon (C): 2.55 - Phosphorus (P): 2.19 - Boron (B): 2.04 - Hydrogen (H): (between B and C, identical to P, so approximately 2.19) Now, we will calculate the electronegativity difference for each bond: 1. P-H: 2.19 - 2.19 = 0 2. O-H: 3.44 - 2.19 = 1.25 3. N-H: 3.04 - 2.19 = 0.85 4. F-H: 3.98 - 2.19 = 1.79 5. C-H: 2.55 - 2.19 = 0.36
02

Rank the bonds based on decreasing electronegativity difference

Now that we have the electronegativity difference for each bond, we can rank them in order of decreasing polarity. A higher electronegativity difference corresponds to a higher polarity. So, we arrange the bonds in the decreasing order of their electronegativity difference: 1. F-H: 1.79 2. O-H: 1.25 3. N-H: 0.85 4. C-H: 0.36 5. P-H: 0
03

Write the final answer

The bonds are ranked in order of decreasing polarity as follows: \(\mathrm{F-H > O-H > N-H > C-H > P-H}\)

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Key Concepts

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

Bond Polarity
In chemistry, bond polarity arises when two atoms in a chemical bond have different electronegativity values. This results in the electrons being more attracted to one atom than the other, creating a partial negative charge on one end and a partial positive charge on the other. This imbalance in electron distribution is what generates electrical dipoles, making a bond polar.
To determine bond polarity, it is crucial to understand which atoms are involved and their electronegativity values. Greater differences in electronegativity result in stronger polarity. Polar bonds are more "directional" in how they align in the environment because one side acts as a partial negative pole, and the other as a positive pole. This concept plays a significant role especially in molecules where multiple polar bonds can contribute to a molecule’s overall dipole moment.
Electronegativity Difference
Electronegativity difference is a fundamental concept for understanding bond polarity. It refers to the difference in electronegativity between two bonded atoms. The greater this difference, the more polar the bond. This is because a larger difference means that one atom has a much stronger pull on the bonding electrons compared to the other.
When the electronegativity difference is significant (greater than 1.7, typically), ionic bonds might form as one atom "steals" electrons from the other. For smaller differences, such as those below 0.5, the bond may be considered nonpolar, with electrons shared more equally between the atoms. Understanding this helps in predicting the nature of the bond whether it would be ionic, covalent, or polar covalent.
Chemical Bonds
Chemical bonds are the forces that hold atoms together in compounds. They form because atoms are more stable when they share or transfer electrons to complete their outer electron shell, also known as valence shell. The main types of chemical bonds are ionic, covalent, and metallic bonds.
  • Ionic bonds occur when electrons are transferred from one atom to another, creating oppositely charged ions that attract each other.
  • Covalent bonds happen when atoms share pairs of electrons. Within covalent bonds, if the sharing is unequal due to differences in electronegativity, polar covalent bonds are formed.
Learning about these bonds helps us understand material properties and reactions. For instance, covalent compounds might act as insulators, while ionic compounds can conduct electricity when dissolved in water.
Hydrogen Bonds
Even though hydrogen bonds are not true bonds like covalent or ionic bonds, they are quite significant in chemistry, especially in biological systems. They occur when a hydrogen atom covalently bonded to an electronegative atom, such as nitrogen, oxygen, or fluorine, interacts with an electronegative atom of another molecule.
Hydrogen bonds are responsible for the unique properties of water, such as its high boiling point and surface tension. They are crucial in the structure and function of biological molecules like DNA and proteins. These interactions, while weaker than covalent within the same molecule, significantly influence the shape and properties of molecules, stabilizing structures essential for life functions.

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Most popular questions from this chapter

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

Write Lewis structures and predict the molecular structures of the following. (See Exercises 115 and 116.) a. \(\mathrm{OCl}_{2}, \mathrm{KrF}_{2}, \mathrm{BeH}_{2}, \mathrm{SO}_{2}\) b. \(\mathrm{SO}_{3}, \mathrm{NF}_{3}, \mathrm{IF}_{3}\) c. \(\mathrm{CF}_{4}, \mathrm{SeF}_{4}, \mathrm{KrF}_{4}\) d. \(\mathrm{IF}_{5}, \mathrm{AsF}_{5}\)

Identify the five compounds of \(\mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) described as follows. For each compound, write a Lewis structure that is consistent with the information given. a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds \(\mathrm{C}\) and \(\mathrm{D}\) are ionic and compound \(\mathrm{B}\) is covalent. b. Nitrogen occurs in its highest possible oxidation state in compounds \(\mathrm{A}\) and \(\mathrm{C}\); nitrogen occurs in its lowest possible oxidation state in compounds \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\). The formal charge on both nitrogens in compound \(\mathrm{C}\) is \(+1\); the formal charge on the only nitrogen in compound \(\mathrm{B}\) is 0. c. Compounds \(\mathrm{A}\) and \(\mathrm{E}\) exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound A are normally 16\(M .\) The commercial, concentrated solution of compound \(E\) is 15\(M .\) d. Commercial solutions of compound E are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent. e. Compound \(D\) is 43.7\(\% \mathrm{N}\) and 50.0\(\%\) O by mass. If compound D were a gas at STP, it would have a density of 2.86 \(\mathrm{g} / \mathrm{L} .\) f. A formula unit of compound \(\mathrm{C}\) has one more oxygen than a formula unit of compound \(\mathrm{D}\). Compounds \(\mathrm{C}\) and \(\mathrm{A}\) have one ion in common when compound \(\mathrm{A}\) is acting as a strong electrolyte. g. Solutions of compound \(\mathrm{C}\) are weakly acidic; solutions of compound \(\mathrm{A}\) are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of 0.726 g compound \(\mathrm{B}\) requires 21.98 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HCl}\) for complete neutralization.

The molecules \(\mathrm{BF}_{3}, \mathrm{CF}_{4}, \mathrm{CO}_{2}, \mathrm{PF}_{5},\) and \(\mathrm{SF}_{6}\) are all nonpolar, even though they contain polar bonds. Why?

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl} \quad\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O} \quad\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

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