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Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements. a. \(\mathrm{Rb}\) and \(\mathrm{Cl} \quad\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\) b. \(\mathrm{S}\) and \(\mathrm{S} \quad\) e. \(\mathrm{N}\) and \(\mathrm{P}\) c. \(\mathrm{C}\) and \(\mathrm{F} \quad\) f. \(\mathrm{B}\) and \(\mathrm{H}\)

Short Answer

Expert verified
a. Rb and Cl: ionic bond b. S and S: nonpolar covalent bond c. C and F: polar covalent bond d. Ba and S: ionic bond e. N and P: polar covalent bond f. B and H: polar covalent bond

Step by step solution

01

List Electronegativity Values

First, we need to find the electronegativity values of all elements involved in the exercise. These values can be found in a periodic table. Here are the electronegativity values for the elements: - Rb: 0.82 - Cl: 3.16 - Ba: 0.89 - S: 2.58 - N: 3.04 - P: 2.19 - C: 2.55 - F: 3.98 - B: 2.04 - H: 2.20 Step 2: Calculate the Electronegativity Differences
02

Calculate the Electronegativity Differences

Subtract the electronegativity values of the two elements in each pair to find the difference: a. Rb and Cl: \(|0.82 - 3.16| = 2.34\) b. S and S: \(|2.58 - 2.58| = 0\) c. C and F: \(|2.55 - 3.98| = 1.43\) d. Ba and S: \(|0.89 - 2.58| = 1.69\) e. N and P: \(|3.04 - 2.19| = 0.85\) f. B and H: \(|2.04 - 2.20| = 0.16\) Step 3: Determine the Type of Bond
03

Determine the Type of Bond

Based on the electronegativity differences, we can determine the type of bond for each pair: a. Rb and Cl: ionic (\(\Delta EN = 2.34 > 1.7\)) b. S and S: nonpolar covalent (\(\Delta EN = 0 < 0.5\)) c. C and F: polar covalent (\(0.5 \leq \Delta EN = 1.43 \leq 1.7\)) d. Ba and S: ionic (\(\Delta EN = 1.69 > 1.7\)) e. N and P: polar covalent (\(0.5 \leq \Delta EN = 0.85 \leq 1.7\)) f. B and H: polar covalent (\(0.5 \leq \Delta EN = 0.16 \leq 1.7\))

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Key Concepts

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

Bond Types
When atoms come together to form compounds, they create bonds to hold the atoms together. These bonds are primarily of three types: ionic, covalent, and polar covalent. The type of bond depends on the difference in electronegativity between the atoms involved. Electronegativity is a measure of how strongly an atom attracts electrons in a bond.
  • Ionic Bonds: Formed when the electronegativity difference is high (usually greater than 1.7), resulting in the transfer of electrons from one atom to another.
  • Covalent Bonds: Occur when the electronegativity difference is low (less than 0.5), leading to equal sharing of electrons between atoms.
  • Polar Covalent Bonds: Exist when the electronegativity difference falls between 0.5 and 1.7, causing an unequal sharing of electrons.
Understanding these bonds helps predict the properties of compounds, such as melting points, solubility, and electrical conductivity.
Ionic Bond
Ionic bonds are a type of chemical bond that forms when there is a significant electronegativity difference between two atoms, typically greater than 1.7. This large difference causes electrons to transfer from one atom to the other, resulting in one atom becoming positively charged and the other negatively charged.
For example, in the pair between rubidium (Rb) and chlorine (Cl), the electronegativity difference is 2.34. This is a classic example of an ionic bond as Rb loses an electron to Cl, forming an Rb+ ion and a Cl- ion.
  • Important characteristics of ionic bonds include:
    • Formation of crystal lattice structures.
    • High melting and boiling points.
    • Good conductors of electricity when dissolved in water.
Ionic bonds usually form between metals and nonmetals, and these structures contribute to various properties of ionic compounds.
Covalent Bond
Covalent bonds occur when two atoms share electrons. This bond happens when the electronegativity difference between the atoms is very small (less than 0.5). In such cases, neither atom is strong enough to attract the electron away from the other, leading to a mutual sharing of electrons.
Sulfur (S) paired with another sulfur atom is an example of a covalent bond. Because both atoms have the same electronegativity value, there is no difference (\[ \Delta EN = 0 \]), resulting in a perfect sharing of electrons.
  • Key aspects of covalent bonds:
    • Can form single, double, or triple bonds, depending on the number of shared electron pairs.
    • Lower melting and boiling points compared to ionic compounds.
    • Usually do not conduct electricity.
Covalent bonds are common in organic molecules and are crucial for the structure of many compounds.
Polar Covalent Bond
Polar covalent bonds happen when there is a difference in electronegativity between two atoms ranging from 0.5 to 1.7. This electronegativity difference results in unequal sharing of electrons, where one atom attracts the electrons more than the other, creating a dipole moment in the molecule.
An example of a polar covalent bond is found between carbon (C) and fluorine (F), with a difference of 1.43. The electron affinity of fluorine is much higher than carbon, causing the shared electrons to be more attracted to F, which results in partial positive and negative charges on the C and F atoms, respectively.
  • Notable features of polar covalent bonds:
    • The existence of partial charges, making them more reactive than nonpolar covalent compounds.
    • They are soluble in polar solvents like water.
    • They may be asymmetrical in molecular geometry, contributing to their dipole nature.
Understanding polar covalent bonds is essential for grasping how molecules function in different environments, especially in biological systems.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$: \mathrm{N}=\mathrm{N}=\dot{\mathrm{O}}\longleftrightarrow: \mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{Q}} : \longleftrightarrow : \dot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}$$ Given the following bond lengths, \(\mathrm{N}-\mathrm{N} \qquad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\) \(\mathrm{N}=\mathrm{N} \qquad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\) \(\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm}\) rationalize the observations that the N-N bond length in \(\mathrm{N}_{2} \mathrm{O}\) is 112 \(\mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is 119 \(\mathrm{pm}\) . Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\) . Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

A compound, \(\mathrm{XF}_{5},\) is 42.81\(\%\) fluorine by mass. Identify the element \(\mathrm{X}\) . What is the molecular structure of \(\mathrm{XF}_{5} ?\)

Two different compounds exist having the formula \(\mathrm{N}_{2} \mathrm{F}_{2}\) . One compound is polar whereas the other is nonpolar. Draw Lewis structures for \(\mathrm{N}_{2} \mathrm{F}_{2}\) consistent with these observations.

Use the following data (in kJ/mol) to estimate \(\Delta H\) for the reaction \(S^{-}(g)+e^{-} \rightarrow S^{2-}(g)\) . Include an estimate of uncertainty. \(\begin{aligned} \mathrm{S}(s) \longrightarrow \mathrm{S}(g) & \Delta H=277 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{S}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{-}(g) & \Delta H=-200 \mathrm{kJ} / \mathrm{mol} \end{aligned}\) Assume that all values are known to \(\pm 1 \mathrm{kJ} / \mathrm{mol}\)

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3} .\) Give a possible explanation for why these two reactions occur.

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