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Which of the following bonds are polar: (a) \(\mathrm{B}-\mathrm{F}\), (b) \(\mathrm{Cl}-\mathrm{Cl}\), (c) \(\mathrm{Se}-\mathrm{O}\), (d) \(\mathrm{H}-\mathrm{I}\) ? Which is the more electronegative atom in each polar bond?

Short Answer

Expert verified
The given bonds B-F, Se-O, and H-I are polar, while Cl-Cl is non-polar. In each polar bond, the more electronegative atom is F (Fluorine) in B-F, O (Oxygen) in Se-O, and I (Iodine) in H-I.

Step by step solution

01

Electronegativity values for the involved atoms

To determine the bond polarities, we need to know the electronegativity values for B, F, Cl, Se, O, H, and I. They are as follows: - B (Boron): 2.0 - F (Fluorine): 3.9 - Cl (Chlorine): 3.2 - Se (Selenium): 2.6 - O (Oxygen): 3.5 - H (Hydrogen): 2.1 - I (Iodine): 2.7
02

Determine the electronegativity difference for each bond

Calculate the electronegativity difference for each bond: (a) B-F: |2.0 - 3.9| = 1.9 (b) Cl-Cl: |3.2 - 3.2| = 0 (c) Se-O: |2.6 - 3.5| = 0.9 (d) H-I: |2.1 - 2.7| = 0.6
03

Identify polar bonds

A bond is considered polar if the electronegativity difference is greater than 0.5. Based on the calculated differences: - B-F (1.9) is polar - Cl-Cl (0) is non-polar - Se-O (0.9) is polar - H-I (0.6) is polar
04

Identify the more electronegative atom

For each polar bond, identify the more electronegative atom: - B-F: Fluorine (F) is more electronegative (3.9 > 2.0) - Se-O: Oxygen (O) is more electronegative (3.5 > 2.6) - H-I: Iodine (I) is more electronegative (2.7 > 2.1)
05

Conclusion

The polar bonds among the given bonds are B-F, Se-O, and H-I. The more electronegative atom in each polar bond is F in B-F, O in Se-O, and I in H-I.

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Key Concepts

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

Polar Bonds
Polar bonds occur when two atoms within a molecule have differing electronegativity values. Electronegativity refers to an atom’s ability to attract electrons in a bond. When atoms with different electronegativities form a bond, the electrons are not shared equally. This unequal sharing creates a dipole moment, resulting in one atom having a partial negative charge and the other a partial positive charge.
For example, consider the bond between boron (B) and fluorine (F). Fluorine is significantly more electronegative than boron. This means fluorine attracts the shared electrons closer to itself, creating a polar bond. Here, fluorine (F) becomes partially negative, and boron (B) becomes partially positive.
  • Polar bonds occur with an electronegativity difference greater than 0.5.
  • The greater the difference in electronegativity, the more polar the bond.

Understanding polar bonds helps explain phenomena such as the solubility of substances and the interactions between molecules, which are crucial in chemistry.
Electronegativity Difference
The concept of electronegativity difference is critical for determining whether a bond is polar or non-polar. This value is calculated by subtracting the electronegativity values of the two atoms involved in the bond. The absolute value is considered because polarity does not depend on the order of the atoms.
For instance, in the bond between selenium (Se) and oxygen (O), the electronegativity of oxygen is 3.5, and that of selenium is 2.6. The electronegativity difference is \(|3.5 - 2.6| = 0.9\). Since this difference is greater than 0.5, the Se-O bond is polar.
  • A difference of 0 indicates a non-polar bond.
  • Differences greater than 0.5 suggest a polar bond.
  • The higher the difference, the greater the bond polarity.

Calculating the electronegativity difference helps in predicting the nature of bonds and the molecular behavior of compounds.
Non-Polar Bonds
Non-polar bonds are characterized by an even distribution of electrons between two atoms. This occurs when the atoms involved have equal or nearly equal electronegativity values, resulting in an electronegativity difference of 0 or very close to it. Consequently, there is no dipole moment created in the bond.
An excellent example of a non-polar bond is the \( ext{Cl-Cl}\) bond. Both chlorine atoms have the same electronegativity of 3.2, leading to an electronegativity difference of \(|3.2 - 3.2| = 0\). This "0" difference indicates an equal sharing of electrons and establishes that the \( ext{Cl-Cl}\) bond is non-polar.
  • Non-polar bonds are often found in diatomic molecules of the same element.
  • Molecules with non-polar bonds may still be polar overall, depending on their geometry.

Recognizing non-polar bonds is essential for understanding molecular interactions, especially in determining whether a molecule is hydrophobic or hydrophilic.

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Most popular questions from this chapter

(a) What are valence electrons? (b) How many valence electrons does a nitrogen atom possess? (c) An atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} .\) How many valence electrons does the atom have?

Barium azide is \(62.04 \%\) Ba and \(37.96 \%\) N. Each azide ion has a net charge of \(1-\). (a) Determine the chemical formula of the azide ion. (b) Write three resonance structures for the azide ion. (c) Which structure is most important? (d) Predict the bond lengths in the ion.

(a) What is the octet rule? (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3}\), how many electrons must it gain to achieve an octet?

(a) Use bond enthalpies to estimate the enthalpy change for the reaction of hydrogen with ethene: $$\mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ (b) Calculate the standard enthalpy change for this reaction, using heats of formation. Why does this value differ from that calculated in (a)?

(a) Determine the formal charge on the chlorine atom in the hypochlorite ion, \(\mathrm{ClO}^{-}\), and the perchlorate ion, \(\mathrm{ClO}_{4}^{-}\), using resonance structures where the \(\mathrm{Cl}\) atom has an octet. (b) What are the oxidation numbers of chlorine in \(\mathrm{ClO}^{-}\) and in \(\mathrm{ClO}_{4}^{-} ?(\mathrm{c})\) Is it uncommon for the formal charge and the oxidation state to be different? Explain. (d) Perchlorate is a much stronger oxidizing agent than hypochlorite. Would you expect there to be any relationship between the oxidizing power of the oxyanion and either the oxidation state or the formal charge of chlorine?

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