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Chapter 8: Problem 147

Classify the bonding in each of the following molecules as ionic, polar covalent, or nonpolar covalent. a. \(\mathrm{H}_{2} \quad\) e. \(\mathrm{HF}\) b. \(\mathrm{K}_{3} \mathrm{P} \quad\) f. \(\mathrm{CCl}_{4}\) c. \(\mathrm{Nal} \quad\) g. \(\mathrm{CF}_{4}\) d. \(\mathrm{SO}_{2} \quad\) h. \(\mathrm{K}_{2} \mathrm{S}\)

Short Answer

Expert verified
a. \(H_2:\) nonpolar covalent b. \(K_3P:\) polar covalent c. \(NaI:\) ionic d. \(SO_2:\) polar covalent e. \(HF:\) ionic f. \(CCl_4:\) polar covalent g. \(CF_4:\) polar covalent h. \(K_2S:\) ionic

Step by step solution

01

Calculate the electronegativity difference for each molecule

For each molecule, find the electronegativity values for the elements involved in the bond from a periodic table or electronegativity chart. Then, calculate the difference between the electronegativity values for each bond. a. \(H_2:\) Electronegativity of H=2.1, Difference = \(|2.1 - 2.1| = 0\) b. \(K_3P:\) Electronegativity of K=0.8 and P=2.1, Difference = \(|0.8 - 2.1| = 1.3\) c. \(NaI:\) Electronegativity of Na=0.9 and I=2.5, Difference = \(|0.9 - 2.5| = 1.6\) d. \(SO_2:\) Electronegativity of S=2.5 and O=3.5, Difference = \(|2.5 - 3.5| = 1.0\) e. \(HF:\) Electronegativity of H=2.1 and F=4.0, Difference = \(|2.1 - 4.0| = 1.9\) f. \(CCl_4:\) Electronegativity of C=2.5 and Cl=3.0, Difference = \(|2.5 - 3.0| = 0.5\) g. \(CF_4:\) Electronegativity of C=2.5 and F=4.0, Difference = \(|2.5 - 4.0| = 1.5\) h. \(K_2S:\) Electronegativity of K=0.8 and S=2.5, Difference = \(|0.8 - 2.5| = 1.7\)
02

Classify the bonding based on electronegativity difference

Use the electronegativity differences calculated in Step 1 to classify each molecule as ionic, polar covalent, or nonpolar covalent using the rules mentioned in the analysis. a. \(H_2:\) Electronegativity difference = 0, Bond type: nonpolar covalent b. \(K_3P:\) Electronegativity difference = 1.3, Bond type: polar covalent c. \(NaI:\) Electronegativity difference = 1.6, Bond type: ionic d. \(SO_2:\) Electronegativity difference = 1.0, Bond type: polar covalent e. \(HF:\) Electronegativity difference = 1.9, Bond type: ionic f. \(CCl_4:\) Electronegativity difference = 0.5, Bond type: polar covalent g. \(CF_4:\) Electronegativity difference = 1.5, Bond type: polar covalent h. \(K_2S:\) Electronegativity difference = 1.7, Bond type: ionic

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Key Concepts

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

Electronegativity
Electronegativity is a key concept in understanding chemical bonding. It measures how strongly an atom attracts the electrons in a bond. The higher the electronegativity, the stronger its pull. Each element has a unique electronegativity value, typically found in a periodic table or electronegativity chart. These values help us predict how atoms will interact when they form bonds.

When two atoms bond, their electronegativity difference can tell us the bond type. If the difference is large, typically greater than 2.0, the bond is likely ionic. If the difference is small, less than 0.4, the bond is usually nonpolar covalent. For moderate differences, typically between 0.5 and 1.9, the bond tends to be polar covalent. Thus, knowing these values enables researchers and students to determine the nature of chemical bonds.
Ionic Bond
An ionic bond occurs when electrons are transferred between atoms. This happens when there is a significant difference in electronegativity between the atoms. The atom with the higher electronegativity will attract and gain electrons, becoming a negative ion (anion). The other atom loses electrons and becomes a positive ion (cation).

Ionic bonds are common in compounds like sodium chloride (NaCl) and are characterized by high melting and boiling points. These bonds form between metals and nonmetals, where metals tend to lose electrons and nonmetals gain them.
  • High electronegativity difference (usually > 2.0)
  • Formation of cations and anions
  • Generally form between metals and nonmetals
Polar Covalent Bond
In a polar covalent bond, electrons are shared unequally between atoms. This is due to a moderate difference in electronegativity between them, typically ranging from 0.5 to 1.9. One atom attracts the shared electrons more strongly, creating a dipole with a slightly positive and slightly negative side. Water (H鈧侽) is a classic example of a polar covalent compound.

These bonds result in molecules with partial electric charges, leading to interesting properties like higher solubility in water. Polar covalent bonds are crucial in biological systems, influencing how molecules interact and bond with each other.
  • Electronegativity difference between 0.5 and 1.9
  • Unequal sharing of electrons
  • Results in partial positive and negative charges
Nonpolar Covalent Bond
A nonpolar covalent bond forms when electrons are shared equally between atoms. This occurs primarily between atoms with identical or very similar electronegativity values, resulting in little to no difference (less than 0.4).

Molecules like nitrogen gas (N鈧�) and hydrogen gas (H鈧�) exemplify nonpolar covalent bonds. These molecules do not exhibit any dipole moments, as their electron distributions are symmetric. As a result, nonpolar covalent compounds generally have lower boiling and melting points compared to ionic or polar covalent counterparts.
  • Electronegativity difference less than 0.4
  • Equal sharing of electrons
  • No partial charges are present

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Most popular questions from this chapter

Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. \(\mathrm{Na}^{+} \quad\) c. \(\mathrm{Al}^{3+}\) b. \(\mathrm{Ca}^{2+} \quad\) d. \(\mathrm{Rb}^{+}\)

Write Lewis structures for the following. Show all resonance structures where applicable. a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{N}_{2} \mathrm{O}_{4} \text { exists as } \mathrm{O}_{2} \mathrm{N}-\mathrm{NO}_{2} .\right)\) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{SCN}^{-} . )\)

Which of the following incorrectly shows the bond polarity? Show the correct bond polarity for those that are incorrect. a. \(^{\delta+} \mathrm{H}-\mathrm{F}^{\delta-} \quad\) d. \(\delta^{+} \mathrm{Br}-\mathrm{Br}^{\delta-}\) b. \(^{\delta+} \mathrm{Cl}-\mathrm{I}^{\delta-} \qquad\) e. \(\quad\) e. \(\quad ^{\delta+}\mathrm{O}-\mathrm{P}^{\delta-}\) c. \(\quad \delta+\mathrm{Si}-\mathrm{S}^{\delta-}\)

Write Lewis structures that obey the octet rule for each of the following molecules and ions. (In each case the first atom listed is the central atom.) a. \(\mathrm{POCl}_{3}, \mathrm{SO}_{4}^{2-}, \mathrm{XeO}_{4}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}\) b. \(\mathrm{NF}_{3}, \mathrm{SO}_{3}^{2-}, \mathrm{PO}_{3}^{3-}, \mathrm{ClO}_{3}^{-}\) c. \(\mathrm{ClO}_{2}-, \mathrm{SCl}_{2}, \mathrm{PCl}_{2}^{-}\) d. Considering your answers to parts a, b, and c, what conclusions can you draw concerning the structures of species containing the same number of atoms and the same number of valence electrons?

The standard enthalpies of formation for \(\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g),\) and \(\mathrm{SF}_{6}(g)\) are \(+278.8,+79.0,-775,\) and \(-1209 \mathrm{kJ} / \mathrm{mol}\) respectively. a. Use these data to estimate the energy of an \(\mathrm{S}-\) F bond. b. Compare your calculated value to the value given in Table \(8.5 .\) What conclusions can you draw? c. Why are the \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) not equal to zero, since sulfur and fluorine are elements?
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