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Chapter 9: Problem 18

Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than the corresponding one in \(\mathrm{PH}_{3}\) ?

Short Answer

Expert verified
The nonbonding electron-pair domain in 狈贬鈧� is smaller as compared to the one in 笔贬鈧�. This is due to the greater electronegativity and smaller atomic orbitals of nitrogen in 狈贬鈧�, which attracts the electron pair more effectively and contracts the size of the nonbonding domain.

Step by step solution

01

Identify the electron-pair domains

First, we need to draw the Lewis structure for both 狈贬鈧� and 笔贬鈧� molecules. In 狈贬鈧�, nitrogen (N) has 5 valence electrons and hydrogen (H) has 1 valence electron, making a total of 8 valence electrons. In the Lewis structure, nitrogen is the central atom and needs to form three single bonds with three hydrogen atoms, leaving one lone pair of electrons (nonbonding electron-pair domain) on nitrogen. Similarly, in 笔贬鈧�, phosphorus (P) has 5 valence electrons and hydrogen (H) has 1 valence electron, making a total of 8 valence electrons. In the Lewis structure, phosphorus is the central atom and needs to form three single bonds with three hydrogen atoms, leaving one lone pair of electrons (nonbonding electron-pair domain) on phosphorus.
02

Compare the size of nonbonding electron-pair domains

Nonbonding electron-pair domains are regions where the electron pair is not involved in any bond formation. The size of the nonbonding electron-pair domain depends on the electronegativity of the central atom and the size of its atomic orbitals. In our case, the electronegativity of nitrogen (3.04) is greater than that of phosphorus (2.19). So, the nitrogen in 狈贬鈧� attracts the electron pair more effectively, and consequently, the nonbonding electron-pair domain is more contracted and smaller in size as compared to the one in 笔贬鈧�. On the other hand, the atomic orbitals of phosphorus are larger than that of nitrogen due to its increased size and the presence of more shells. Thus, the electron cloud is more dispersed for 笔贬鈧� and the size of the nonbonding electron-pair domain is greater for 笔贬鈧� than 狈贬鈧�.
03

Conclusion

After comparing the electronegativity and size of atomic orbitals, we can conclude that the nonbonding electron-pair domain in 狈贬鈧� will be smaller in size than the corresponding one in 笔贬鈧� due to the greater electronegativity and smaller atomic orbitals of nitrogen.

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Key Concepts

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

Lewis structure
Understanding Lewis structures is crucial in predicting molecular shapes and reactivity. Lewis structures are diagrams that show the bonding between atoms and the lone pairs of electrons that may exist in the molecule. In drawing a Lewis structure:
  • Determine the total number of valence electrons for the molecule.
  • Use a pair of electrons to form a bond between each pair of bound atoms.
  • Distribute the remaining electrons to satisfy the octet rule, which means that atoms share electrons until they each have eight electrons in their valence shell.
For example, using the molecule 狈贬鈧�: - Nitrogen has five valence electrons; each hydrogen has one. - This gives a total of eight valence electrons. - We place nitrogen in the center, establishing three single bonds with hydrogen, resulting in one lone pair of electrons on nitrogen.
Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Generally, electronegativity values increase across a period and decrease down a group in the periodic table. This concept is critical when comparing two different molecules. For instance, nitrogen (N) in 狈贬鈧� has a higher electronegativity value (3.04) than phosphorus (P) in 笔贬鈧� (2.19). This implies that nitrogen can attract the nonbonding pair of electrons more strongly, leading to a more compact and contracted electron cloud around nitrogen than around phosphorus. As a result, nonbonding domains (lone pairs) are smaller in 狈贬鈧� compared to 笔贬鈧�.
Atomic orbitals
Atomic orbitals represent regions in an atom where the probability of finding electrons is highest. They are described by quantum numbers and have different shapes, such as s, p, d, and f orbitals. In the context of molecules like 狈贬鈧� and 笔贬鈧�, the size and shape of atomic orbitals can significantly affect the properties of the molecules: - Nitrogen uses its smaller 2p orbitals to form bonds and hold the lone pair tightly in 狈贬鈧�. - Phosphorus, which is larger, uses its 3p orbitals, resulting in a looser hold on the lone pair in 笔贬鈧�. Due to the larger orbitals of phosphorus, the nonbonding electron pair in 笔贬鈧� is more spread out, while in 狈贬鈧�, it is more compact due to the smaller size of nitrogen's orbitals.
狈贬鈧�
Ammonia, with the chemical formula 狈贬鈧�, is a simple nitrogen hydride. Its structure features nitrogen atom at the center with three hydrogen atoms bonded to it and one lone pair of electrons. The presence of the lone pair affects its molecular geometry: - The molecule adopts a trigonal pyramidal shape due to the lone pair, which repels the bonded electrons, influencing the angle between hydrogen atoms. - 狈贬鈧� is a polar molecule, owing to the difference in electronegativity and the geometry caused by the lone pair. This polarity contributes to hydrogen bonding, giving 狈贬鈧� higher boiling and melting points compared to other similar-sized molecules without hydrogen bonds.
笔贬鈧�
Phosphine, designated as 笔贬鈧�, is structurally similar to ammonia, with phosphorus as the central atom bonded to three hydrogen atoms, plus one lone pair of electrons. The main difference from 狈贬鈧� arises from the difference in electronegativity and orbital sizes: - 笔贬鈧� is also trigonal pyramidal, but due to phosphorus' lower electronegativity and larger orbitals, the lone pair is more spread out. - Consequently, 笔贬鈧� is less polar than 狈贬鈧�, leading to weaker forces between its molecules. As a result, phosphine's boiling point is lower than ammonia's. It is less soluble in water and generally characterized by its lack of hydrogen bonding capabilities.

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Most popular questions from this chapter

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. ( b) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) have? Would they be expected to have dipole moments?

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{H}_{2}^{+}, \mathrm{B}_{2}, \mathrm{N}_{2}^{+}, \mathrm{F}_{2}^{+},\) and \(\mathrm{Ne}_{2}\) .

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{Fbond}\) angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\operatorname{AsF}_{3},(\mathbf{b}) \mathrm{CH}_{3}^{+},(\mathbf{c}) \operatorname{Br} \mathrm{F}_{3},(\mathbf{d}) \mathrm{ClO}_{3},(\mathbf{e}) \mathrm{XeF}_{2}\) \((\mathbf{f}) \mathrm{BrO}_{2}^{-}\)
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