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

Short Answer

Expert verified
The non-bonding electron-pair domain in NH₃ (ammonia) would be smaller in size compared to the corresponding one in PH₃ (phosphine) due to the higher electronegativity of nitrogen and the smaller size of its 2p orbitals in comparison to phosphorus's 3p orbitals.

Step by step solution

01

Identify the non-bonding electron-pair domain in each molecule

In ammonia (NH₃), there are three N-H bonds and one lone pair of electrons on the nitrogen atom. Likewise, in phosphine (PH₃), there are three P-H bonds and one lone pair of electrons on the phosphorus atom. The non-bonding electron-pair domains in both molecules are the lone pairs on the central atoms.
02

Consider the electronegativity of nitrogen and phosphorus

Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond. Nitrogen (N) has an electronegativity value of 3.04, while Phosphorus (P) has a lower electronegativity value of 2.19. This means that Nitrogen has a higher tendency to attract electrons towards itself compared to Phosphorus.
03

Consider the size of nitrogen and phosphorus atomic orbitals

Nitrogen and phosphorus belong to Group 15 in the periodic table. Nitrogen is in period 2, while phosphorus is in period 3. The larger the period number, the larger the atomic size and the atomic orbitals involved. In the case of NH₃, nitrogen's lone pair occupies a 2p orbital, whereas, in PH₃, phosphorus's lone pair occupies a 3p orbital. Since 3p orbitals are larger than 2p orbitals, the size of phosphorus's atomic orbitals is larger than that of nitrogen.
04

Combine factors to determine the size of the non-bonding electron-pair domain

Since electronegativity and atomic orbital size are inversely related, we can deduce that the higher electronegativity of nitrogen in NH₃ will cause the non-bonding electron-pair domain to be smaller and closer to the central atom. On the other hand, the larger atomic orbitals of phosphorus in PH₃ will result in a larger non-bonding electron-pair domain that is more spread out.
05

Conclusion:

The non-bonding electron-pair domain in NH₃ (ammonia) would be smaller in size compared to the corresponding one in PH₃ (phosphine).

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Key Concepts

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

Electronegativity
Electronegativity is a fundamental concept in chemistry that describes an atom's ability to attract and hold onto electrons when it forms a chemical bond. Think of it as a measure of the atom's electron greed: the higher the electronegativity value, the more it wants electrons for itself.

In our NH3 vs PH3 comparison, nitrogen exhibits a stronger pull on electrons due to its higher electronegativity value compared to phosphorus. This stronger pull essentially shrinks the space that the non-bonding electron-pair domain occupies, as the electrons are held more tightly towards the nitrogen atom's nucleus.

Understanding how electronegativity affects the distribution of electrons is crucial for predicting the behavior of molecules, like knowing how acidic or basic a substance is, or forecasting the types of reactions it may undergo.
Atomic Orbitals
Atomic orbitals are regions in an atom where there is a high probability of finding electrons. These orbitals come in various shapes and sizes, such as s, p, d, and f orbitals, and are organized in different energy levels or 'shells'. When picturing them, think of fuzzy clouds where electrons hang out.

In our exercise, we differentiate between the 2p orbital of nitrogen in NH3 and the 3p orbital of phosphorus in PH3. Due to being on a higher energy level, the 3p orbital is larger and can accommodate the lone pair of electrons more loosely than the 2p orbital. This looser fit makes the non-bonding electron-pair domain in PH3 spread out further from the atom's nucleus compared to NH3's more snug 2p orbital fitting.
Periodic Table Trends
The periodic table is not just a table of elements; it's a roadmap that shows how elements behave based on their position. For instance, as you move from left to right across a period, electronegativity generally increases. But as you move down a group, atomic size and orbital sizes increase.

These trends help us understand why the lone pair of electrons in NH3 is different from that in PH3. Nitrogen resides in the second period of the table, which makes it smaller with a higher electronegativity than phosphorus located in the third period. This movement down the group accounts for the increasing atomic orbital size, spreading the non-bonding electron-pair domain out further in PH3 compared to NH3. Periodic table trends offer invaluable insight into predicting and explaining various chemical behaviors and interactions.

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Most popular questions from this chapter

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S}_{\text {, }}\) (b) \(\mathrm{BCl}_{3}\), (c) \(\mathrm{CH}_{3} \mathrm{I}_{,}\)(d) \(\mathrm{CBr}_{4}\), (e) \(\mathrm{TeBr}_{4}\).

Predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{CCl}_{4}\), (b) \(\mathrm{NH}_{3}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{XeF}_{4}\), (e) \(\mathrm{CH}_{3} \mathrm{Br}\), (f) \(\mathrm{GaH}_{3}\).

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{N}_{2}\) . A closely related substance is hydrazobenzene, $\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2}$ . The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the N atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? (c) Predict the \(N-N-C\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) \(\mathrm{HCN}^{\text {, }}\) (c) \(\mathrm{SO}_{3}\), (d) \(\mathrm{TeCl}_{2}\).

{An} \mathrm{} \mathrm{AB}_{3}$ molecule is described as having a trigonal- bipyramidal electron-domain geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) trigonal planar, (ii) trigonal pyramidal, (iii) T-shaped, or (iv) tetrahedral?

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