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An \(\mathrm{AB}_{3}\) molecule is described as having a trigonalbipyramidal electron-domain geometry. How many nonbonding domains are on atom A? Explain.

Short Answer

Expert verified
In an AB3 molecule with a trigonal bipyramidal electron-domain geometry, there are 2 nonbonding electron domains on atom A. This is because there are 3 bonding domains (B atoms) and a total of 5 electron domains required for the trigonal bipyramidal geometry, which results in the difference of 2 nonbonding electron domains for the central atom A.

Step by step solution

01

Understand the trigonal bipyramidal geometry

First, let's have a quick understanding of trigonal bipyramidal geometry. In this geometry, there are five electron domains around the central atom (A). Three of these domains lie in the same plane and are referred to as the equatorial domains, while the other two are perpendicular to this plane and lie above and below (referred to as axial domains).
02

Determine the total electron domains for atom A

In an AB3 molecule with a trigonal bipyramidal geometry, there are three bonding domains (B atoms) connected to the central atom A. This implies that there are at least three electron domains around atom A.
03

Determine the number of nonbonding electron domains on atom A

To reach five electron domains (the number required for a trigonal bipyramidal geometry), we need to consider the presence of nonbonding electron domains. The difference between the total number of electron domains in trigonal bipyramidal geometry (5) and the number of bonding electron domains we have in AB3 (3) will give us the number of nonbonding electron domains. 5 total electron domains - 3 bonding electron domains = 2 nonbonding electron domains
04

Conclusion

There are 2 nonbonding electron domains on atom A in this AB3 molecule with a trigonal bipyramidal electron-domain geometry. The presence of these nonbonding electron domains is necessary to achieve the desired geometry around the central atom (A).

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Most popular questions from this chapter

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

(a) Does \(\mathrm{SCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{BeCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S} ;\) (b) \(\mathrm{HCN}\); (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{PF}_{3}\), (b) \(\mathrm{CH}_{3}{ }^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{4}^{-}(\mathrm{e}) \mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

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