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Chapter 22: Problem 77

(a) How does the structure of diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) differ from that of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?(\mathbf{b})\) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?

Short Answer

Expert verified
The structure of diborane (B2H6) differs from that of ethane (C2H6) in that two hydrogen atoms bridge between both boron atoms, forming a planar arrangement with 3c-2e bonds, while ethane has a direct carbon-carbon bond with a 2c-2e single bond. Diborane adopts its unique geometry due to its electron-deficient nature, which results in the formation of 3c-2e bonds with bridging hydrogen atoms, stabilizing the molecule. The term "hydridic" hydrogen atoms in diborane signifies their role in the unusual 3c-2e bonds and their more negative charge, which plays a crucial role in the reactivity, stability, and properties of diborane and related compounds.

Step by step solution

01

Describe the structure of diborane

To answer part (a), we need to know the molecular structure of diborane and how it differs from that of ethane. Diborane is a compound consisting of 2 boron atoms and 6 hydrogen atoms. The structure of diborane can be represented as the following: Boron atoms are bridged by two hydrogen atoms, while each boron is also bonded to two additional hydrogen atoms, forming a planar arrangement. The two boron atoms are connected by two bridging hydrogen atoms that participate in electron-deficient, three-center two-electron (3c-2e) bonds.
02

Describe the structure of ethane

Ethane is a compound composed of two carbon atoms connected by a single bond and 6 hydrogen atoms bonded to the carbon atoms. Its structure is as follows: Each carbon atom is bonded to 3 hydrogen atoms in a tetrahedral geometry, with the H-C-H bond angle being approximately 109.5°. The carbon-carbon bond is a conventional two-center two-electron (2c-2e) single bond.
03

Compare the structures of diborane and ethane

Diborane and ethane both possess 2 central atoms (B and C, respectively) in their structures. However, the structures differ in the following aspects. 1. In diborane, two hydrogen atoms bridge between both boron atoms, forming a planar arrangement with 3c-2e bonds. In ethane, there is a direct carbon-carbon bond with a 2c-2e single bond. 2. The bond angles in diborane are different from those of ethane. The bond angle around boron in diborane is approximately 120°, while the H-C-H bond angle in ethane is approximately 109.5°.
04

Explain why diborane adopts its geometry

In answer to part (b), diborane adopts its unique geometry due to its electron-deficient nature. Boron atoms in diborane have only three valence electrons, which means they cannot form conventional 2c-2e bonds with all hydrogen atoms. As a result, the 3c-2e bonds with bridging hydrogen atoms are formed, which allow boron to accommodate fewer than an octet of electrons in its valence shell. This is a type of multicenter bonding that stabilizes the molecule.
05

Discuss the significance of "hydridic" hydrogen atoms

In response to part (c), when hydrogen atoms in diborane are referred to as "hydridic," it signifies their role in the unusual 3c-2e bonds. These "hydridic" hydrogen atoms have a more negative charge than those in typical covalently bonded hydrogen atoms, thus they have a greater tendency to donate electrons. This characteristic of hydridic hydrogen plays a crucial role in the reactivity, stability, and properties of diborane and related compounds.

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Most popular questions from this chapter

Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) nitric oxide, (b) hydrazine, (c) potassium cyanide, (d) sodium nitrite, (e) ammonium chloride, (f) lithium nitride.

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) HNO \(_{2},(\mathbf{b}) \mathrm{N}_{3},(\mathbf{c}) \mathrm{N}_{2} \mathrm{H}_{5}^{+},(\mathbf{d}) \mathrm{NO}_{3}^{-} .\)

The physical properties of \(\mathrm{D}_{2} \mathrm{O}\) differ from those of \(\mathrm{H}_{2} \mathrm{O}\) because (a) \(\mathrm{D}\) has a different electron configuration than O. (b) \(\mathrm{D}\) is radioactive. (c) \(\mathrm{D}\) forms stronger bonds with O than \(\mathrm{H}\) does.(\boldsymbol{d} ) D is much more massive than H .

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