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In diborane, the two \(\mathrm{H}-\mathrm{B}-\mathrm{H}\) angles are nearly (a) \(60^{\circ}, 120^{\circ}\) (b) \(95^{\circ}, 150^{\circ}\) (c) \(95^{\circ}, 120^{\circ}\) (d) \(120^{\circ}, 180^{\circ}\)

Short Answer

Expert verified
The correct angles in diborane are \(95^{\circ}, 120^{\circ}\).

Step by step solution

01

Understanding the Structure of Diborane

Diborane (\(\mathrm{B}_2\mathrm{H}_6\)) is a molecule consisting of two boron atoms and six hydrogen atoms. It contains two types of hydrogen atoms: terminal and bridging. The structure includes a unique arrangement known as three-center two-electron (3c-2e) bonds or banana bonds.
02

Analyzing the B-H-B Angles

The focus is on the angles involving the bridging hydrogen atoms in the structure of diborane. The significant interaction is the B-H-B bond where the two hydrogen atoms bridge between the two boron atoms. Due to the geometry of these bridges, the angle between the atoms is smaller compared to traditional angles.
03

Valuing the Experimental H-B-H Bond Angles

Experimentally, it is found that the bridging hydrogen atoms in diborane create an H-B-H bond angle that is approximately \(95^{\circ}\). This is due to the spatial constraints and electron pair sharing involving the bridge structure.
04

Identifying Terminal B-H Bond Angles

The terminal B-H bonds tend to be closer to the expected tetrahedral bond angle of \(120^{\circ}\) when viewed in the planar form of each boron tetrahedron fragment. These larger angles account for the rest of the molecule's arrangement.
05

Comparing Given Options with Known Angles

Compare the known bonding angles to the given choices:1. \(60^{\circ}, 120^{\circ}\) is incorrect due to unrealistic angles.2. \(95^{\circ}, 150^{\circ}\) does not match any known angles.3. \(95^{\circ}, 120^{\circ}\) matches the known angles of the bridging and terminal hydrogen bonds.4. \(120^{\circ}, 180^{\circ}\) includes an impossible angle for molecules like diborane.

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Key Concepts

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

Three-Center Two-Electron (3c-2e) Bonds
In chemistry, many students find the concept of three-center two-electron (3c-2e) bonds intriguing and a little puzzling at first. These bonds, which are also affectionately known as "banana bonds," are crucial for the stability and structure of certain complex molecules like diborane (\(\mathrm{B}_2\mathrm{H}_6\)). Unlike typical bonds that involve just two atoms and two electrons, 3c-2e bonds spread two electrons across three atoms. To visualize this, think of a banana-shaped cloud of electrons that bend around the three atoms, in this case, two boron atoms and one bridging hydrogen atom.
  • This bonding situation arises because boron, with only three valence electrons, does not have enough electrons to form conventional two-center bonds with all the hydrogen atoms in diborane.
  • Thus, the electrons are shared between three atoms, allowing the molecule to remain stable despite the electron deficiency.
For beginner chemists, understanding this bond develops a deeper appreciation of molecular dynamics and the creative ways in which atoms bond to form stable structures.
H-B-H Bond Angles
The H-B-H bond angles in diborane are a fascinating consequence of its unique structure. These angles arise between the hydrogen atoms that act as bridges in the diborane molecule. Experimentally, it has been determined that these angles are approximately \(95^{\circ}\).

This seemingly unusual angle is primarily due to the spatial arrangement forced by the three-center two-electron bonding. Since the electron cloud has to engage three atoms, the geometry is different from more conventional molecules, where bond angles might be larger or simpler to understand.
  • The small angle reduces strain by fitting the electron cloud appropriately around all involved atoms while maintaining overall molecular stability.
  • Understanding these unique angles is important for predicting molecule behavior and interactions with other compounds.
By comparing this angle with typical tetrahedral or planar arrangements, it emphasizes how diborane represents an important study in exceptions to basic bonding rules.
Boron Chemistry
Boron chemistry is a subject that opens the door to the study of elements that don't always conform to the standard rules, bringing fascinating exceptions into play. As a group 13 element, boron plays a unique role with its electron deficiency, leading to structures like diborane.

Due to its ability to form bonds like three-center two-electron bonds, boron often demonstrates flexibility and creativity in chemical bonding that isn't seen with other elements, especially those with a complete octet in their valence shell.
  • Because of its versatility, boron chemistry is essential for understanding compounds with unique properties, engaging in areas such as materials science and organic chemistry.
  • Furthermore, boron-containing compounds are used in areas ranging from organic synthesis to pharmaceuticals.
As you delve deeper into boron chemistry, it becomes clear how fundamental it is to overcoming electron deficiencies and generating stable molecular structures, displaying the elegance and complexity of chemistry.

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Most popular questions from this chapter

Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for this is that (a) strong affinity of \(\mathrm{HCl}\) gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke. (b) oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas (c) due to strong affinity for water, concen-trated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud. (d) concentrated hydrochloric acid emits strongly smelling HCl gas all the time.

\(\mathrm{XeO}_{3}\) can be prepared by: (a) \(\mathrm{XeF}_{2}\) hydrolysis (b) \(\mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow\) (c) \(\mathrm{XeF}_{4}\) hydrolysis (d) \(\mathrm{XeF}_{6}\) hydrolysis

Which of the following is the wrong statement? (a) Ozone is violet black in solid state (b) Ozone is diamagnetic gas (c) ONCl and \(\mathrm{ONO}^{-}\)are isoelectronic (d) \(\mathrm{O}_{3}\) molecule is bent

Borax is converted into B by following steps Borax \(\stackrel{\mathrm{A}}{\longrightarrow} \mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{B}_{2} \mathrm{O}_{3} \quad \mathrm{~B} \stackrel{\mathrm{B}}{\longrightarrow}\) Product Reagents \(\mathrm{A}\) and \(\mathrm{B}\) are (a) acid, \(\mathrm{Fe}\) (b) acid, \(\mathrm{Mg}\) (c) acid, \(\mathrm{Sn}\) (d) acid, Al

The bond dissociation energy of \(\mathrm{B}-\mathrm{F}\) in \(\mathrm{BF}_{3}\) is 646 \(\mathrm{kJ} \mathrm{mol}^{-1}\) whereas that of \(\mathrm{C}-\mathrm{F}\) in \(\mathrm{CF}_{4}\) is \(515 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The correct reason for higher B - F bond dissociation energy as compared to that of \(\mathrm{C}-\mathrm{F}\) is: (a) stronger bond between \(\mathrm{B}\) and \(\mathrm{F}\) in \(\mathrm{BF}_{3}\) as compared to that between \(\mathrm{C}\) and \(\mathrm{F}\) is \(\mathrm{CF}_{4}\) (b) significant \(\mathrm{p}-\mathrm{p}\) interaction between \(\mathrm{B}\) and \(\mathrm{F}\) in \(\mathrm{BF}_{3}\) whereas there is no possibility of such interaction between \(\mathrm{C}\) and \(\mathrm{F}\) in \(\mathrm{CF}_{4}\) (c) lower degree of \(\mathrm{p}-\mathrm{p}\) interaction between \(\mathrm{B}\) and \(\mathrm{F}\) \(\mathrm{BF}_{3}\) than that between \(\mathrm{C}\) and \(\mathrm{F}\) in \(\mathrm{CF}_{4}\) (d) smaller size of \(\mathrm{B}\) - atom as compared to that of C- atom

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