Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

Select the correct statements about diborane: (a) \(\mathrm{H}_{\mathrm{b}} \ldots \mathrm{B} \ldots \mathrm{H}_{\mathrm{b}}\) bond angle is \(122^{\circ}\) (b) All hydrogens in \(\mathrm{B}_{2} \mathrm{H}_{6}\) lie in the same plane (c) \(\mathrm{B}_{2} \mathrm{H}_{6}\) has three centered bond (d) Each boron atom lies in sp \(^{3}\) hybrid state

Step-by-Step Solution

Answer

Only statements (c) and (d) are correct.

Step 1: Understand Diborane Structure

Diborane, denoted as \( \mathrm{B}_2\mathrm{H}_6 \), is a compound composed of two boron atoms and six hydrogen atoms. It features a unique bonding system with two types of hydrogen atoms: terminal hydrogens connected directly to a single boron atom and bridging hydrogens which form bridge bonds between the boron atoms.

Step 2: Analyze Statement (a)

Statement (a) claims the \( \mathrm{H}_{\mathrm{b}} \ldots \mathrm{B} \ldots \mathrm{H}_{\mathrm{b}} \) bond angle is \( 122^{\circ} \). In diborane, the \( \mathrm{H}_{\mathrm{b}} \ldots \mathrm{B} \ldots \mathrm{H}_{\mathrm{b}} \) bridges have bond angles close to \( 90^{\circ} \), not \( 122^{\circ} \). Therefore, this statement is incorrect.

Step 3: Evaluate Statement (b)

This statement suggests all hydrogens in \( \mathrm{B}_{2} \mathrm{H}_{6} \) lie in the same plane. In reality, diborane's structure is such that the bridging hydrogen atoms create a non-planar, banana-shaped bond, making this statement incorrect.

Step 4: Consider Statement (c)

The structure of diborane includes two hydrogen atoms forming three-center two-electron bonds (\( \mathrm{B}-\mathrm{H}-\mathrm{B} \)), which are indeed three-centered bonds. Thus, statement (c) is correct.

Step 5: Verify Statement (d)

Statement (d) suggests each boron atom in diborane is in an \( ext{sp}^3 \) hybrid state. In diborane, each boron atom forms a total of four bonds, which is consistent with \( ext{sp}^3 \) hybridization. Thus, this statement is correct.

Key Concepts

Three-Center Two-Electron Bond

In the world of chemistry, the concept of the three-center two-electron bond is quite intriguing, especially when exploring molecules like diborane (\( \mathrm{B}_2\mathrm{H}_6 \)). In a typical two-center bond, two atoms share an electron pair, making it straightforward and common. However, when it comes to diborane, something unique happens. The three-center two-electron bond involves three atoms sharing just two electrons. Here, two boron atoms and a bridging hydrogen atom share this bond format.

These bonds are not just any bonds—they're crucial to maintaining the structure of diborane with its often-discussed, fascinating "banana" shape. This structure accommodates the need for boron to achieve stability while having an incomplete octet. Essentially:

• The boron atoms lack sufficient electrons to form full octet bonds independently.
• The bridging hydrogen atoms essentially "bridge the gap" by participating in the electron-sharing.

In diborane, two sets of these bonds occur, keeping the entire molecule stable and illustrating the often unusual, yet brilliantly practical, solutions that chemical structures can manifest.

sp3 Hybridization

Hybridization is a foundational concept in chemistry that explains the merging of atomic orbitals to form new, equivalent hybrid orbitals. For boron atoms in diborane, they undergo \( \mathrm{sp}^3 \) hybridization, as each forms a total of four bonds, consistent with this hybrid state.

What exactly does \( \mathrm{sp}^3 \) hybridization entail? Let's break it down:

• One s orbital mixes with three p orbitals of boron to form four \( \mathrm{sp}^3 \) hybrid orbitals.
• These hybrid orbitals are equivalent and arranged in a tetrahedral geometry, explaining the overall bonding and angles seen in certain parts of the diborane molecule.

Despite the complicated look of diborane’s structure, hybridization simplifies our understanding of bond formations and molecular geometry.

For boron within diborane, even with the unusual three-center two-electron bonds, the concept of \( \mathrm{sp}^3 \) hybridization provides insight into how boron can bond with four other atoms (whether they be hydrogen in terminal or bridging positions) effectively within the molecule.

Molecular Geometry of B2H6

The molecular geometry of diborane (\( \mathrm{B}_2\mathrm{H}_6 \)) is quite uncommon in chemistry, primarily due to its significant reliance on those three-center two-electron bonds that depart from the norm of planar or straightforward geometries. Instead, imagine a shape closer to the folded wings of a butterfly or a seesaw, that’s often described as "banana-shaped" for its bridging bonds.

Diborane's molecular structure is characterized by:

• The four terminal hydrogen atoms that align in two pairs, each bonding directly to one of the boron atoms.
• The two bridging hydrogens that connect the boron atoms over the top in a curved, graceful arc.
• The bond angles of these bridges are close to \( 90^{\circ} \), deviating significantly from expected angles seen in typical molecules with regular tetrahedral or planar bond formations.

This configuration is not merely for aesthetic variety but rather out of chemical necessity, effectively allowing boron to address its electron deficiency and form a stable molecular structure within known chemical parameters. Through studying diborane's geometry, chemists can appreciate the lengths to which atoms will go, balancing complexity with the quest for stability and the subtle dances of electron sharing.