Animated Solution for Chemistry - s and p-Block Elements: The correct statement about ${\text{B}}_2{\text{H}}_6$ is

Diborane (${\text{B}}_2{\text{H}}_6$) is one of the most fascinating molecules in inorganic chemistry. At first glance, it looks like it should have a structure similar to ethane (${\text{C}}_2{\text{H}}_6$). However, boron only has three valence electrons, meaning there simply aren't enough electrons to form a standard single bond between the two boron atoms while keeping all the hydrogen atoms attached.

To solve this electron deficiency, nature employs a brilliant workaround: the 3-center 2-electron (3c-2e) bond, affectionately known as the "banana bond." In diborane, two hydrogen atoms act as bridges between the two boron atoms. These bridging hydrogens share a single pair of electrons across three atoms (B-H-B), creating a highly unique and stable geometry.

Angles and Lengths

Let's systematically break down the given statements to find the truth.

Option (a) claims all B-H-B angles are $120^{\circ }$. This is a common misconception. While the terminal H-B-H angle is indeed close to $120^{\circ }$, the bridging B-H-B angle is much more acute, measuring approximately $83^{\circ }$. The geometric constraints of the bridge force this angle to be smaller. Thus, option (a) is incorrect.

Option (b) suggests the two B-H-B bonds are of different lengths. In reality, the diborane molecule is highly symmetric. The two bridging B-H bonds are perfectly equivalent, both measuring around $134\text{ pm}$. Interestingly, these bridging bonds are longer and weaker than the terminal B-H bonds ($119\text{ pm}$) because the electron density is smeared over three atoms instead of two. Option (b) is also incorrect.

Option (d) states that the ${\text{BH}}_3$ fragment behaves as a Lewis base. Let's recall the definition of a Lewis base: an electron-pair donor. If we look at the ${\text{BH}}_3$ monomer, the central boron atom is surrounded by only six valence electrons (three from boron, three from the hydrogens). It has an incomplete octet!

Because it is electron-deficient, ${\text{BH}}_3$ desperately wants to accept a pair of electrons to achieve stability. Therefore, it acts as a classic Lewis acid, not a Lewis base. Option (d) is incorrect.

This brings us to Option (c), which compares the p-character of the terminal and bridging bonds. To understand this, we must invoke Bent's Rule.

Bent's rule provides a profound insight into hybridization: Atomic s-character concentrates in orbitals directed toward electropositive substituents, or more simply, larger bond angles correspond to a higher fractional s-character.

Let's apply this to diborane: 1. The terminal H-B-H angle is large ($\sim 120^{\circ }$). A larger angle implies that the hybrid orbitals forming these bonds have more s-character (closer to $s{p}^2$ hybridization, which is 33% s and 66% p). 2. The bridging H-B-H angle is smaller ($\sim 97^{\circ }$). A smaller angle implies less s-character and, consequently, more p-character (closer to $s{p}^3$ hybridization, which is 25% s and 75% p).

Since the terminal bonds have more s-character, they must mathematically have less p-character compared to the bridging bonds.

By meticulously analyzing the geometry and electronic structure of diborane, we have proven that terminal B-H bonds indeed possess less p-character than their bridging counterparts. Statement (c) stands as the sole correct assertion, showcasing the beautiful interplay between molecular geometry and orbital hybridization.