Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The decreasing values of bond angles from ${\text{NH}}_3$ ($107^{\circ }$) to ${\text{SbH}}_3$ ($91^{\circ }$) down group-15 of the periodic table is due to

Imagine you are looking at the molecular structures of the Group 15 hydrides. You start at the top with ammonia (${\text{NH}}_3$) and notice a bond angle of $107^{\circ }$. But as you travel down the periodic table to stibine (${\text{SbH}}_3$), the angle shrinks drastically to just $91^{\circ }$. Why does this happen? What invisible forces are at play here?

To solve this, we need to dive into the Valence Shell Electron Pair Repulsion (VSEPR) theory. According to VSEPR, the geometry of a molecule is dictated by the repulsions between electron pairs in the valence shell of the central atom. However, the intensity of this repulsion depends heavily on exactly where those electron pairs are located.

The position of the shared bond pairs is governed by a fundamental property: electronegativity. Electronegativity is an atom's ability to attract shared electrons in a chemical bond. As we move down Group 15 from Nitrogen ($\text{N}$) to Antimony ($\text{Sb}$), the atomic size increases, and the electronegativity steadily decreases.

This creates a fascinating tug-of-war for the electrons between the central atom and the hydrogen atoms.

High Electronegativity

Nitrogen is highly electronegative. In an ${\text{NH}}_3$ molecule, Nitrogen pulls the shared bond pairs of electrons strongly towards itself.

Because these bond pairs are pulled so close to the central Nitrogen atom, they become crowded in a very small volume of space. This intense crowding leads to a strong bond pair-bond pair (bp-bp) repulsion. To minimize this repulsive energy, the bonds are forced to spread apart, widening the $\text{H-N-H}$ angle to $107^{\circ }$.

Low Electronegativity

Now, let's look at Antimony in ${\text{SbH}}_3$. Antimony is much larger and far less electronegative than Nitrogen. It cannot pull the shared electrons as effectively.

As a result, the bond pairs sit further away from the central Antimony atom, closer to the Hydrogen atoms. Because they are spread out over a larger distance from the center, the distance between the bond pairs themselves is greater. This significantly reduces the bp-bp repulsion.

With less resistance from the bond pairs, the lone pair sitting on top of the Antimony atom can easily push the bonds closer together, shrinking the $\text{H-Sb-H}$ angle down to $91^{\circ }$.

Therefore, the decreasing values of bond angles down Group 15 are primarily due to the decreasing electronegativity of the central atom, which in turn decreases the bond pair-bond pair repulsion. This makes option (d) the correct answer.

As an advanced insight, the fact that the angles for ${\text{PH}}_3$, ${\text{AsH}}_3$, and ${\text{SbH}}_3$ are so close to $90^{\circ }$ is also beautifully explained by Drago's Rule. This rule suggests that for elements in Period 3 and below, when bonded to less electronegative atoms like Hydrogen, hybridization is practically absent. The central atom uses almost pure, unhybridized $p$-orbitals for bonding, which are naturally oriented at $90^{\circ }$ to each other!