Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Bond angle of $109^{\circ }28^{\prime }$ is found in

When you see the angle $109^{\circ }28^{\prime }$ (or $109.5^{\circ }$), your mind should immediately jump to one specific geometric shape: the perfect regular tetrahedron.

In the realm of chemical bonding, this angle is the hallmark of an atom that is ${\text{sp}}^3$ hybridized and is surrounded symmetrically by four identical bond pairs, with absolutely zero lone pairs to distort the symmetry. Let's embark on a journey through the given options to see how lone pairs act as the invisible bullies of molecular geometry.

VSEPR Theory

Before we analyze the molecules, we must recall the fundamental rule of the Valence Shell Electron Pair Repulsion (VSEPR) theory. Lone pairs of electrons are held by only one nucleus, making their electron clouds fatter and more expansive than those of bond pairs.

Because they take up more space, they push the adjacent bond pairs closer together. The order of repulsion is strictly:

This means that for every lone pair you add to an ${\text{sp}}^3$ hybridized atom, the ideal $109.5^{\circ }$ angle will shrink.

Let's put our options under the microscope:

1. Ammonia (${\text{NH}}_3$) Nitrogen in ammonia is ${\text{sp}}^3$ hybridized. It forms three bonds with hydrogen and keeps one lone pair for itself. This single lone pair pushes the three $\text{N-H}$ bonds downward, compressing the bond angle from $109.5^{\circ }$ to approximately $107^{\circ }$. The resulting shape is a trigonal pyramid.

2. Water (${\text{H}}_2\text{O}$) Oxygen in water is also ${\text{sp}}^3$ hybridized, but it is even greedier—it holds onto two lone pairs! The intense repulsion between these two lone pairs forces the $\text{O-H}$ bonds even closer together. The angle shrinks drastically to about $104.5^{\circ }$, giving water its characteristic bent or V-shape.

3. The Mythical ${\text{CH}}_5^{+}$ This option is a classic trap! Carbon belongs to the second period and has only four valence electrons. It can form a maximum of four bonds to complete its octet. To form ${\text{CH}}_5^{+}$, carbon would need to form five bonds, which violates the octet rule because it lacks d-orbitals to expand its valency. Therefore, under normal conditions, this species simply does not exist.

Ammonium Ion

Finally, we arrive at the ammonium ion, ${\text{NH}}_4^{+}$.

How is it formed? Ammonia (${\text{NH}}_3$) donates its lone pair to a proton (${\text{H}}^{+}$) via a coordinate covalent bond. Once formed, all four $\text{N-H}$ bonds are completely identical.

Nitrogen is now ${\text{sp}}^3$ hybridized with four bond pairs and zero lone pairs. Without any lone pairs to bully the bonds, the four hydrogen atoms spread out as far apart as geometrically possible in 3D space. They settle perfectly at the corners of a regular tetrahedron, locking the bond angle exactly at $109^{\circ }28^{\prime }$.

Thus, the correct answer is undeniably ${\text{NH}}_4^{+}$.