Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct order of bond angles (smallest first) in ${\text{H}}_2\text{S}$, ${\text{NH}}_3$, ${\text{BF}}_3$ and ${\text{SiH}}_4$ is

Imagine you are an architect, but instead of buildings, you are designing molecules. The foundation of molecular architecture is the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states a very simple but profound truth: electron pairs around a central atom hate each other. They want to stay as far apart as physically possible to minimize repulsion.

However, not all electron pairs are created equal. We have bond pairs (electrons shared between two atoms) and lone pairs (electrons chilling exclusively on the central atom). Because lone pairs are pulled by only one nucleus, they spread out more in space and exert a stronger repulsive force. The golden rule of VSEPR is the repulsion order: $\text{lp-lp}>\text{lp-bp}>\text{bp-bp}$. The more lone pairs you have, the more they squeeze the bond angles together.

Let's start with the molecules that have zero lone pairs. These are the "perfect" geometries where all electron pairs repel each other equally.

Take ${\text{BF}}_3$ (Boron Trifluoride). Boron has 3 valence electrons and forms 3 bonds with Fluorine. With 3 bond pairs and 0 lone pairs, the molecule adopts a flat, symmetrical trigonal planar geometry. The bonds spread out evenly in a circle, giving a maximum bond angle of $120^{\circ }$.

Next, look at ${\text{SiH}}_4$ (Silicon Hydride). Silicon, just like Carbon, has 4 valence electrons and forms 4 bonds. With 4 bond pairs and 0 lone pairs, the molecule forms a beautiful 3D tetrahedral shape. The bonds are perfectly spaced out, resulting in the classic tetrahedral bond angle of $109^{\circ }28^{\prime }$.

Now, let's introduce some lone pairs and watch the geometry distort.

Consider ${\text{NH}}_3$ (Ammonia). Nitrogen has 5 valence electrons. It uses 3 to bond with Hydrogen, leaving 1 lone pair. This lone pair acts like a bulky cloud, pushing down on the three $\text{N-H}$ bonds. Because $\text{lp-bp}$ repulsion is stronger than $\text{bp-bp}$ repulsion, the bonds are squeezed closer together. The perfect tetrahedral angle shrinks down to $107^{\circ }$, giving the molecule a trigonal pyramidal shape.

Finally, we arrive at ${\text{H}}_2\text{S}$ (Hydrogen Sulfide). Sulfur has 6 valence electrons. It forms 2 bonds with Hydrogen, leaving 2 lone pairs. Now we have the ultimate repulsive force: $\text{lp-lp}$ repulsion! These two lone pairs aggressively push the $\text{S-H}$ bonds together. Furthermore, according to Drago's Rule, elements in the third period and below (like Sulfur) tend not to hybridize when bonded to less electronegative atoms. They use their pure, unhybridized $p$-orbitals for bonding, which are naturally at $90^{\circ }$ to each other. Thus, the bond angle in ${\text{H}}_2\text{S}$ is drastically reduced to approximately $90^{\circ }$.

By analyzing the repulsive forces and the number of lone pairs, we can easily rank these molecules. The more lone pairs, the smaller the angle.

- ${\text{H}}_2\text{S}$: 2 lone pairs $\to$ $90^{\circ }$ - ${\text{NH}}_3$: 1 lone pair $\to$ $107^{\circ }$ - ${\text{SiH}}_4$: 0 lone pairs (Tetrahedral) $\to$ $109^{\circ }28^{\prime }$ - ${\text{BF}}_3$: 0 lone pairs (Trigonal Planar) $\to$ $120^{\circ }$

Therefore, the correct increasing order of bond angles is ${\text{H}}_2\text{S}<{\text{NH}}_3<{\text{SiH}}_4<{\text{BF}}_3$.