Q.Apart from tetrahedral geometry, another possible geometry for $CH_4$ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why $CH_4$ is not square planar?

Concept: VSEPR Theory (Valence Shell Electron Pair Repulsion)

Carbon in $CH_4$ has four valence electrons that form four equivalent $\ce{C-H}$ single bonds. These four bonding pairs arrange themselves to minimize electron-pair repulsion.

In a tetrahedral geometry, the bond angles are $109.5°$, which maximizes the distance between electron pairs in three-dimensional space. This is the most stable arrangement for four equivalent bonds.

In a square planar geometry, the bond angles would be $90°$ between adjacent bonds and $180°$ between opposite bonds. The $90°$ angles place electron pairs much closer together than in the tetrahedral case, creating significantly greater repulsion and higher energy. …

The question asks us to use fundamental principles of molecular geometry to rule out a square planar structure for methane. The key lies in understanding how electron pairs arrange themselves in space.

Why VSEPR Theory governs molecular shape

Valence Shell Electron Pair Repulsion (VSEPR) theory rests on a simple physical idea: electron pairs—whether bonding or lone pairs—are negatively charged clouds that repel one another. To minimize this repulsion, they spread out as far apart as possible in three-dimensional space. The geometry that maximizes the distance between electron pairs is the one nature adopts.

For a central atom surrounded by four electron pairs, we need to ask: what arrangement in 3D space keeps them maximally separated?

Comparing the two geometries

Let's examine both proposed structures for $CH_4$:

1. Tetrahedral geometry

   The carbon atom sits at the center of a regular tetrahedron, with the four hydrogen atoms at the vertices. Every $\ce{H-C-H}$ bond angle is $109.5°$. This is the geometry of four points on a sphere that are as far apart as possible—it's the solution to maximizing mutual distances in three dimensions.

2. Square planar geometry

   The carbon atom lies at the center of a square, with the four hydrogen atoms at the corners. All atoms are coplanar. The $\ce{H-C-H}$ bond angles are $90°$ between adjacent hydrogens and $180°$ between opposite hydrogens.

Now compare the electron-pair repulsions:

In the square planar arrangement, the four $\ce{C-H}$ bonding pairs are compressed into a single plane. Adjacent bonding pairs are separated by only $90°$, which is significantly smaller than the tetrahedral angle of $109.5°$. This means the electron clouds are much closer together, leading to much stronger repulsion.

3. Energy consideration …