Question

The geometry of \(\mathrm{CH}_{4}\) could be square planar, with the four \(\mathrm{H}\) atoms at the corners of a square and the \(\mathrm{C}\) atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral \(\mathrm{CH}_{4}\) molecule.

Short answer

The tetrahedral geometry of \(CH_4\) is more stable than the hypothetically proposed square planar shape due to less repulsion between bonding electron pairs resulting from larger H-C-H bond angles in the tetrahedral shape.

Step-by-step solution

Sketching the Square Planar Structure

Start by drawing the Carbon (C) atom at the center and four Hydrogen (H) atoms at the four corners of the square. Each hydrogen atom should be connected to the carbon atom by a straight line representing the bond.

Analyzing the Stability of the Square Planar Structure

In a square planar structure, the angle between bonds (H-C-H bond angle) will be 90 degrees. In terms of stability, there could be higher repulsion between the bonding electron pairs due to the smaller bond angles, leading to a less stable molecule.

Sketching and Analyzing the Tetrahedral Structure of \(CH_4\)

Draw a tetrahedron, with the Carbon atom at the center and the hydrogen atoms at the corners. The bond angles in a tetrahedral structure are approximately 109.5 degrees. This larger angle lessens electron pair repulsion, thus providing more stability.

Comparing the Square Planar and Tetrahedral Structure

Comparing both structures, the tetrahedral configuration, which has larger H-C-H bond angles and less electron pair repulsion, is more stable than the square planar configuration.