Question

The four covalent bonds in methane ($\text{CH}_4$) are arranged around carbon to give which one of the following geometries?

• A. Linear
• B. Tetrahedral
• C. Trigonal planar
• D. Trigonal pyramidal

Hint:

Any central atom with four single covalent bonds and zero lone pairs (4 steric number) will always adopt a perfect tetrahedral geometry to maximize the distance between the bonding electron pairs.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The spatial arrangement of covalent bonds around a central atom is determined by the hybridization of its valence orbitals and the Valence Shell Electron Pair Repulsion (VSEPR) theory.
VSEPR theory states that electron pairs around a central atom arrange themselves to minimize electrostatic repulsion.

Step 2: Detailed Explanation:
In methane ($\text{CH}_4$), the central carbon atom has four valence electrons.
It undergoes $\text{sp}^3$ hybridization, combining one $2\text{s}$ and three $2\text{p}$ orbitals to form four equivalent $\text{sp}^3$ hybrid orbitals.
These four hybrid orbitals overlap with the $1\text{s}$ orbitals of four hydrogen atoms to form four equivalent $\text{C--H}$ $\sigma$-bonds.
To minimize electron-electron repulsion between these four bonding pairs, the bonds project toward the four corners of a regular tetrahedron.
The bond angles between the adjacent $\text{C--H}$ bonds are exactly $109.5^{\circ}$.
A linear geometry corresponds to $\text{sp}$ hybridization, trigonal planar corresponds to $\text{sp}^2$ hybridization, and trigonal pyramidal is typical of molecules with one lone pair (such as ammonia, $\text{NH}_3$).

Step 3: Final Answer:
Thus, the geometry of the four covalent bonds in methane is tetrahedral, corresponding to option (B).