Explain the tetrahedral structure of $CH_4$ (methane) using its bonding structure.

(N/A) In $CH_4$,the ground state electronic configuration of $C$ is $[He] 2s^2 2p^2$. One electron from $2s^2$ is promoted to the empty $2p$ orbital to form an excited state carbon atom $(C^*)$. The energy required for this excitation is compensated by the energy released during bond formation.
This excited carbon atom has four half-filled orbitals. These four half-filled orbitals $(2s, 2p_x, 2p_y, 2p_z)$ undergo $sp^3$ hybridization to form four $sp^3$ hybrid orbitals.
The four $sp^3$ hybrid orbitals are arranged in a tetrahedral orientation at an angle of $109.5^{\circ}$ pointing towards the four corners of a tetrahedron.
These four half-filled hybrid orbitals overlap with the half-filled $1s$ orbitals of four $H$ atoms along the internuclear axis to form four $C-H$ $\sigma$ bonds.
As a result,in $CH_4$,all four $C-H$ bonds are arranged in a tetrahedral geometry in three-dimensional space with a bond angle of $109.5^{\circ}$. Thus,$CH_4$ has a tetrahedral shape.