Explain the formation of four sigma bonds by $sp^3$ hybrid orbitals with an example.

(N/A) In methane $(CH_4)$,the ground state electron configuration of carbon is $[He] \ 2s^2 \ 2p^2$. In the excited state,one electron from the $2s$ orbital is promoted to the $2p_z$ orbital. The energy required for this excitation is compensated by the energy released during bond formation.
Carbon ground state: $C: [He] \ 2s^2 \ 2p_x^1 \ 2p_y^1 \ 2p_z^0$
Carbon excited state: $C^*: [He] \ 2s^1 \ 2p_x^1 \ 2p_y^1 \ 2p_z^1$
The four atomic orbitals $(2s, 2p_x, 2p_y, 2p_z)$ of the excited carbon atom undergo $sp^3$ hybridization to form four equivalent $sp^3$ hybrid orbitals.
These four $sp^3$ hybrid orbitals are directed towards the four corners of a regular tetrahedron,with an inter-orbital angle of $109.5^{\circ}$.
Each of these four $sp^3$ hybrid orbitals overlaps axially with the $1s$ orbital of a hydrogen atom to form four $C-H$ sigma $(\sigma)$ bonds.
As a result,the $CH_4$ molecule adopts a tetrahedral geometry with $H-C-H$ bond angles of $109.5^{\circ}$.