Explain the hybridization and bonding in methane (CH₄) and ethyne (C₂H₂).

Carbon has an electron configuration of $1s^2 2s^2 2p^2$. To form four bonds, one electron from the $2s$ orbital is promoted to an empty $2p$ orbital, resulting in $1s^2 2s^1 2p^3$.

The carbon atom mixes one $2s$ orbital and three $2p$ orbitals to form four equivalent $sp^3$ hybrid orbitals.

Each of the four $sp^3$ hybrid orbitals on carbon contains one electron and overlaps head-on with the $1s$ orbital of a hydrogen atom (which also contains one electron) to form a strong sigma ($\sigma$) bond. This results in four C-H sigma bonds. The $sp^3$ hybridization leads to a tetrahedral geometry around the carbon atom with bond angles of $109.5^\circ$.

Each carbon atom has an electron configuration of $1s^2 2s^2 2p^2$. For bonding, one electron from the $2s$ orbital is promoted to an empty $2p$ orbital, resulting in $1s^2 2s^1 2p^3$.

Each carbon atom is bonded to one hydrogen and one other carbon atom, forming two electron domains. Therefore, each carbon atom undergoes $sp$ hybridization, mixing one $2s$ orbital and one $2p$ orbital to form two $sp$ hybrid orbitals. Two $2p$ orbitals remain unhybridized.

• One $sp$ hybrid orbital on each carbon overlaps with the $1s$ orbital of a hydrogen atom to form a C-H sigma ($\sigma$) bond.
• The other $sp$ hybrid orbital on each carbon overlaps head-on with the $sp$ hybrid orbital of the adjacent carbon atom to form a C-C sigma ($\sigma$) bond.
• The two unhybridized $2p$ orbitals on each carbon (oriented perpendicular to each other and to the $sp$ orbitals) overlap side-by-side with the corresponding unhybridized $2p$ orbitals on the adjacent carbon atom to form two pi ($\pi$) bonds. This results in a triple bond between the carbon atoms (one $\sigma$ and two $\pi$ bonds) and two C-H sigma bonds. The $sp$ hybridization leads to a linear geometry around each carbon atom with bond angles of $180^\circ$.

2. Valence Bond Theory and Bonding in NH₃

Nitrogen has an electron configuration of $1s^2 2s^2 2p^3$. It has one lone pair in the $2s$ orbital and three unpaired electrons in the $2p$ orbitals.

While the $2p$ orbitals could theoretically overlap with hydrogen $1s$ orbitals to form three bonds, this would predict bond angles of $90^\circ$. However, the experimentally observed bond angle in ammonia is approximately $107^\circ$. To account for this, Valence Bond Theory uses the concept of hybridization. The nitrogen atom undergoes $sp^3$ hybridization.

The nitrogen atom mixes its one $2s$ orbital and three $2p$ orbitals to form four equivalent $sp^3$ hybrid orbitals.

• One of these $sp^3$ hybrid orbitals contains the lone pair of electrons.
• The other three $sp^3$ hybrid orbitals each contain one unpaired electron and overlap head-on with the $1s$ orbital of a hydrogen atom (each containing one electron) to form three N-H sigma ($\sigma$) bonds.

The four $sp^3$ hybrid orbitals arrange themselves in a tetrahedral electron geometry. However, because one position is occupied by a lone pair, the molecular geometry of NH₃ is trigonal pyramidal. The lone pair exerts greater repulsion on the bonding pairs, compressing the H-N-H bond angles from the ideal $109.5^\circ$ to approximately $\boxed{\text{107}^\circ}$.