Explain the hybridization and bonding in methane (CH4) and ethyne (C2H2), including their molecular geometries.

The ground state electron configuration of carbon is $1s^2 2s^2 2p^2$. To form four bonds, one electron from the $2s$ orbital is promoted to an empty $2p$ orbital, resulting in an excited state configuration of $1s^2 2s^1 2p^3$.

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. Each $sp^3$ hybrid orbital 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 $\boxed{\text{109.5}^\circ}$.

Each carbon atom has an excited state electron configuration of $1s^2 2s^1 2p^3$.

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 $\boxed{\text{180}^\circ}$.

2. Valence Bond Theory and Bonding in NH₃

Nitrogen has an electron configuration of $1s^2 2s^2 2p^3$. In its valence shell, it has one lone pair in the $2s$ orbital and three unpaired electrons in the $2p$ orbitals.

To explain the observed bond angles and geometry, the nitrogen atom undergoes $sp^3$ hybridization. It 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 $\boxed{\text{107}^\circ}$.

3. Bonding in Benzene (C₆H₆)

Since you mentioned benzene in your notes, here's a brief explanation:

Each carbon atom in benzene is bonded to two other carbon atoms and one hydrogen atom, forming three electron domains. Therefore, each carbon atom undergoes $sp^2$ hybridization, mixing one $2s$ orbital and two $2p$ orbitals to form three $sp^2$ hybrid orbitals. One $2p$ orbital remains unhybridized.

• Each $sp^2$ hybrid orbital on a carbon atom forms a sigma ($\sigma$) bond: two with adjacent carbon atoms and one with a hydrogen atom. This forms the hexagonal ring structure and the C-H bonds.
• The unhybridized $2p$ orbital on each carbon atom (perpendicular to the plane of the ring) overlaps side-by-side with the unhybridized $2p$ orbitals of its neighboring carbon atoms. This forms a delocalized pi ($\pi$) electron system above and below the plane of the ring, giving benzene its characteristic stability. The $sp^2$ hybridization leads to a trigonal planar geometry around each carbon atom with bond angles of $\boxed{\text{120}^\circ}$.