Which statement best explains, using hybridization, why carbon can form exactly four bonds in its most common compounds?

• A. Carbon's $2_s$ and $2_p$ orbitals remain unhybridized, allowing it to form only two bonds.
• B. Carbon forms four bonds because it has four electrons in its $1_s$ orbital.
• C. Carbon uses only its $p$ orbitals for bonding, which limits it to three bonds.
• D. Carbon undergoes ${\mathrm{sp}}^3$ hybridization, creating four equivalent orbitals that can each form a sigma bond with another atom.

Answer: (D)

Verified step by step guidance

1. Recall that carbon's ground-state electron configuration is 1s² 2s² 2p², meaning it has four valence electrons in the second shell: two in the 2s orbital and two in the 2p orbitals.
2. Understand that in order to form four equivalent bonds, carbon undergoes sp³ hybridization, where one 2s orbital mixes with three 2p orbitals to form four new equivalent hybrid orbitals.
3. Each of these four sp³ hybrid orbitals contains one unpaired electron, allowing carbon to form four sigma (σ) bonds with other atoms by overlapping these orbitals with orbitals from other atoms.
4. Recognize that this hybridization explains why carbon forms exactly four bonds in its most common compounds, as it has four equivalent orbitals available for bonding.
5. Note that the other options are incorrect because they either ignore hybridization or misrepresent the electron configuration and bonding capacity of carbon.