After ${\mathrm{sp}}^3$ hybridization, the carbon atom has:

• A. four equivalent ${\mathrm{sp}}^3$ hybrid orbitals arranged tetrahedrally
• B. two equivalent ${\mathrm{sp}}^2$ hybrid orbitals and two unhybridized $p$ orbitals
• C. three equivalent ${\mathrm{sp}}^2$ hybrid orbitals and one unhybridized $p$ orbital
• D. two equivalent ${\mathrm{sp}}^1$ hybrid orbitals and two unhybridized $p$ orbitals

Answer: (A)

Verified step by step guidance

1. Recall that hybridization involves mixing atomic orbitals to form new equivalent hybrid orbitals suitable for bonding.
2. For sp³ hybridization, one s orbital mixes with three p orbitals, resulting in four equivalent sp³ hybrid orbitals.
3. These four sp³ hybrid orbitals arrange themselves in a tetrahedral geometry to minimize electron repulsion, with bond angles of approximately 109.5°.
4. Since all four orbitals are hybridized, there are no remaining unhybridized p orbitals on the carbon atom in sp³ hybridization.
5. Therefore, the carbon atom after sp³ hybridization has four equivalent sp³ hybrid orbitals arranged tetrahedrally.