Final answer:
An atom with sp² hybridization forms sigma bonds with sp² orbitals and pi bonds with the remaining unhybridized p orbital, thus making the statement true.
Step-by-step explanation:
The statement that an atom with sp² hybridization has sigma bonds and p orbital(s) unchanged that form pi bonds is true. sp² hybridization involves the combination of one s orbital and two p orbitals, resulting in three sp² hybrid orbitals that lie in a single plane. The remaining unhybridized p orbital is oriented perpendicular to this plane and is responsible for forming pi (π) bonds through side-by-side overlap with an adjacent unhybridized p orbital. As seen in molecules like C₂H₄ (ethylene), the sigma (σ) bonds are formed by the end-to-end overlap of sp² hybrid orbitals with other orbitals, while the pi bonds are formed by the side-by-side overlap of the unhybridized 2p orbitals.