1. A topological order places each vertex before every vertex it points to.
2. If there were a directed cycle, each of its vertices would have to precede the next and so, going round, precede itself — impossible.
3. Conversely, an acyclic directed graph always has a vertex with no incoming edge; removing it repeatedly produces a valid order.
4. Connectivity is irrelevant, since a graph in several pieces can still be ordered. So a topological order exists exactly when the graph contains no directed cycle.

Answer: contains no directed cycle.

Topological sorting and cycle detection are the same computation: Kahn's algorithm outputs an order, and any vertices it cannot output lie on a cycle.

Shortcut. A topological order exists exactly for DAGs: a directed cycle would force a vertex to come before itself, and connectivity plays no part.

Common trap. "is strongly connected" sounds rigorous, but a strongly connected graph with two or more vertices always contains a cycle, so it rules an order out.