Given the graph below which one of the following edges cannot be added in that…

Key idea: Kruskal's algorithm adds edges in non-decreasing order of weight. Any order in which edges are added must respect that ordering (edges with smaller weights are considered before larger ones).

• Edge A: a–b (weight 1)

• Edge B: d–f (weight 1)

• Edge C: b–f (weight 2)

• Edge D: d–c (weight 2)

• Edge E: d–e (weight 3)

Therefore, any valid order must place both weight-1 edges (A and B) before the weight-2 edges (C and D), and both weight-2 edges before the weight-3 edge (E).

1. A, B, C, D, E — weights: 1, 1, 2, 2, 3 → valid (non-decreasing).

2. A, B, D, C, E — weights: 1, 1, 2, 2, 3 → valid.

3. B, A, C, E, D — weights: 1, 1, 2, 3, 2 → invalid. This order puts a weight-3 edge (E) before a weight-2 edge (D), violating Kruskal's required non-decreasing processing order, so it cannot occur.

4. B, A, D, C, E — weights: 1, 1, 2, 2, 3 → valid.

Conclusion: The only order that cannot be produced by Kruskal's algorithm is B, A, C, E, D because it violates the non-decreasing weight order requirement.