Explanation

(d)
Concentrated sulphuric acid, being a strong acid, oxidises bromides and iodides but not chlorides and fluorides since, the later are more electronegative. Hence it can be reduced only by NaBr among the given options.
+6 -1 +6 -1
H_{2} SO_{4} + NaBr ⟶ NaHSO_{4} + HBr
-1 +6 0 +4
2HBr + H_{2}SO_{4} ⟶2H_{2}O + Br_{2} + SO_{2}
_{reduction}