Step 1: Understanding the Concept:
An oxidizing agent gets reduced during a chemical reaction. A compound tends to act as a strong oxidizing agent when its central metal atom is in a highly unstable maximum oxidation state.
Step 2: Key Formula or Approach:
Determine the oxidation state of each metal and assess its relative stability based on its position in the d-block groups. Heavier transition metals stabilize higher oxidation states better than lighter ones.
Step 3: Detailed Explanation:
1. Chromium (VI) compounds: In $CrO_{3}$, $CrO_{4}^{2-}$, and $Cr_{2}O_{7}^{2-}$, Chromium is in its highest group oxidation state ($+6$). In the first transition series (3d), this $+6$ state is quite unstable and eagerly accepts electrons to reduce to the stable $Cr^{3+}$ state, making these species very strong oxidizing agents.
2. Molybdenum (VI) compound: Molybdenum is in the same group as Chromium but in the second transition series (4d). A key trend in the d-block is that heavier elements form much more stable higher oxidation states compared to the 3d elements. Therefore, $MoO_{3}$ (where Mo is $+6$) is thermodynamically very stable and does not readily act as an oxidizing agent.
Step 4: Final Answer:
$MoO_{3}$ will not act as a typical oxidizing agent under normal conditions compared to the Cr(VI) species.