Final answer:
The chromium compounds K₂Cr₂O₇ and BaCrO₄ can act only as oxidants since chromium is in the high oxidation state of +6. Cr(OH)₂ and [Cr(H₂O)₆]Cl₃ can act as reducing agents or undergo disproportionation as chromium is in lower oxidation states. The transition from HAsO₄² to AsH₃ involves a reduction of arsenic from +5 to -3 in oxidation state.
Step-by-step explanation:
To determine the oxidation states of chromium in the given compounds, we need to understand the general rules for assigning oxidation states. In general, for a compound, the sum of the oxidation states equals the overall charge of the compound.
K₂Cr₂O₇: Potassium has an oxidation state of +1, and oxygen has an oxidation state of -2. Since the compound is neutral, the two chromium atoms together must have an oxidation state such that when added to the potassium and oxygen oxidation states, the total is zero. This leads to an oxidation state of +6 for each Cr atom.
[Cr(H₂O)₆]Cl₃: The complex is neutral, and the chloride ions are outside the complex, so they do not affect the oxidation state of Cr, which is +3 here.
Cr(OH)₂: Hydroxide has an oxidation state of -1, and since there are two of them, Cr must be +3 here as well to make the compound neutral.
BaCrO₄: Barium has an oxidation state of +2, oxygen -2; thus, chromium must be +6 to balance the compound to neutral.
For the redox behavior of these compounds:
Compounds where Cr is in its highest oxidation state, +6 (K₂Cr₂O₇ and BaCrO₄), can act only as oxidants.
Compounds where Cr is in a lower oxidation state (Cr(OH)₂ and [Cr(H₂O)₆]Cl₃) can act as reducing agents or undergo disproportionation.
Regarding the transition from HAsO₄² to AsH₃:
As changes from +5 (HAsO₄²) to -3 (AsH₃) in oxidation state, which means it is being reduced (gain of electrons). Therefore, the correct answer is B) Reduction.