The correct answer is (4) Acidic, basic, and amphoteric.
1. \(Mn_2O_7\) (Manganese Heptoxide):
\(Mn_2O_7\) is an oxide of manganese. In this compound, each manganese atom is in the +7 oxidation state. It's important to note that the oxidation state of the metal in an oxide can give us clues about its acidic or basic properties.
- Manganese (Mn) in \(Mn_2O_7\) is in a high oxidation state (+7).
- High oxidation state metals tend to form acidic oxides.
So, \(Mn_2O_7\) is acidic because it can release hydrogen ions (\(H^+\)) when dissolved in water, making it an acid.
2. \(CrO\) (Chromium(II) Oxide):
\(CrO\) is an oxide of chromium. In this compound, each chromium atom is in the +2 oxidation state. Again, let's consider the oxidation state of the metal to determine its properties.
- Chromium (Cr) in \(CrO\) is in a low oxidation state (+2).
- Low oxidation state metals tend to form basic oxides.
So, \(CrO\) is basic because it can accept hydrogen ions (\(H^+\)) when dissolved in water, making it a base.
3. \(V_2O_5\) (Vanadium Pentoxide):
\(V_2O_5\) is an oxide of vanadium. In this compound, each vanadium atom is in the +5 oxidation state. As before, let's consider the oxidation state of the metal.
- Vanadium (V) in \(V_2O_5\) is in an intermediate oxidation state (+5).
- Intermediate oxidation-state metals can exhibit amphoteric behavior. Amphoteric compounds can act as both acids and bases depending on the conditions.
So, \(V_2O_5\) is amphoteric because it can either donate or accept hydrogen ions (\(H^+\)) depending on the pH of the solution in which it's dissolved.
In summary:
- \(Mn_2O_7\) is acidic due to the high oxidation state of manganese.
- \(CrO\) is basic due to the low oxidation state of chromium.
- \(V_2O_5\) is amphoteric due to the intermediate oxidation state of vanadium.
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