The compound with the lowest oxidation state is:

Options

A. `Fe_(2)O`
B. `Cr^(2+)`
C. `Ti^(4+)`
D. `Mn^(2+)`

Answer

To determine the compound with the lowest oxidation state, we will analyze the oxidation states of the central metal atoms in the given compounds. Let's break it down step by step.

### Step 1: Analyze the first compound, Fe2O
- We denote the oxidation state of iron (Fe) as \( X \).
- Since there are two iron atoms, the total contribution from iron is \( 2X \).
- Oxygen (O) typically has an oxidation state of \(-2\).
- The overall charge of the compound is neutral (0), so we can set up the equation:
\[ 2X + (-2) = 0 \]
- Solving for \( X \):
\[ 2X - 2 = 0 \\ 2X = 2 \\ X = +1 \]
- Therefore, the oxidation state of Fe in Fe2O is +1.

### Step 2: Analyze the second compound, Cr2+
- The oxidation state of chromium (Cr) is given directly as +2.
- Therefore, the oxidation state of Cr in Cr2+ is +2.

### Step 3: Analyze the third compound, Ti4+
- The oxidation state of titanium (Ti) is given directly as +4.
- Therefore, the oxidation state of Ti in Ti4+ is +4.

### Step 4: Analyze the fourth compound, Mn2+
- The oxidation state of manganese (Mn) is given directly as +2.
- Therefore, the oxidation state of Mn in Mn2+ is +2.

### Step 5: Compare the oxidation states
- We have calculated the following oxidation states:
- Fe in Fe2O: +1
- Cr in Cr2+: +2
- Ti in Ti4+: +4
- Mn in Mn2+: +2

### Conclusion
- The compound with the lowest oxidation state is Fe2O, where the oxidation state of Fe is +1.

Thus, the answer is **Fe2O**.

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