The pair of compounds having metals in their highest oxidation state is

Options

A. `MnO_(2) "and "CrO_(2)Cl_(2)`
B. `[Fe(CN)_(6)]^(3-) "and" [Cu(CN)_(4)]^(-2)`
C. `[NiCl_(4)]^(2-) "and" [CoCl_(4)]^(2-)`
D. `[FeCl_(4)]^(-)"and"Co_(2)O_(3)`

Answer

To determine the pair of compounds that have metals in their highest oxidation states, we will analyze each option provided in the question step by step.

### Step 1: Analyze Option A - MnO2 and CrO2Cl2
1. **Calculate the oxidation state of Mn in MnO2:**
- Oxygen (O) has an oxidation state of -2.
- In MnO2, there are 2 oxygen atoms:
\[ \text{Total oxidation state of O} = 2 \times (-2) = -4 \]
- Let the oxidation state of Mn be \( x \):
\[ x + (-4) = 0 \implies x = +4 \]
- Therefore, the oxidation state of Mn in MnO2 is +4.

2. **Calculate the oxidation state of Cr in CrO2Cl2:**
- Oxygen (O) has an oxidation state of -2 and there are 2 oxygen atoms:
\[ \text{Total oxidation state of O} = 2 \times (-2) = -4 \]
- Chlorine (Cl) has an oxidation state of -1 and there are 2 chlorine atoms:
\[ \text{Total oxidation state of Cl} = 2 \times (-1) = -2 \]
- Let the oxidation state of Cr be \( x \):
\[ x + (-4) + (-2) = 0 \implies x - 6 = 0 \implies x = +6 \]
- Therefore, the oxidation state of Cr in CrO2Cl2 is +6.

### Step 2: Analyze Option B - Fe(CN)6^3- and Cu(CN)4^2-
1. **Calculate the oxidation state of Fe in Fe(CN)6^3-:**
- Cyanide (CN) has an oxidation state of -1 and there are 6 cyanide ions:
\[ \text{Total oxidation state of CN} = 6 \times (-1) = -6 \]
- Let the oxidation state of Fe be \( x \):
\[ x + (-6) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Therefore, the oxidation state of Fe in Fe(CN)6^3- is +3.

2. **Calculate the oxidation state of Cu in Cu(CN)4^2-:**
- Again, cyanide (CN) has an oxidation state of -1 and there are 4 cyanide ions:
\[ \text{Total oxidation state of CN} = 4 \times (-1) = -4 \]
- Let the oxidation state of Cu be \( x \):
\[ x + (-4) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Therefore, the oxidation state of Cu in Cu(CN)4^2- is +2.

### Step 3: Analyze Option C - NiCl4^2- and CoCl4^2-
1. **Calculate the oxidation state of Ni in NiCl4^2-:**
- Chlorine (Cl) has an oxidation state of -1 and there are 4 chlorine atoms:
\[ \text{Total oxidation state of Cl} = 4 \times (-1) = -4 \]
- Let the oxidation state of Ni be \( x \):
\[ x + (-4) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Therefore, the oxidation state of Ni in NiCl4^2- is +2.

2. **Calculate the oxidation state of Co in CoCl4^2-:**
- Similarly, for Co:
\[ x + (-4) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Therefore, the oxidation state of Co in CoCl4^2- is +2.

### Step 4: Analyze Option D - FeCl4^- and Co2O3
1. **Calculate the oxidation state of Fe in FeCl4^-:**
- Chlorine (Cl) has an oxidation state of -1 and there are 4 chlorine atoms:
\[ \text{Total oxidation state of Cl} = 4 \times (-1) = -4 \]
- Let the oxidation state of Fe be \( x \):
\[ x + (-4) = -1 \implies x - 4 = -1 \implies x = +3 \]
- Therefore, the oxidation state of Fe in FeCl4^- is +3.

2. **Calculate the oxidation state of Co in Co2O3:**
- Oxygen (O) has an oxidation state of -2 and there are 3 oxygen atoms:
\[ \text{Total oxidation state of O} = 3 \times (-2) = -6 \]
- Let the oxidation state of Co be \( y \):
\[ 2y + (-6) = 0 \implies 2y - 6 = 0 \implies 2y = 6 \implies y = +3 \]
- Therefore, the oxidation state of Co in Co2O3 is +3.

### Conclusion:
After analyzing all options, we find:
- **Option A:** MnO2 (Mn +4) and CrO2Cl2 (Cr +6) have the highest oxidation states among the given compounds.

Thus, the final answer is:
**Option A: MnO2 and CrO2Cl2.**