The metal with the highest oxidation state present in `K_2CrO_4, NbCl_5 and MnO_2` is

Options

A. `Nb`
B. Mn
C. K
D. Cr

Explanation

To determine the metal with the highest oxidation state present in the compounds \( K_2CrO_4 \), \( NbCl_5 \), and \( MnO_2 \), we will calculate the oxidation states of the metals in each compound step by step.

### Step 1: Calculate the oxidation state of chromium in \( K_2CrO_4 \)

1. The formula for potassium chromate is \( K_2CrO_4 \).
2. Potassium (K) has an oxidation state of +1. Since there are 2 potassium atoms, their total contribution is \( 2 \times (+1) = +2 \).
3. Oxygen (O) typically has an oxidation state of -2. There are 4 oxygen atoms, so their total contribution is \( 4 \times (-2) = -8 \).
4. Let the oxidation state of chromium (Cr) be \( x \).
5. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ +2 + x + (-8) = 0 \]
6. Simplifying the equation:
\[ x - 6 = 0 \implies x = +6 \]

### Step 2: Calculate the oxidation state of niobium in \( NbCl_5 \)

1. The formula for niobium pentachloride is \( NbCl_5 \).
2. Chlorine (Cl) has an oxidation state of -1. Since there are 5 chlorine atoms, their total contribution is \( 5 \times (-1) = -5 \).
3. Let the oxidation state of niobium (Nb) be \( y \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ y + (-5) = 0 \]
5. Simplifying the equation:
\[ y - 5 = 0 \implies y = +5 \]

### Step 3: Calculate the oxidation state of manganese in \( MnO_2 \)

1. The formula for manganese dioxide is \( MnO_2 \).
2. Again, oxygen (O) has an oxidation state of -2. There are 2 oxygen atoms, so their total contribution is \( 2 \times (-2) = -4 \).
3. Let the oxidation state of manganese (Mn) be \( z \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ z + (-4) = 0 \]
5. Simplifying the equation:
\[ z - 4 = 0 \implies z = +4 \]

### Step 4: Compare the oxidation states

Now we have the oxidation states:
- Chromium in \( K_2CrO_4 \): +6
- Niobium in \( NbCl_5 \): +5
- Manganese in \( MnO_2 \): +4

The highest oxidation state among these is +6 from chromium in \( K_2CrO_4 \).

### Conclusion

The metal with the highest oxidation state present in \( K_2CrO_4 \), \( NbCl_5 \), and \( MnO_2 \) is **chromium**.

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