`K_(2)Cr_(2)O_(7)` react with hydrazins to form product. The oxidation state of Cr in the product will be :-

Options

A. `+4`
B. `+3`
C. `+5`
D. `+2`

Answer

To determine the oxidation state of chromium in the product formed when potassium dichromate \((K_2Cr_2O_7)\) reacts with hydrazine \((N_2H_4)\), we can follow these steps:

### Step 1: Identify the Reactants and Products
The reaction involves potassium dichromate \((K_2Cr_2O_7)\) reacting with hydrazine \((N_2H_4)\) in the presence of sulfuric acid \((H_2SO_4)\). The products of this reaction are chromium sulfate \((Cr_2(SO_4)_3)\), nitrogen gas \((N_2)\), potassium sulfate \((K_2SO_4)\), and water \((H_2O)\).

### Step 2: Determine the Oxidation State of Chromium in \(K_2Cr_2O_7\)
1. Let the oxidation state of chromium be \(x\).
2. The formula for potassium dichromate is \(K_2Cr_2O_7\).
3. Potassium (K) has an oxidation state of +1, and there are 2 potassium atoms, contributing +2.
4. Oxygen (O) has an oxidation state of -2, and there are 7 oxygen atoms, contributing -14.
5. The overall charge of the compound is 0.

Setting up the equation:
\[ 2(+1) + 2x + 7(-2) = 0 \]
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \implies x = +6 \]

Thus, the oxidation state of chromium in \(K_2Cr_2O_7\) is +6.

### Step 3: Determine the Oxidation State of Chromium in \(Cr_2(SO_4)_3\)
1. In the product \(Cr_2(SO_4)_3\), we again let the oxidation state of chromium be \(y\).
2. The sulfate ion \((SO_4)^{2-}\) has an oxidation state of -2, and there are 3 sulfate ions, contributing -6.
3. The overall charge of the compound is 0.

Setting up the equation:
\[ 2y + 3(-2) = 0 \]
\[ 2y - 6 = 0 \]
\[ 2y = 6 \implies y = +3 \]

Thus, the oxidation state of chromium in \(Cr_2(SO_4)_3\) is +3.

### Conclusion
The oxidation state of chromium in the product formed from the reaction of \(K_2Cr_2O_7\) with hydrazine is +3.

### Final Answer
The oxidation state of Cr in the product is +3.

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