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2 mole of N(2)H(4) loses 16 mole of elec...

2 mole of `N_(2)H_(4)` loses 16 mole of electron is being converted to a new compound `X`. Assuming that all of the `N` appears in the new compound. What is the oxidation state of `'N'` in `X ` ?

A

`-1`

B

`-2`

C

`+2`

D

`+4`

Text Solution

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The correct Answer is:
To find the oxidation state of nitrogen in the new compound \( X \) after the conversion of hydrazine (\( N_2H_4 \)), we can follow these steps: ### Step 1: Determine the initial oxidation state of nitrogen in \( N_2H_4 \) 1. The molecular formula of hydrazine is \( N_2H_4 \). 2. Let the oxidation state of nitrogen be \( x \). 3. The oxidation state of hydrogen is +1. 4. The equation for the overall charge of the molecule is: \[ 2x + 4(+1) = 0 \] 5. Simplifying this gives: \[ 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \] Thus, the oxidation state of nitrogen in \( N_2H_4 \) is -2. ### Step 2: Calculate the total number of electrons lost 1. According to the problem, 2 moles of \( N_2H_4 \) lose 16 moles of electrons. 2. Therefore, 1 mole of \( N_2H_4 \) loses: \[ \frac{16 \text{ moles of electrons}}{2} = 8 \text{ moles of electrons} \] ### Step 3: Determine the number of nitrogen atoms in \( N_2H_4 \) 1. Each mole of \( N_2H_4 \) contains 2 moles of nitrogen atoms. 2. Therefore, 1 mole of \( N_2H_4 \) has 2 nitrogen atoms. ### Step 4: Calculate the number of electrons lost per nitrogen atom 1. Since 1 mole of \( N_2H_4 \) loses 8 moles of electrons and contains 2 moles of nitrogen, the number of electrons lost per nitrogen atom is: \[ \frac{8 \text{ moles of electrons}}{2 \text{ moles of nitrogen}} = 4 \text{ moles of electrons} \] ### Step 5: Determine the new oxidation state of nitrogen in compound \( X \) 1. The initial oxidation state of nitrogen was -2. 2. Losing 4 electrons means that the oxidation state increases by 4 units (since losing negative charge increases the oxidation state). 3. Therefore, the new oxidation state \( y \) of nitrogen in \( X \) is: \[ y = -2 + 4 = +2 \] ### Conclusion The oxidation state of nitrogen in the new compound \( X \) is +2. ---
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