One mole on iron [55.8 gm], when oxidised to `+2` oxidation state gives up :

Options

A. `1N_(A)` electron
B. `2N_(A)` electron
C. `3N_(A)` electron
D. `0.5` mole of electron

Answer

To solve the problem of how many electrons are given up when one mole of iron (Fe) is oxidized to a +2 oxidation state, we can follow these steps:

### Step 1: Understand the oxidation process
When iron is oxidized from its elemental state (0 oxidation state) to a +2 oxidation state, it loses electrons. Specifically, for each atom of iron that is oxidized, it loses 2 electrons.

### Step 2: Determine the number of atoms in one mole of iron
One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) atoms. Therefore, one mole of iron also contains \(6.022 \times 10^{23}\) atoms.

### Step 3: Calculate the total number of electrons lost
Since each atom of iron loses 2 electrons upon oxidation, we can calculate the total number of electrons lost when one mole of iron is oxidized:

\[ \text{Total electrons lost} = \text{Number of atoms} \times \text{Electrons lost per atom} \]

Substituting in the values we have:

\[ \text{Total electrons lost} = 6.022 \times 10^{23} \text{ atoms} \times 2 \text{ electrons/atom} \]

### Step 4: Perform the calculation
Calculating the total number of electrons lost:

\[ \text{Total electrons lost} = 6.022 \times 10^{23} \times 2 = 1.2044 \times 10^{24} \text{ electrons} \]

### Conclusion
Therefore, when one mole of iron is oxidized to a +2 oxidation state, it gives up approximately \(1.2044 \times 10^{24}\) electrons.