The number of electrons needed to reduce 3gm of `Mg^(2+)` to Mg are

Options

A. N
B. `(N)/(2)`
C. `(N)/(4)`
D. `(N)/(8)`

Explanation

To solve the problem of determining the number of electrons needed to reduce 3 grams of \( \text{Mg}^{2+} \) to Mg, we can follow these steps:

### Step 1: Calculate the number of moles of magnesium
The number of moles can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of magnesium = 3 g
- Molar mass of magnesium (Mg) = 24 g/mol

Substituting the values:

\[ \text{Number of moles} = \frac{3 \, \text{g}}{24 \, \text{g/mol}} = \frac{1}{8} \, \text{moles} \]

### Step 2: Determine the number of electrons required for reduction
The reduction of \( \text{Mg}^{2+} \) to Mg involves the gain of 2 electrons. Therefore, for 1 mole of \( \text{Mg}^{2+} \), 2 moles of electrons are required.

### Step 3: Calculate the number of electrons for \( \frac{1}{8} \) moles of magnesium
Since we have \( \frac{1}{8} \) moles of magnesium, we can calculate the number of electrons required as follows:

\[ \text{Electrons required} = 2 \times \left(\frac{1}{8}\right) = \frac{2}{8} = \frac{1}{4} \, \text{moles of electrons} \]

### Step 4: Convert moles of electrons to number of electrons
To find the total number of electrons, we multiply the moles of electrons by Avogadro's number (\( N \)), which is approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \):

\[ \text{Number of electrons} = \frac{1}{4} \times N = \frac{N}{4} \]

### Final Answer
Thus, the number of electrons needed to reduce 3 grams of \( \text{Mg}^{2+} \) to Mg is:

\[ \frac{N}{4} \]

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