The total number of electrons in 4.2 g of `N^(3-)` ion is (`N_(A)` is the Avogadro's number)

Options

A. `2.1 N_(A)`
B. `4.2 N_(A)`
C. `3 N_(A)`
D. `3.2 N_(A)`

Answer

To determine the total number of electrons in 4.2 g of the `N^(3-)` ion, we can follow these steps:

### Step 1: Calculate the number of moles of `N^(3-)`
To find the number of moles, we will use the formula:

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

For nitrogen, the atomic weight is approximately 14 g/mol. Since the `N^(3-)` ion has 1 nitrogen atom, the molar mass of `N^(3-)` is also 14 g/mol.

\[ \text{Number of moles of } N^{3-} = \frac{4.2 \text{ g}}{14 \text{ g/mol}} = 0.3 \text{ moles} \]

### Step 2: Determine the number of electrons in one `N^(3-)` ion
A neutral nitrogen atom has 7 electrons. Since the `N^(3-)` ion has gained 3 additional electrons, the total number of electrons in one `N^(3-)` ion is:

\[ \text{Total electrons in } N^{3-} = 7 + 3 = 10 \text{ electrons} \]

### Step 3: Calculate the total number of electrons in 0.3 moles of `N^(3-)`
Using Avogadro's number (\(N_A = 6.022 \times 10^{23}\) mol\(^{-1}\)), we can find the total number of electrons in 0.3 moles of `N^(3-)`.

\[ \text{Total electrons} = \text{Number of moles} \times \text{Electrons per mole} \]

The number of electrons in 1 mole of `N^(3-)` is:

\[ 10 \times N_A = 10 \times 6.022 \times 10^{23} \text{ electrons} \]

Thus, for 0.3 moles:

\[ \text{Total electrons} = 0.3 \text{ moles} \times 10 \times N_A = 0.3 \times 10 \times 6.022 \times 10^{23} \]

Calculating this gives:

\[ \text{Total electrons} = 3 \times N_A = 3 \times 6.022 \times 10^{23} \text{ electrons} \]

### Final Answer
\[ \text{Total number of electrons in 4.2 g of } N^{3-} = 3 N_A \]