If `N_(A)` is Avogadro's number then number of valence electrons in 4.2 g of nitride ions `(N^(3-))`

Options

A. `2.1 N_(A)`
B. `4.2 N_(A)`
C. `1.6 N_(A)`
D. `3.2 N_(A)`

Explanation

To solve the problem of finding the number of valence electrons in 4.2 g of nitride ions (N³⁻), we can follow these steps:

### Step 1: Calculate the number of moles of nitride ions
To find the number of moles, we use the formula:

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

The molar mass of nitrogen (N) is approximately 14 g/mol. Therefore, for nitride ions (N³⁻), we have:

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

### Step 2: Calculate the number of nitride ions
Using Avogadro's number (Nₐ = \(6.022 \times 10^{23}\) ions/mol), we can find the total number of nitride ions:

\[ \text{Number of nitride ions} = \text{Number of moles} \times N_a = 0.3 \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mol} \]

Calculating this gives:

\[ \text{Number of nitride ions} = 0.3 \times 6.022 \times 10^{23} \approx 1.81 \times 10^{23} \text{ ions} \]

### Step 3: Determine the number of valence electrons in each nitride ion
Each nitrogen atom has 5 valence electrons (as it is in group 15 of the periodic table). Since the nitride ion (N³⁻) has gained 3 additional electrons, the total number of valence electrons in each nitride ion is:

\[ \text{Valence electrons in N}^{3-} = 5 + 3 = 8 \text{ valence electrons} \]

### Step 4: Calculate the total number of valence electrons in all nitride ions
Now, we multiply the number of nitride ions by the number of valence electrons per ion:

\[ \text{Total valence electrons} = \text{Number of nitride ions} \times \text{Valence electrons per ion} \]

Substituting the values we have:

\[ \text{Total valence electrons} = 1.81 \times 10^{23} \text{ ions} \times 8 \text{ electrons/ion} \approx 1.45 \times 10^{24} \text{ electrons} \]

### Final Answer
The total number of valence electrons in 4.2 g of nitride ions (N³⁻) is approximately \(1.45 \times 10^{24}\) electrons.

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