The total number of valence electrons in `4. 2g` of ` N_3^-` ion are :

Options

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

Answer

To find the total number of valence 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 calculate the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass = 4.2 g
- Molar mass of \( N_3^- \) = 3 × 14 g/mol (since the molar mass of nitrogen (N) is approximately 14 g/mol)
\[ \text{Molar mass of } N_3^- = 3 \times 14 = 42 \text{ g/mol} \]
Now, substituting the values:
\[ \text{Number of moles} = \frac{4.2 \text{ g}}{42 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Calculate the total number of valence electrons in one \( N_3^- \) ion
Each nitrogen atom contributes 5 valence electrons. Since there are three nitrogen atoms in \( N_3^- \):
\[ \text{Valence electrons from nitrogen} = 3 \times 5 = 15 \]
Additionally, the \( N_3^- \) ion has a negative charge, which means there is one extra valence electron:
\[ \text{Total valence electrons in } N_3^- = 15 + 1 = 16 \]

### Step 3: Calculate the total number of valence electrons in 0.1 moles of \( N_3^- \)
Using Avogadro's number (\( N_A = 6.02 \times 10^{23} \) molecules/mol), we can find the total number of \( N_3^- \) ions in 0.1 moles:
\[ \text{Number of } N_3^- \text{ ions} = 0.1 \text{ moles} \times N_A = 0.1 \times 6.02 \times 10^{23} \text{ ions} \]
Now, to find the total number of valence electrons in 0.1 moles:
\[ \text{Total valence electrons} = \text{Number of } N_3^- \text{ ions} \times \text{Valence electrons per ion} \]
\[ = (0.1 \times 6.02 \times 10^{23}) \times 16 \]
Calculating this gives:
\[ = 1.6 \times 6.02 \times 10^{23} \text{ valence electrons} \]

### Final Answer
The total number of valence electrons in 4.2 g of \( N_3^- \) ion is:
\[ 1.6 \times 6.02 \times 10^{23} \text{ valence electrons} \]