Question

find the total valence electrons in 4.2gm of azide ion

Answer 1

$9.6352 \times 10^{23}$

Answer 2

Explanation of the Solution:

1. Calculate Valence Electrons per Ion:

   • Azide ion is $N_3^-$.
   • Each Nitrogen (N) atom has 5 valence electrons.
   • Total valence electrons from 3 N atoms = $3 \times 5 = 15$.
   • The -1 charge indicates the gain of 1 electron.
   • Total valence electrons in one $N_3^-$ ion = $15 + 1 = 16$ valence electrons.

2. Calculate Molar Mass of Azide Ion:

   • Atomic mass of N = 14 g/mol.
   • Molar mass of $N_3^-$ = $3 \times 14 = 42$ g/mol.

3. Calculate Moles of Azide Ion:

   • Given mass = 4.2 g.
   • Number of moles = $\frac{\text{Mass}}{\text{Molar Mass}} = \frac{4.2 \text{ g}}{42 \text{ g/mol}} = 0.1 \text{ mol}$.

4. Calculate Number of Azide Ions:

   • Using Avogadro's number ($N_A = 6.022 \times 10^{23}$ ions/mol):
   • Number of ions = Moles $\times N_A = 0.1 \text{ mol} \times 6.022 \times 10^{23} \text{ ions/mol} = 6.022 \times 10^{22}$ ions.

5. Calculate Total Valence Electrons:

   • Total valence electrons = Number of ions $\times$ Valence electrons per ion
   • Total valence electrons = $6.022 \times 10^{22} \text{ ions} \times 16 \text{ valence electrons/ion}$
   • Total valence electrons = $96.352 \times 10^{22} = 9.6352 \times 10^{23}$ valence electrons.