if Na is the avagadro number then the number of valence electrons in 4.2g of nitride ion is

Question

N_a

Answer 1

The nitride ion is N^3-.  Accepting that the mass of an electron is negligible relative to that of the proton and neutron, then:

 

4.2 g N^3- x 1 mole N^3-/14 g = 0.3 moles N^3-

# of N^3- ions in 0.3 moles N^3- = 0.3 x N_a

# valence electrons in one N^3- ion = 8 (5 in neutral N + 3 extra for the 3- charge)

# valence electrons in 0.3N_a nitride ions = 8 x 0.3N_a = 2.4N_a

 

If you want the actual number of valence electrons is would be:

0.3 moles N^3- x 6.02x10²³ atoms/mole = 1.806x10²³ atoms

1.806x10²³ atoms N^3- x 8 valence electrons/atom N^3- = 1.4x10²⁴ valence electrons