If Na is Avogardros number then number of valence electron in 4.2 g of nitride ions (N3- is
Molecular wt of N
3
−
=3×14=42 g
One ion of N
3
−
has valence electrons = 3×5+1=16
42 g N
3
−
=N
A
N
3
−
ion
valence electrons in N
A
ions = 16N
A
(where N
A
= Avogadro's number)
So, 1 g of N
3
−
will have valence electrons = 16N
A
/42
therefore,4.2g of N
3
−
ion will have valence electrons = 16N
A
×4.2/42=1.6N
A
.
Molecular wt of N
3
−
=3×14=42 g
One ion of N
3
−
has valence electrons = 3×5+1=16
42 g N
3
−
=N
A
N
3
−
ion
valence electrons in N
A
ions = 16N
A
(where N
A
= Avogadro's number)
So, 1 g of N
3
−
will have valence electrons = 16N
A
/42
therefore,4.2g of N
3
−
ion will have valence electrons = 16N
A
×4.2/42=1.6N
A
.
Hi jainam,
No of moles =4.2/ 14 ( 4.2 given. , .14 for Na)
=0.3 mol
0.3mol = 0.3 × Na electrons
Total valence electrons in N3- ion = 5+3 =8
No of valence electrons = 8 × 0.3 Na
=2.4Na.
So the answer is 2.4 Na.
