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.