1.

The total number of valence electrons in 4.2 g of N_(3)^(-)ions is (N_Ais the Avogadro number):

Answer»

`1.6 N_A`
`3.2 N_A`
`2.1 N_A`
`4.2 N_A`

Solution :Each N- atom has 5 VALENCE ELECTRONS
` therefore ` Total number of electrons in `N_(3)^(-)` ion ` = 5 xx 3 + 1 = 16`
MOLES of `N_(3)^(-)` ions present ` = (4.2)/(42) = 0.1 `(mol. Wt.of `N_(3)^(-)1 = 42 )`
` therefore ` No. of valenceelectrons in 4.2 g`= 0.1 xx 16 xx N_A =1.6 N_A`


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