The total no. Of valence electrons in 4.2 g of n3- ion is
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Answer:
Explanation:
Given mass= 4.2g
Molar mass of N3^- (azide ion) = 14*3= 42
Therefore, No. Of moles of azide ion = given mass/ molecular mass= 4.2/42= 0.1 moles
No. of valence electrons in one nitrogen atom = 5
No. of valence electrons in azide ion (N3^-) ( N=N=N ) is =(5*3)+1=16electrons
Therefore, total no. of valence electrons in 4.2g of azide (N3^-) ion = no. of moles of azide ion * no. of valence electrons in one azide ion * Avogadro's constant
= 0.1* 16* 6.022*10^23
=1.6* 6.022*10^ 23
=9.6352* 10^23 valence electrons
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