How many total no. of electrons are present in nitrarte ions?

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Solution

$N i t r a t e i o n = N {O_{3}}^{-} N u m b e r o f e l e c t r o n s i n N = 7 N u m b r o f e l e c t r o n s i n O = 8 F o r 3 o x y g e n = 3 \times 8 = 24 O n e n e g a t i v e c h a r g e = 1 e l e c t r o n T o t a l n u m b e r o f e l e c t r o n s = 7 + 24 + 1 = 32 e l e c t r o n s$

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