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Question

The hybridization of atomic orbitals of nitrogen in $$ NO^+_2 , NO^-_3 $$ and $$ NH^+_4 $$ are :


A
sp,sp3 and sp2 respectively
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B
sp,sp2 and sp3 respectively
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C
sp2,sp and sp3 respectively
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D
sp2,sp3 and sp respectively
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Solution

The correct option is B $$ sp , sp^2$$ and $$sp^3 $$ respectively
$$ NO^+_2 $$
$$sp$$ hybridisation and linear shape 
Bond angle $$ 180^0 .$$
$$ NO^-_3 $$
$$ sp^2 $$ hybridisation and triangular planar shape
Bond angle $$ 120^0 $$
$$ NH^+_4 $$
$$ sp^3 $$ hybridisation and tetrahedral shape
Bond angle $$ 109^028'$$

Chemistry

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