Hybridization of nitrogen in pyrrole is:

{s p}^{3}

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s p

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{s p}^{2}

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{s p}^{3} d

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Solution

The correct option is B ${s p}^{2}$
In the structure of pyrrole, the bond uses the $s p^{2}$ orbital and the lone pair is in the orbital. In general, if there is a double bond drawn to the nitrogen atom, then the bond uses the orbital and the lone pair is in the $s p^{2}$ orbital and does not count toward aromaticity.

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Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

L i H + {B - -}_{2} H_{6} \to 2 L i B - - H_{4}

N

N (C H_{3})_{3}

N (S i H_{3})_{3}

L i H + {B - -}_{2} H_{6} \to 2 L i B - - H_{4}

B e H_{2}

C H_{2} B r_{2}

P F_{6}^{-}

B F_{3}

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