The hybridisation of nitrogen atom in pyridine is:

s p^{3}

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s p^{2}

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s p

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s p^{3} d

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Solution

The correct option is B $s p^{2}$
The nitrogen in pyridine has a $s p^{2}$ hybridization because it is in de-localized and/or aromatic system & the hybridization allows the participation of one unhybridized p-orbital in a pi system of the aromatic ring.

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Similar questions

N O_{3}^{-}, N O_{2}^{+}, N H_{4}^{+}

A l H_{3}

A l H_{4}^{-}

C (1) H_{2} = C (2) H - C (3) \equiv C H

N O_{3}^{-}, N O_{2}^{+} a n d N H_{4}^{+}

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

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