Hybridisation of nitrogen I and II in the following compound is:

s p^{2}

s p^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p, s p^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p^{2}, s p^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s p^{2}, s p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $s p^{2}, s p^{2}$
nitrogen's lone pair in 1 is in resonance with double bond to make the compound aromatic therefore only 3 bond pairs hence sp2 hybridised
nitrogen's lone pair in 2 is in sp2 orbital as it doesnt participate in resonance

Suggest Corrections

Similar questions

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

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

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

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

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

Join BYJU'S Learning Program