sp^3 hybridized nitrogen is present in | vedclass

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Question

(C) $A)$ Pyridine: The nitrogen atom is $sp^2$ hybridized as it is part of the aromatic ring and has one lone pair in the $sp^2$ orbital.$B)$ Pyrrole:

Vedclass · Accepted Answer

$CH_2 = CH - NH_3^+$

Vedclass · Answer

(C) $A)$ Pyridine: The nitrogen atom is $sp^2$ hybridized as it is part of the aromatic ring and has one lone pair in the $sp^2$ orbital.$B)$ Pyrrole: The nitrogen atom is $sp^2$ hybridized because its lone pair participates in the aromatic sextet of the ring.$C)$ $CH_2 = CH - NH_3^+$: The nitrogen atom is bonded to four atoms (three $H$ atoms and one $C$ atom) with no lone pair,resulting in $sp^3$ hybridization.$D)$ $CH_2 = CH - NH_2$: The nitrogen atom is $sp^2$ hybridized due to conjugation of its lone pair with the adjacent double bond.Therefore,the correct option is $C$.

$sp^3$ hybridized nitrogen is present in

Similar Questions

Choose the molecules in which hybridization occurs in the ground state?
$(i)$ $BCl_3$ $(ii)$ $NH_3$
$(iii)$ $PCl_3$ $(iv)$ $BeF_2$
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The geometry of the ammonia molecule and the hybridization of the nitrogen atom involved in it are:

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