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 $C$.

List-$I$ (Species)	List-$II$ (Hybrid Orbitals)
$(a)$ $SF_{4}$	$(i)$ $sp^{3}d^{2}$
$(b)$ $IF_{5}$	$(ii)$ $d^{2}sp^{3}$
$(c)$ $NO_{2}^{+}$	$(iii)$ $sp^{3}d$
$(d)$ $NH_{4}^{+}$	$(iv)$ $sp^{3}$
	$(v)$ $sp$

$sp^3$ hybridized nitrogen is present in

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Match List-$I$ with List-$II$ :

List-$I$ (Species) List-$II$ (Hybrid Orbitals)

$(a)$ $SF_{4}$ $(i)$ $sp^{3}d^{2}$

$(b)$ $IF_{5}$ $(ii)$ $d^{2}sp^{3}$

$(c)$ $NO_{2}^{+}$ $(iii)$ $sp^{3}d$

$(d)$ $NH_{4}^{+}$ $(iv)$ $sp^{3}$

$(v)$ $sp$

Choose the correct answer from the options given below :

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