What is the hybridisation state of the centra | vedclass

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(C) The conjugate base of the ammonium ion $(NH_4^+)$ is obtained by removing a proton $(H^+)$: $NH_4^+ \rightarrow NH_3 + H^+$. In the ammonia molecu

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$sp^3$,pyramidal

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(C) The conjugate base of the ammonium ion $(NH_4^+)$ is obtained by removing a proton $(H^+)$: $NH_4^+ ightarrow NH_3 + H^+$. In the ammonia molecule $(NH_3)$,the nitrogen atom $(N)$ is bonded to three hydrogen atoms and has one lone pair of electrons. According to $VSEPR$ theory,the steric number is $3 	ext{ (bond pairs)} + 1 	ext{ (lone pair)} = 4$,which corresponds to $sp^3$ hybridisation. Due to the presence of one lone pair,the geometry is distorted from tetrahedral to pyramidal.

What is the hybridisation state of the central atom and the shape of the conjugate base of the ammonium ion?

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