Q.Determine the hybridisation of the nitrogen atom in $\text{NH}_3$ and explain how one of the four $sp^3$ hybrid orbitals accommodates the lone pair.

[!TLDR] Nitrogen in $\text{NH}_3$ is $sp^3$ hybridised (steric number 4: 3 bond pairs + 1 lone pair). [!ANSWER] Three of nitrogen's four $sp^3$ hybrid orbitals form N–H sigma bonds; the fourth $sp^3$ …

Nitrogen in $\text{NH}_3$ has a steric number of 4 (3 bonding pairs to hydrogen + 1 lone pair), the same total electron-pair count as carbon in $\text{CH}_4$, so nitrogen also undergoes $sp^3$ hybridisation: one $2s$ and three $2p$ orbitals mix to give four equivalent $sp^3$ hybrid orbitals directed toward the corners of a tetrahedron. Because hybridisation depends on the total number of electron pairs regardless of whether each pair is used for bonding or held as a lone pair, all four of nitrogen's $sp^3$ hybrid orbitals are chemically equivalent orbitals — but only three of them are used to form sigma bonds, one to each hydrogen $1s$ orbital. The fourth $sp^3$ hybrid orbital is not used for bonding at all; it simply holds nitrogen's lone pair of electrons. This is exactly why $\text{NH}_3$'s molecular shape (defined only by where the atoms are) is trigonal pyramidal rather than trigonal planar — the lone pair occupies real space …

Determine the steric number on N (3 bond pairs + 1 lone pair = 4), identify sp3 hybridisation, and explain that hybridisation applies to all fou …

Assuming only the three bonded orbitals are hybridised while the lone pair sits in an unhybridised orbital — in sp3 hybridisation all four o …