In ammonia the N atom because it is period 2 can only realistically hybridize as sp, sp2 or sp3. The geometry is trigonal pyramidal, so sp3 hybridisation is the best fit, the other options would not fit the geometry.
In actuality the bond angle of 1070 shows that as the angle is closer to 90 than in "true" sp3 (1090)the bonds have slightly more p character (porbitals are at 900), leaving the lone pair with slightly more s character..
The hybridization of PH3 is sp3
The hybridization is sp3d.
CCl4 features all single covalent bonds, so the hybridization is sp3.
sp3
sp3
The nitrogen atom undergoes sp3 hybridization in ammonia.
Sp3 hybridization because of the three Hydrogens coming off of the Nitrogen plus one lone pair of electrons on the Nitrogen to satisfy it's octet rule.
The hybridization of PH3 is sp3
The hybridization is sp3d.
CCl4 features all single covalent bonds, so the hybridization is sp3.
sp3
sp3
sp3
sp3
sp3
Sp3
sp3