Need help with bonding structures of ammonia and phosphorus trihydride?

Question

ammonia (NH3) has a trigonal pyramidal structure shape whilst phosphorus trihydride (PH3) has a trigonal planar shape.

why is this so?

when i look at this i think

- both phosphorus and nitrogen have 5 electrons in their outermost shell.

- both are in group 5.

when i done some research i found that ammonia is the only of all the group 5 elements to form a bond with 3 hydrogen with a trigonal pyramidal shape, all the rest (e.g. arsenic, antimony) have trigonal planar shape.

I am guessing that maybe ammonia is covalently bonded, whilst all the rest such as phosphorus, arsenic and antimony are ionic bonding.

Possibly something to do with hydrogens unique properties.

Answer 1

the shape changes due to the presence of hydrogen bonding in ammonia while the rest of compounds doesnt have them and also as u said ionic nature increases down and also nitrogen is the smallest of all

the main thing is that decrease increased bond angle due to repulsions between the pairs

Answer 2

You need to do more research, then, because your premise is false, and you're spending time trying to explain something that isn't real.

Phosphine, arsine, and stibine are NOT trigonal planar, they are all pyramidal -- you don't even need spectroscopy to determine that, the fact that all these molecules have dipole moments tells you they cannot be planar.  In fact, they are MORE pyramidalized than NH3.

Bond angles: NH3 107º, PH3 93º, AsH3 92º, SbH3 91º.  Different sources will give you values that vary by plus/minus a degree, depending on technique and whether its in the liquid/gas phase.

You could explain that trend just based on the size of the central atom, resulting in a larger lone pair and less relative steric repulsion of the H atoms, but that doesn't give you a quantitative rationale, and doesn't explain the sudden jump from near-tetrahedral to near-orthogonal.

Valence bond theory explains bond angles in terms of hybridization of orbitals. N has a small energy difference between the 2s and 2p orbitals, so it's not a great energy penalty to hybridize and form sp3 hybrids to both hold the lone pair and form the bonds -- indeed, the overlap is stronger with the hybrids, the bonds are stronger, so you get a big benefit by doing so. But not true with P. P-H is weaker than N-H, so the benefit to the bonds you can get from hybridization is reduced. P is bigger, Hs farther apart, steric benefit by avoiding steric repulsion is also reduced. The s-p gap is much much greater in P, so the energy penalty to hybridize is greater -- moving the lone pair out of a pure s into an ap3 hybrid isn't worth it. So you don't hybridize. PH3 uses three atomic p orbitals for bonds, keeps a (nearly) 90º angle, and puts the LP in a pure s orbital. As and Sb same argument but moreso.

There's a different (and even better) argument using molecular orbitals, but it invokes symmetry mixing and a Walsh diagram. Boils down to the s-p gap and ease of mixing the s and the pz AOs to form the 2a1 HOMO. Most people stop caring long before you finish that description.