PH3 has lesser bond angle than that of NH3.Why?

Question

PH3 has lesser bond angle than that of NH3.Why?

Answer 1

The problems with John Lee's answer are that it's purely steric in origin, and it neglects the reason that the molecule is pyramidal in the first place -- repulsion of the bond pairs with the lone pair.  If NH3 has increased BP-BP repulsion it should also have increased BP-LP repulsion.  If the BP-LP repulsion is greater, the angle should be narrower!

An explanation is no good if you could use exactly the same underlying premise to explain exactly the opposite result.

OK, which theory do you want to use, because we can explain this lots of different ways.

Lewis structure has eight electrons at P or N, three in bond pairs, one in lone pair.  VSEPR designation: AX3E, predicts trigonal pyramidal geometry, C3v symmetry, exactly as observed.  The usual VSEPR explanation for the narrower angle in PH3 is that of John's, plus consideration of the lone pair:  In PH3 the H atoms are farther away from each other because a P-H bond is longer because P is a larger atom, and the P lone pair is also much bigger because P is a larger atom, which means the BP-BP repulsions are less and the BP-LP repulsions are *greater*, and so the bonds get pushed closer together.

You asked a qualitative question, there's a perfectly good qualitative answer.

More interesting: NH3 is 107º, PH3 is 93º, AsH3 is 92º, SbH3 91º.  Why the sudden and dramatic jump from almost tetrahedral angle to almost orthogonal?  Why does the angle never drop below 90?  Can't explain that with Lewis.

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 tue 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 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 one.

Answer 2

Due to larger atomic size of P the electron density on it in PH3 is relatively small which provides lesser repulsion resulting in lower value of bond angle.

Answer 3

This can be answered by considering three things:

1) Steric effects of single bonds

2) Electronegativity

3) Atomic Size

1) When single bonds are close to each other they repel each other due to steric effects.

2) Electronegativity tells us that the greater the electronegativity the greater the ability to attract shared electrons.

Electronegativity of N = 3.0

Electronegativity of P = 2.1

You would suspect more electron density from all single bonds to be closer to Nitrogen in NH3 causing an increase in repulsion => increase in bond angle

You would suspect not as much electron density from all single bonds to be closer to Phosphorus in PH3 causing a decrease in repulsion => decrease in bond angle

3) If the atomic size is greater you would suspect the single bonds to be more spread apart therefore less repulsion; Phosphorus>Nitrogen in atomic size

Therefore PH3<NH3