Why the HPH angles in PH3 are 94 instead of 107?

Answer 1

The usual, easy, and unsatifying explanation that you'll ususally see is a clunky application of VSEPR with some handwaving steric and inductive electronic arguments thrown in. N to P results in a larger central atom, so the P lone pair is larger and repels the bond pairs more, and the H atoms are farther away from one another and so repel each other less, thus the angle contracts.

Feh.

There are many things this model does *not* explain. Why does the angle change so dramatically N to P, but then exhibit almost no further change P to As to Sb? The size of the atom keeps going up (covalent radii 110 125 145 pm) but the angle barely responds (94, 92, 91º). Same trend in Gr 16: H20 is 104, but then H2S H2Se H2Te are 92 91 90. If it's just a qualitative push-more-push-less argument, there should be no limit on it, and no arbitrary "minimum angle" one might find, but there is no Gr 15 EX3 or Gr 16 EX2 compound (X = any of H F Cl Br CH3) that has an angle less than 90º. Why should that be the case? Why coincidentally stop at exactly 90º?

Why does the angle decrease NH3 to NF3 (107 to 103) and H2O to OF2 (104 to 103) but increase PH3 to PF3 (94 to 97) and H2S to SF2 (92 to 98)? F is always bigger than H, right?

A much more useful, general, and quantitative explanation comes from valence bond theory. The basic argument is that a near-tetrahedral angle, as observed in NH3 and H2O, can only be explained by invoking hybridization of atomic orbitals at the central atom, specifically the use of sp3 hybrids to form the bonds and hold the lone pairs. In contrast, PH3 AsH3 SbH3 are all nearly 90º, which requires no hybridization. These molecules use (nearly) pure p-orbitals to form their bonds, leaving the lone pair to the lower-E unhybridized s-orbital.

Why? Because as you move down the p-block, the s-p energy gap increases, and hybridization becomes less energetically favourable if there are lone pairs. sp3 hybrids are higher energy than an s-orbital, so hybridization means raising the energy of the lone pair.

As the s-p gap goes up, so does that destabilization cost. As well, you have weaker bonds, so hybridizing them means less energetic savings, and there's no steric repulsion due to the larger atom. Thus, PH3 AsH3 SbH3 have ~90º angles because hybridization would pay more to get less, and it's more stable to just use p-orbitals to bond.

It's a bit more subtle than that, of course, sp3 and pure p are just two extreme points on a continuum. You can use any linear combination of s and p orbitals to form the bonds, and the angles adjust accordingly. (i.e. NH3 at 107º is 23% s in each bond, 32% in the lone pair, while NF3 at 103º is 18% s in each bond, leaving 45% in the lone pair.) The observed angle in any of these molecules is a balance of the s-p gap, the steric repulsion among the groups, and the electronegativity difference across the bonds.

Of course, if you really want to know what's going on, you need molecular orbitals, and a Walsh diagram. But that takes two hours to explain.

Answer 2

This question is all about the influence of the lone pair on the central atom. As you go down Group 5 (aka Group 15) the lone pair gets smaller and smaller relative to the size of the atom as a whole. It is therefore less repulsive, and so are the bonding pairs. The bonding pairs gravitate towards their unhybridised bond angles, rather than the expected sp3 angle of 109.5.

Answer 3

in NH3 the HNH is nearly 107

in PH3 HPH angle is 94 because as we move down the group,the size of central atom goes on increasing and its electronegativity goes on decreasing.As a result the bond pairs of electrons tend to lie away and away from central atom as we move from NH3 to SbH3 .

due to this the FORCE OF REPULSION between adjacent bond pairs is maximum in NH3 and minimum in SbH3. consequently the bond angle goes on decreasing from NH3 to SbH3.


the angles are

NH3 (107.8)
PH3(93.6)
AsH3(91.8)
SbH3(91.3)

Answer 4

Lone pair of electrons on the Phos repels the bonded H