Question

"Bond angle in NH3 is greater than in NF3 but bond angle in PH3 is less than that in PF3" Why? Please explain. (the one clears this to me will get voted brainliest)

Answer 1

Both NH3 and NF3 form pyramidal shape. But the angle between these molecules differ due to the electronegativity difference between them. The electronegativity of Nitrogen is more than that of H , therefore the bond cloud tries to attract to the nitrogen atom. Therefore the whole electron cloud attract to the central atom. Then the each electron pair come closer to the each and due to this electron tries to repel each other. This leads to increase in the bond angle of NH3 but in NF3 the bond pairs attract to the Fluorine atom, (terminal atoms). The distance b/w electron pairs increase and hence, repulsion decreases. So, NF3 has smaller bond angle.

While the bond angle of PH3 is less than that of PF3. Both PH3 and PF3 are pyramidal in shape. Both have one lone pair on phosphorous. But PF3 has partial double bond character due to back π-donation of electrons from p orbitals of F atom to empty d orbitals of P. This results in large repulsion between P-F bonds and hence bond angle is large. There is no partial double bond character possible in PH3​(due to lack of p-orbitals in H).

The bond angle is more in PF​3 due to lone pair-bond pair repulsion. As flourine is smaller in size and more electronegative than hydrogen, the bond pair of electrons is towards F.

Answer 2

Explanation:

In NH3, the bond dipoles reinforce the effect of the unshared pair, so NH3 is very polar (u = 1.47 D).

In NF3, the bond dipoles opposes the effect of the unshared pair, so NF3 is only slightly polar (u = 0.23 D)

We can now use this information to explain, the bond angles observed in NF3 and NH3.

Because of the direction of the bond dipoles in NH3, the electron-rich end of each N-H bond is at the central atom, N.

On the other hand, the fluorine end of each bond in NF3 is the electron rich end.

As a result, the lone pair can more closely approach the N in NF3 than in NH3. In NF3 the lone pair therefore exerts greater repulsion towards the bonded pairs than in NH3.

In addition, the longer N−F bond length makes the bp-bp distance greater in NF3, than in NH3, so that the bp/bp repulsion in NF3 is less that in NH3.

The net effect is that the bond angles are reduced more in NF3.

*Bonded pair repulsions are weaker in NF3 than in NH3 due to the longer N-F bond.*

“We might expect the larger F atoms (r = 0.72 A) to repel each other more strongly than the H atoms (r = 0.37 A), leading to larger bond angles in NF3 than in NH3. This is not the case, however, because the N−F bond is longer than the N−H bond. The N−F bond is farther from the N than the N−H bond density.”