[darkv0id]

Greetings!

Can anyone help me compare the bond angles in PF₃, PCl₃ and PBr₃. In my opinion the bond angles should be in the order PF₃ > PCl₃ > PBr₃.

My explanation:

Let's take PF₃ as our reference molecule. Now, in PCl₃, due to lower electronegativity of Cl, the bond pair electrons will be closer to Phosphorus, (and consequently, towards the lone pair), therefore increasing lp-bp repulsion and thus decreasing the Cl-P-Cl bond angle w.r.t. the F-P-F bond angle.

A similar argument can be provided for the PBr₃ molecule.

However, a textbook of mine, gives the exactly opposite order of bond angles. While it presents the the same logic, it says that since " the electron density near P is least in case of PF₃, the lone pair will have "strongest influence" on molecular shape, "squeezing" the molecule, and reducing the bond angle vis-a-vis PCl₃ ".

Can someone illustrate the flaw in my logic, and provide a suitable explanation for the answer provided in my book?

[AWK]

http://en.wikipedia.org/wiki/Phosphorus_halides

[darkv0id]

http://en.wikipedia.org/wiki/Phosphorus_halides

Ok. Apparently I'm wrong.

Still, can anyone point out the flaw in my argument and give an explanation as to why the bond angle orderis PF3 < PCl3 < PBr3, and not the other way round ?

[Rutherford]

Nice question.
There are two types of electron repulsions that shape these molecules:
1.lone pair-bond pair which decrease the angle
2.bond pair-bond pair which increase the angle
The lone pair-bond pair repulsion increases slowly with the decrease in electronegativity of the bounded element, but the bond pair-bond pair repulsion increases by the increase in size of the bounded element (bigger atom means bigger electron density around it). The bond pair-bond pair repulsion increases more than the lone pair-bond pair repulsion when going down a group.

[Vidya]

Radeford is right
here the dominating factor is bond pair - bond pair repulsion.As I is the biggest atom with maximum no of electrons and hence will max bond angles .

[darkv0id]

Nice question.
There are two types of electron repulsions that shape these molecules:
1.lone pair-bond pair which decrease the angle
2.bond pair-bond pair which increase the angle
The lone pair-bond pair repulsion increases slowly with the decrease in electronegativity of the bounded element, but the bond pair-bond pair repulsion increases by the increase in size of the bounded element (bigger atom means bigger electron density around it). The bond pair-bond pair repulsion increases more than the lone pair-bond pair repulsion when going down a group.

Radeford is right
here the dominating factor is bond pair - bond pair repulsion.As I is the biggest atom with maximum no of electrons and hence will max bond angles .

Thank you everyone, for the prompt responses.

I presume then that this

the electron density near P is least in case of PF₃, the lone pair will have "strongest influence" on molecular shape, "squeezing" the molecule, and reducing the bond angle vis-a-vis PCl₃

is factually and logically incorrect?

[AWK]

Radeford is right
here the dominating factor is bond pair - bond pair repulsion.As I is the biggest atom with maximum no of electrons and hence will max bond angles .

lone pair- bond pair repulsion is dominating! - Thus spoke Ratherford.

[Rutherford]

Yes, the lone pair-bond pair repulsion is dominating, but the influence of bond pair-bond pair repulsion has a bigger increase when going down the group.
I didn't say bond pair-bond pair repulsion is dominating.

[Sophia7X]

I presume then that this

the electron density near P is least in case of PF₃, the lone pair will have "strongest influence" on molecular shape, "squeezing" the molecule, and reducing the bond angle vis-a-vis PCl₃

is factually and logically incorrect?

Not necessarily. It is true that the electron density near P in case of PF3 is the least since fluorine is the most electronegative (thus having a greater ability to draw the electrons closer to itself and away from phosphorus).


Here's an easy thing to remember when dealing when comparing bond angles:

Generally,
When comparing something like XY3 vs XZ3, the compound with the more electronegative attachments has the lower bond angle. (Electronegative attachments decrease the center's electron density so you have this compressing effect)
When comparing something like XY3 vs ZY3, the compound with the lower electronegative central atom has the lower bond angle. (If the center isn't very electronegative, then the electron density is lower compared to if the center was very electronegative)


Here's an example of a compound with the same central atom.
 H2O > H2S > H2Se > H2Te
Oxygen is the most electronegative so the electron density will be concentrated in the center atom, so highest bond angle.

[Vidya]

Radeford is right
here the dominating factor is bond pair - bond pair repulsion.As I is the biggest atom with maximum no of electrons and hence will max bond angles .

lone pair- bond pair repulsion is dominating! - Thus spoke Ratherford.

it is true that LP-LP > LP-BP >BP-BP
however is our question P is the same atom and only size of the atom bonded to P is changing.In this example the deciding factor is BP-BP