[shafaifer]

PCl₅ has D_3h symmetry. Clicking on the following link will allow you to see the caracter table for this symmetry: https://www.google.dk/search?q=d3h+character+table&biw=1242&bih=585&source=lnms&tbm=isch&sa=X&ei=-MwtVNX-JYjhywPcsYGACQ&sqi=2&ved=0CAYQ_AUoAQ#facrc=_&imgdii=_&imgrc=Ud1v3QhZboBZ2M%253A%3BD0FQxBA43PWSSM%3Bhttp%253A%252F%252Ffaculty.uml.edu%252Fndeluca%252F84.334%252Ftopics%252FD3htable.jpg%3Bhttp%253A%252F%252Ffaculty.uml.edu%252Fndeluca%252F84.334%252Ftopics%252Ftopic4.htm%3B617%3B194

In the attachment you can see how the coordinate axeses are situated.

Exercise question:

Which of the 3s, 3p and 3d orbitals on the central P atom can interact with the 3p_z orbitals of the 3 equatorial Cl-atoms (creating binding / antibinding MO's)? I am not sure what key point here is. Has it something to do with the earlier determination of the distribution of the irreps of the 3p_z orbitals of the 3 equatorial Cl-atoms that I completede beforehand?



Sincerely,

Best regards,

Shafaifer. 

[shafaifer]

Sorry, I forgot to note: The vertical axis is z axis.

[Corribus]

What you usually would do is determine the symmetry representations of each of the atomic orbitals on the central atom, and then the symmetry representation(s) of the 3 chlorine atom Pz orbitals (together, as a triplet), and see which of the latter match with those of the former. Does that make sense?

[shafaifer]

I will take it into consideration more thoroughly. Thank you so long.

[shafaifer]

Okay. I have been thinking of it thoroughly now. D3h symmetry has 12 symmetry operations: E, 2C3, 3C2, sigma(h), 2S3 and 3sigma(v). In the earlier question I got the following symmetry representation for the three 3pz orbitals of the three equatorial Cl atoms:

E: 3
C3: 0
C2: -1
sigma(h): -3
S3: 0
sigma(v): 1

So now I should do the same with the phosphorus atom and compare? Should I look at the orbitals of the central P atom as a whole like in the former question? Otherwise how would I get a comparable identity value of 3?

With great thanks,

Shafaifer.

[Irlanur]

first, you have to decompose (=reduce) this irreducible representation you got.

So now I should do the same with the phosphorus atom and compare? Should I look at the orbitals of the central P atom as a whole like in the former question? Otherwise how would I get a comparable identity value of 3?

if you have a proper charachter table, the symmetry properties of the orbitals can usually be found on the right-hand side.

[shafaifer]

first, you have to decompose (=reduce) this irreducible representation you got.

So now I should do the same with the phosphorus atom and compare? Should I look at the orbitals of the central P atom as a whole like in the former question? Otherwise how would I get a comparable identity value of 3?

if you have a proper charachter table, the symmetry properties of the orbitals can usually be found on the right-hand side.

Thank you very much. Do you mean:

Gamma = (1/3)A1'' + (2/3)A2'' + E''? If this is correct, can I do something that will help now?

[Corribus]

You can't have fractions of an irrep. Do you know the proper formula to use to reduce a reducible representation into a sum of irreducible representations? And do you know how to assign symmetry representations to the three p-orbitals on the central atom (hint: they transform like the linear axes x, y and z)? What about the s and d orbitals?

[shafaifer]

Thank you very much.

Do you know the proper formula to use to reduce a reducible representation into a sum of irreducible representations?

For this question, yes: You multiply 1/order of the point group by the sum up the products of symmetri representations and all of the 6 different rows of irreps for this point group, D3h, so it will look like the following:

(1/n) * Σ "chi"(R) R_b. Sorry for poor notation; I am not used to this site yet.

Is this correct at all?

With great love,

Shafaifer.

[Corribus]

Notations vary quite a bit from book to book. The best way to tell is to use your formula to reduce your reducible representation into a sum of irreps. Can you do that?

[shafaifer]

Thank you again, Corribus. I just write the sums:

For A1': 0 ---- since 3 times 1 (Ag) = 3, 0 times 1 = 0, -1 times 1 = -1, -3 * 1 = -3, 0 times 1 = 0, 1 times 1 = 1, 0 times 1 = 0. This gives me a sum of 0.
For A'2: 0
E': 0
A1'': 4
A2'': 8
E'': 12, so:

Gamma = (4/12)A1'' + (8/12)A2'' + E'' = (1/3)A1'' + (2/3)A2'' + E''. I know I stated this before and you said that you cannot have a fraction in the irrep, but the metod is correct?

Best regards.

[Corribus]

I'm not quite able to reproduce your error, but usually when students are getting fractions in their representation reductions, it's because they are forgetting that there are multiple symmetry elements per class. For example, there are 2 C₃ rotations, 3 C₂ rotations, and so forth. You have to make sure you are multiplying the characters by the number of operations in each class.

E.g., for A'₁ it is (1/12)*(3*1*1 + 0*2*1 + (-1)*3*1 + (-3)*1*1 + 0*2*1 + 1*3*1) = 0

The first number in each triad is the character of your reducible representation, the second number is the number of operations in the symmetry class, and the third is the character in the character table for that symmetry class.

See if this doesn't help you reduce your representation to an integer sum of irreps.

[shafaifer]

Thanks.

Now I got:

Gamma = A''2 + E''.

So the three 3pz orbitals of the three equatorial Cl atoms span E'' + A''2. So this leads to the topic question: How do I find out which irrep(s), respectively, the 3s, 3p and 3d orbitals of the central P atom span in this PCl5 molecule? ? When this is determined, I will be able to answer the true question of MO interaction by using the character table to properly compare symmetri properties.

 Yes, p orbitals transform like x, y and z. s orbitals transform like x^2, y^2 and z^2. d orbitals: 3z^2 − r^2, x^2 − y^2 (for an example of d orbital transformation, see the identity, E, in a Td character table).

[Corribus]

The  s-orbital always transforms as totally symmetric A₁ representation (or the equivalent) because it never changes with respect to any symmetry operation.

The p-orbitals, as pointed out earlier, transform as the x, y, and z linear bases. In D_3h, these correspond to the E'' representation for the (x,y)-oriented p-orbitals (they transform together) and the A''₂ representation for the z-oriented p-oribtials.

The d-orbitals transform as the (xy), (xz), (yz), (x²-y²), and z² quadratic bases, which you should be able to look up in a good D_3h character table.

So now you should be able to find commonalities between the two sets of representations.

[shafaifer]

Please see this attachment (attachment of the D3h character table in the back of Peter Atkin's book: Physical Chemistry, is it good?)

-Since the s orbitals span A'1, they will not create MO's with the orbitals of the three equatorial Cl atoms in question.
-All of the 3p orbitals will do: (xz,yz) appear in E'' along with (Rx,Ry) and z appears in A''2. I am very unsure about this because at E'' I see only xz, yz Rx, Ry and not x,y,z. However, I see z at A''2, so my thinking was only the 3pz orbital will do. But you mentioned that they transformed together; that is why I stated that all of the 3 3p orbitals, 3px, 3py and 3pz will do.
-Only the 3dxz and 3dxy orbitals will do.   

Is this a right way?

With great thanks,

Shafaifer