銀月遊俠 2012-12-13 07:18 AM

Actually, they can both act as lewis and bronsted acid.

And if simply ask which acid is stringer how to answer?

I have big problems in assigning vibrational modes of molecules by group theory ( if that's what I am using)

I have only superficial knowledge of character table that is I only know the meaning of 1 and -1 means, -2,0,+2 already out of my range.

[/size][size=4]Is it right to do the following in assigning vibrational modes for a molecule.[/size]

[size=4][/size][size=4]1. work out the point group of the molecule at hand.[/size]

[size=4][/size][size=4]2. work out the MOs of the molecule[/size]

[size=4][/size][size=4]3. make analogy of the MOs to the vibrational modes by symmetry[/size]

[size=4][/size][size=4]4. determine the IR activity (having x,y,z symmetry) Raman activity (having xy, yz, xz, quartic component)[/size]

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Is that the formula for calculating vibrational modes (3N-6 , 3N-5) is not useful? Because I noticed that for a BH3 molecule the formula will predict 6 vibrational modes present, but by drawing MOs the electrons only fill the a1' and e' so it should have three vibrational modes. what is wrong with my concepts?

Is it true that MOs or vibrational modes having the same symmetry must degenerate? I noticed that it is not, right?

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[[i] 本帖最後由 銀月遊俠 於 2012-12-13 09:56 AM 編輯 [/i]]

jmlo 2012-12-13 12:58 PM

We know that going down the group elements get more basic. It means that Al is more basic than B, and thus B(OH)3 is more acidic than Al(OH)3. Moreover, the better pi-orbital overlap between B and O favors the delocalization of the negative charge of the conjugate base.

[quote] Actually, they can both act as lewis and bronsted acid.

And if simply ask which acid is stringer how to answer? [/quote]

In general, when you are asked to compare the acidity of two species, we are expected to compare their Bronsted acidity unless Lewis acidity is clearly stated.

[quote] I have big problems in assigning vibrational modes of molecules by group theory ( if that's what I am using)

I have only superficial knowledge of character table that is I only know the meaning of 1 and -1 means, -2,0,+2 already out of my range.

Is it right to do the following in assigning vibrational modes for a molecule.

1. work out the point group of the molecule at hand.

2. work out the MOs of the molecule

3. make analogy of the MOs to the vibrational modes by symmetry

4. determine the IR activity (having x,y,z symmetry) Raman activity (having xy, yz, xz, quartic component) [/quote]

No, the procedures are not right. There is no one-to-one correspondence between MO's and vibrational modes. For example, both H-F and H-Cl have one vibrational mode, but HCl has more occupied MO's than HF.

The general procedures of finding the symmetry of the vibrational modes are as follows. It is always assumed that the character tables are available; otherwise, no analysis can be done.

(1) Determine the point group of the molecule and locate the appropriate character table.

(2) Find the characters of the coordinate axes; this is just the sum of the characters for the rows corresponding to x, y and z.

(3) Count the number of stationary atoms under each symmetry operation.

(4) Calculate the total reducible representations by multiplying (2) and (3) above.

(5) Subtract the translational representations (i.e., x, y and z rows) from the total representations.

(6) Subtract the rotational representations (i.e., Rx, Ry and Rz rows) from the total representations.

(7) The remaining is the vibrational representations; it can be decomposed into various irreducible representations of the vibrational modes.

(8) Assign each vibrational mode to the appropriate irreducible representation by checking the corresponding characters under symmetry operations. This step can be done by intuition (in other words, guessing) or step-by-step analysis.

[quote] Is that the formula for calculating vibrational modes (3N-6 , 3N-5) is not useful? Because I noticed that for a BH3 molecule the formula will predict 6 vibrational modes present, but by drawing MOs the electrons only fill the a1' and e' so it should have three vibrational modes. what is wrong with my concepts? [/quote]

As said above, your approach is wrong since there is no direct relationship between occupied MO's and vibrational modes.

If the procedures described above are carried out properly, you should see that the vibrational representation is composed of A1' + A2' + 2E', in total six vibrational modes. This number agrees with that predicted by 3N-6 rule.

(Here I skip the procedures how vibrational representation is decomposed. You can test yourself by working out the answers. It is not difficult.)

[quote] Is it true that MOs or vibrational modes having the same symmetry must degenerate? I noticed that it is not, right? [/quote]

No. Only MO's or vibrational modes belonging to the multi-dimensional representations (e.g. E, T) are necessarily degenerate.

銀月遊俠 2012-12-13 04:10 PM

I understand there is no correlation between vibrational modes and MOs.

Actually, I haven't learnt the procedure to reduce the representation.

And at this stage I just know no. of bonds = no. of stretching. And then assign the LGOs with symmetry and look for IR/Raman activity from character table.

I still got a problem how to assign the stretching modes for C1 point group? In character table it has only E for operation and it has a C1 rotational axis does that mean it have got no dipole moment in the directions perpendicular to the axis? so all its stretching should be inacitve in IR? how about Raman then?

[[i] 本帖最後由 銀月遊俠 於 2012-12-13 07:06 PM 編輯 [/i]]

jmlo 2012-12-13 11:59 PM

Note that boric acid combines with H2O to produce H+ and metaborate through the following equilibrium:

B(OH)3 + H2O <----> [B(OH)4]- + H+

and metaborate ions polymerize when concentration goes up in the solution. Moreover, at concentrated boric acid solution, the following equilibrium exists:

3B(OH)3 <----> [B3O3(OH)4]- + H+ + 2H2O

which has the pKa = 6.8

Such reactions are not seen for Al(OH)3.

[quote] I understand there is no correlation between vibrational modes and MOs.

Actually, I haven't learnt the procedure to reduce the representation.

And at this stage I just know no. of bonds = no. of stretching. And then assign the LGOs with symmetry and look for IR/Raman activity from character table. [/quote]

Are you learning how IR spectra are assigned for metal complexes? And how to predict numbers of IR bands for compounds such as metal carbonyls in various geometries?

[quote] I still got a problem how to assign the stretching modes for C1 point group? In character table it has only E for operation and it has a C1 rotational axis does that mean it have got no dipole moment in the directions perpendicular to the axis? [/quote]

The assignment of the C1 axis for molecules belonging to this symmetry group is arbitray. For sure, those molecules should have non-zero dipole moment.

[quote] so all its stretching should be inacitve in IR? how about Raman then? [/quote]

All vibrational modes for a molecule in C1 symmetry are both IR and Raman active.

銀月遊俠 2012-12-14 12:33 AM

I have got a few exam questions from my course I hope you may explain them to me.

1.Which of the following sets of symmetry operations forms a mathematical group?

a) {E, C4, 4σv} b){E, C4, C43, 4σv}

c){E, C4, C2, C4^3, 4σv} d){E, C3, C3^2, 2σv}

I only know why d is wrong

2. which of the following statements is false?

a) For molecules with a Cnv point group, all the normal vibration modes are both IR and Raman active.

b) For molecules with a Oh point group, if the normal vibration modes are IR active, they must be Raman inactive.

c) For molecules with a Oh point group, some normal vibration modes can be both IR and Raman active.

d) For molecules with a C1 point group, all the normal vibration modes are both IR and Raman active.

why cnv point group have all normal modes both active?

3.For a solution of NaA/HA, where HA is a monoprotic weak acid, which one of the following statements is correct when the solution is in equilibrium? [c]

a) the pH of the solution is equal to pKa.

b) the concentration of the weak acid is greater than that of the conjugate base.

c) pH-pKa is equal to zero when both the acid and its conjugate base have the same concentration.

d) The concentration of the conjugate base is greater than that of the weak acid

I thought b is right

[img]http://i1189.photobucket.com/albums/z428/AnthonyCChem/final2003.png[/img]

Why (b) Ir is 3+ and (c) Fe is 3+?

jmlo 2012-12-14 06:47 AM

a) {E, C4, 4σv} b){E, C4, C43, 4σv}

c){E, C4, C2, C4^3, 4σv} d){E, C3, C3^2, 2σv}

I only know why d is wrong [/quote]

There are four fundamental rules a set of symmetry elements have to fulfill in order for them to form a group. They are:

(i) Closure: The sum of any two elements is an element of the group

(ii) Associativity: The product of three elements should fulfill the following relationship - A(BC) = (AB)C

(iii) Identity: There exists an identity element which when multiplied with any other element yields the element. i.e., EA = A

(iv) Inverse: Each element should have an inverse which is also an element of the group.

When you check the four given sets, the first two do not fulfill (i). For example, the product C4 C4 = C4^2 = C2 is not included in these sets. It means that these sets are not closed and do not form a proper group. For the last set, there is a missing σv elements as the highest order of rotation is 3.

[quote] 2. which of the following statements is false?

a) For molecules with a Cnv point group, all the normal vibration modes are both IR and Raman active.

b) For molecules with a Oh point group, if the normal vibration modes are IR active, they must be Raman inactive.

c) For molecules with a Oh point group, some normal vibration modes can be both IR and Raman active.

d) For molecules with a C1 point group, all the normal vibration modes are both IR and Raman active.

why cnv point group have all normal modes both active? [/quote]

The reasonings are as follows:

(a) For Cnv point groups, z transforms into the same representation as z^2. In the meantime, (x,y) transform into the same representation as (xz,yz). This is due to the absence of an inversion center for Cnv groups. Therefore, all vibrational modes, which belong to the representations dictated by x,y and z, are IR active and Raman active.

(b) The Oh point group has an element of inversion. It implies that (x,y,z) and (z^2, x^2, y^2, xy, xz, yz) transform into the representations of opposite symmetry. In other words, (x,y,z) set is in ungerade symmetry while (x^2, y^2, z^2, xy, xz, yz) set is in gerade symemtry. Since the ground state of a system in Oh point group has a gerade symmetry, the vibrational modes which are IR active have to be in ungerade symmetry. On the contrary, those modes which are Raman active should be in gerade symmetry. Therefore, IR active vibrational bands are Raman inactive.

(c) Due to the complementary properties of IR and Raman active bands for systems in Oh point group, it is not possible to have vibrational modes which are both IR and Raman active.

(d) Since (x,y,z) and (z^2,x^2,y^2,xy,xz,yz) tranform into A representation in C1 point group, all vibrational modes are IR and Raman active.

[quote] 3.For a solution of NaA/HA, where HA is a monoprotic weak acid, which one of the following statements is correct when the solution is in equilibrium? [c]

a) the pH of the solution is equal to pKa.

b) the concentration of the weak acid is greater than that of the conjugate base.

c) pH-pKa is equal to zero when both the acid and its conjugate base have the same concentration.

d) The concentration of the conjugate base is greater than that of the weak acid

I thought b is right [/quote]

This question is about the Henderson-Hasselbalch equation:

pH = pKa - log [HA]/[A-]

which is defined for a buffer solution at equilibrium. According to this equation, (C) is correct since the logarithmic term vanishes when [HA] = [A-]; this results in pH = pKa or pH - pKa = 0. Depending upon the recipe of the buffer preparation, [A-] may be greater or smaller than [HA]. Therefore, (A), (B) and (D) are not necessarily right.

[quote] Why (b) Ir is 3+ and (c) Fe is 3+? [/quote]

(b) The complex is by itself neutral. Since H, PPh2 and benzyl are all considered -1 anionic ligands, Ir has to have the oxidation state of +3.

(c) Note that SCH3 is an -1 anionic ligand while S is an -2 anionic ligand, for electroneutrality, Fe has to have +3 oxidation state. i.e.,

4*Fe + 4*(-1) + 4*(-2) = 0

Fe = 3

銀月遊俠 2012-12-14 02:23 PM

Thank you for all your explanations.

One last question to end this post.

Is H2C=C=CHCl chiral or not because I cannot find any Sn axis but by viewing the mirror image it is superimpossible.

One last question to end this post.

Is H2C=C=CHCl chiral or not because I cannot find any Sn axis but by viewing the mirror image it is superimpossible.

jmlo 2012-12-14 10:18 PM

H2C=C=CHCl is not chiral. It has a Cs symmetry. For allene-based molecules, the units at two ends, i.e., CH2 and CHCl, are perpendicular. It follows that there exists a mirror plane of symmetry containing CHCl that bisects the molecule.

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