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So I have a question on the form of the $T_2$ SALCs of methane. Below I show the $T_d$ character table and the accompanying reducible representation for the sigma framework (using, in the case of methane, the $4$ $1s$ H orbitals).

It is easy to show, and many people have, that the irreducible representations contained therein are $A_1$ and $T_2$.

If we then perform the projection operation for all of the elements of $T_2$, I get the following:

enter image description here

(The X's are intended to be "Chi's" here, signifying the $4$ $1s$ H orbitals that go into the procedure).

Meaning that, if we take the first line for instance, and perform the sum in the projection operator, we get: $$ 3\chi_1-\chi_2-\chi_3-\chi_4 $$

As our first un-normalized $\phi_t$ SALC. I can easily see by intuition how we can obtain the (again, not normalized) known SALCs for a $\sigma$ framework tetrahedron from this: $$ 3\chi_1-\chi_2-\chi_3-\chi_4 = $$ $$ \chi_1-\chi_2+\chi_3-\chi_4 $$ $$ \chi_1+\chi_2-\chi_3-\chi_4 $$ $$ \chi_1-\chi_2-\chi_3+\chi_4 $$

What I do not understand, is how we can derive this result without intuition. If I perform any number of Schmidt orthoganalizations using the other 3 rows of the projection operator table, I always end up with an exceedingly strange SALC, regardless of normalization and inclusion of overlap integrals $\chi_i{\times}\chi_j = S_{ij}$.

Is there any mathematical tool that can be used to generate these known SALCs from the one above?

Again, let me stress, I can clearly see the intuitive path, but I always rest easier when I know there is a reproducible, rigorous theory that leads there.

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  • $\begingroup$ This is worked out in David Bishop's book "Group Theory and Chemistry" on p. 238 and 239, which you might be able to view in Google Books. It's a bit tedious to write out. The gist is to set up each of the combinations you have (eg 3x1-x2-x3-x4) equal to a linear combination of the three p orbitals (since they are the T2 AO's). Then use the linearity of the symmetry operators to write equations to solve for px, py and pz in terms of x1, x2, x3 and x4. The solutions will be the three combinations you figured out intuitively. If you can't find it, I can type it out at some point. $\endgroup$ – Andrew Feb 5 at 21:32
  • $\begingroup$ Thank you Andrew! I've got that book nearby. I've noticed that accepting my own answers when others have provided citations is frowned upon, would you like to write either this, or the actual solution as an answer so I can accept it? Or would you rather I do so? $\endgroup$ – Yajibromine Feb 5 at 21:35
  • $\begingroup$ I can write something up. In case the page numbers in my version are different, look for equation 11-5.13 in the section "hybrid orbitals". That's the final answer you're looking for, and its derivation is the preceding few pages. $\endgroup$ – Andrew Feb 5 at 21:40
  • $\begingroup$ I don't think there's any stigma against self-answering based on resources suggested by others, as long as you give appropriate credit (which isn't so much of an intellectual property thing, but more of a courtesy, I think). $\endgroup$ – orthocresol Feb 6 at 1:52
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UPDATE: added more detail for those without access to the referenced text.

As noted in the comments, the details of this process can be found in "Group Theory and Chemistry" by David M. Bishop (Courier Corporation, 1993) p. 238-9. Here's a summary:

Following the same process that you started, you can get two other projections as (using your notation):

$-\chi_1+3\chi_2−\chi_3−\chi_4$

and

$-\chi_1−\chi_2+3\chi_3−\chi_4$

All three of these normalize with a factor of $\frac{1}{\sqrt{12}}$.

The challenge with degenerate orbitals is that there is an infinite number of linearly independent combinations that will combine to give the desired $T_2$ symmetry properties. Schmidt orthogonalization can be used to generate an orthogonal set, but there is an infinite number of those as well.

To prove that the canonical set is valid, one need only show that all three are linear combinations of the above projections and that they are orthogonal to each other. But to derive the canonical set without prior knowledge, we use the procedure below.

We want a set that is orthonormal and matches up with the central atomic orbitals. That is, each has symmetry that matches a member of the $p$ orbital set of the central atom. First, we set the $t_2$ central atom orbitals as $p_x$, $p_y$ and $p_z$. Then we consider that we must be able to make the hybrid orbital that coincides with each ligand bond from a linear combination of the $s$ and $p$ atomic orbitals. This is essentially a statement of the principle underlying the concept of hybridization. Furthermore, since the $s$ orbital only adds magnitude, not direction, an unnormalized hyrbid orbital can be made from a linear combination of $p$ orbitals only. Finally, any member of the $t_2$ ligand group orbitals must similarly be able to be constructed from a linear combination of the central atom $p$ orbitals.

Thus, $(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12} = (a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)$ for some $a_1, b_1, c_1$ and likewise for the other two projections (adding a minus sign where necessary for directionality with respect to the positive lobes of the $p$ orbitals).

We can then take advantage of the linearity of the symmetry operators to set up equations and solve for $p_x$, $p_y$, and $p_z$ in terms of $\chi_1$, $\chi_2$, $\chi_3$ and $\chi_4$, and the coefficients of those solutions are the coefficients for the H1s orbitals in each of the three $t_2$ ligand group orbitals.

For example, let $\chi_1$ be the hybrid orbital oriented in the positive $x,y$ and $z$ octant of Cartesian space, and let the $C_{3a}$ axis run through it. Thus,

$O_{C3a}(\chi_1)=\chi_1$

$O_{C3a}(\chi_2)=\chi_3$

$O_{C3a}(\chi_3)=\chi_4$

$O_{C3a}(\chi_4)=\chi_2$

Because the operator is linear, we now have that

$O_{C3a}[(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}]=(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}$, i.e. this projection is unchanged by that transformation because $\chi_2$, $\chi_3$, and $\chi_4$ are equivalent in the expression.

Therefore, we can further state that

$O_{C3a}[(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)]=(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)$

We can also determine by inspection that $O_{C3a}(p_x)=p_z$,$O_{C3a}(p_y)=p_x$, and $O_{C3a}(p_z)=p_y$. Therefore, we also have that

$O_{C3a}[(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)]=(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_z+b_1p_x+c_1p_x)$

and we can conclude that $a_1=b_1=c_1$. Thus,

$(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}=(p_x+p_y+p_z)/\sqrt{3}$.

Repeating this for the other symmetry transformations (for example $C_{3b}$ and $C_{3c}$) gives a system of equations that can be solved to yield:

$p_x=\frac12 (\chi_1−\chi_2+\chi_3−\chi_4)$

$p_y=\frac12 (\chi_1−\chi_2-\chi_3+\chi_4)$

$p_z=\frac12 (\chi_1+\chi_2-\chi_3−\chi_4)$

The coefficients of $\chi_1$, $\chi_2$, $\chi_3$ and $\chi_4$ in these three expressions are the coefficients of the H1s orbitals to each canonical $t_2$ ligand group orbital.

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  • $\begingroup$ A clarifying question: If the ligand orbitals are offset from the central atom in a tetrahedron, would it not be impossible to combine the central atomic p orbitals in such a way as to be equivalent the SALCs? That seems to be the content of the equation with $a_1$, $b_1$, and $c_1$ coefficients. Put another way, because there is some non-perfect overlap between the ligand SALCs and the atomic orbitals, does this not imply that they cannot be made exactly equivalent? Why does the requirement that the SALCs and the AOs coincide imply an algebraic equality instead of just equal symmetry? $\endgroup$ – Yajibromine Feb 6 at 13:59
  • $\begingroup$ I believe it works because the hybrid bonding orbital and the ligand orbital are both treated as a unit length position vector extending from the origin (ie the central atom) in the direction of the ligand. Conceptually, the ligand is a point at the end of the vector and the bond is the vector itself, but mathematically they are the same object. Does that make any sense? $\endgroup$ – Andrew Feb 6 at 14:55
  • $\begingroup$ Not entirely, but I'll keep digging. Thanks so much for your response! $\endgroup$ – Yajibromine Feb 6 at 15:09
  • $\begingroup$ I fixed some obvious typos, hope you don't mind. I think the equality symbols in the last $4$ equations might better be replaced by a symbol that means 'transforms like'. Also it's not clear to me from the discussion where the $\chi_2$, $\chi_3$, and $\chi_4$ orbitals lie, although from the last $3$ equations they seem to be in the $\langle-1,-1,1\rangle$, $\langle1,-1,-1\rangle$, and $\langle-1,1,-1\rangle$ directions respectively. Also I think $O_{C3a}(p_x)=p_y$ and so on: you seem to have written the operator's inverse. $\endgroup$ – user5713492 Feb 8 at 20:08
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I feel like the hard approach to this problem is the easy one here. First consider the partners of an irreducible representation $\mu$ of a group $G$ of degree $n_{\mu}$ consisting of the $n_{\mu}\times n_{\mu}$ matrices $D^{(\mu)}(R)$ for all $R\in G$: $$R\phi_j^{\mu}(\vec r)=\phi_j^{(\mu)}(R^{-1}\vec r)=\sum_{k=1}^{n_{\mu}}\phi_k^{(\mu)}(\vec r)D_{kj}^{(\mu)}(R)$$ Where the $\phi_j^{(\mu)}(\vec r)$, $1\le j\le n_{\mu}$ are the partners that engender the irreducible representation $D^{(\mu)}$. Notice that we have to rotate coordinates backwards to rotate functions forwards! This is a tricky point that must always be kept in mind. For example, consider what happens if $\vec r=\langle2.3,1.7,-2.9\rangle$, $\phi_1^{(T_2)}(\vec r)=x$, and $R=C_{3(1\bar1\bar1)}$, which means that if we set our viewpoint at $\langle1,-1,-1\rangle$ and looked towards the origin, functions would be rotated $120°$ counterclockwise about a line containing our viewpoint and the origin. Rotating coordinates backwards, we have $R^{-1}\langle2.3,1.7,-2.9\rangle=\langle-1.7,-2.9,-2.3\rangle$ and picking out the $x$-component we get $R\phi_1^{T_{\mu}}(\vec r)=-1.7$, that is, $Rx=-y$ as it should be.

Now let's make sure that group multiplication works. For any $R,S\in G$ $$\begin{align}RS\phi_j^{(\mu)}(\vec r)&=R\phi_j^{(\mu)}(S^{-1}\vec r)=R\xi(\vec r)=\xi(R^{-1}\vec r)=\phi_j^{(\mu)}\left(S^{-1}(R^{-1}\vec r)\right)=\phi_j^{(\mu)}\left((RS)^{-1}\vec r\right)\\ &=R\sum_{k=1}^{n_{\mu}}\phi_k^{(\mu)}(\vec r)D_{kj}^{(\mu)}(S)=\sum_{\ell=1}^{n_{\mu}}\sum_{k=1}^{n_{\mu}}\phi_{\ell}^{\mu}(\vec r)D_{\ell k}^{(\mu)}(R)D_{\ell j}^{(\mu)}(S)\\ &=\sum_{\ell=1}^{n_{\mu}}\phi_{\ell}^{(\mu)}(\vec r)\left(\sum_{k=1}^{n_{\mu}}D_{lk}^{\mu}(R)D_{kj}^{(\mu)}(S)\right)=\sum_{\ell=1}^{n_{\mu}}\phi_{\ell}^{(\mu)}(\vec r)D_{\ell j}^{(\mu)}(RS)\end{align}$$ And on comparing coefficients of $\phi_{\ell}^{(\mu)}(\vec r)$ we see that the representation matrices $D^{(\mu)}(R)$ satisfy the group homomorphism. In the above we made the substitution $\xi(\vec r)=\phi_j^{(\mu)}(S^{-1}\vec r)$ to make clear where $R^{-1}$ should act and noted that $$(S^{-1}R^{-1})(RS)=S^{-1}(R^{-1}R)S=S^{-1}ES=S^{-1}S=E$$ so $(RS)^{-1}=S^{-1}R^{-1}$. Now we are ready to show that for any function $\phi(\vec r)$ and $j,k\in(1,n_{\mu})$, $$\sum_{S\in G}S\phi(\vec r)D_{jk}^{(\mu)}(S^{-1})$$ transforms as the $k^{th}$ partner of the $\mu^{th}$ irreducible representation of $G$: note that $S^{-1}=S^{-1}E=S^{-1}R^{-1}R=(RS)^{-1}R$ so for any $R\in G$ $$\begin{align}R\sum_{S\in G}S\phi(\vec r)D_{jk}^{(\mu)}(S^{-1})&=\sum_{S\in G}RS\phi(\vec r)D_{jk}^{(\mu)}\left((RS)^{-1}R\right)\\ &=\sum_{T\in G}T\phi(\vec r)D_{jk}^{(\mu)}(T^{-1}R)\\ &=\sum_{T\in G}T\phi(\vec r)\sum_{\ell=1}^{n_{\mu}}D_{j\ell}^{(\mu)}(T^{-1})D_{\ell k}^{(\mu)}(R)\\ &=\sum_{\ell=1}^{n_{\mu}}\left(\sum_{S\in G}S\phi(\vec r)D_{j\ell}^{(\mu)}(S^{-1}\right)D_{\ell k}^{(\mu)}(R)\end{align}$$ Where we have used the fact that as $S$ varies over all elements of $G$ so do all of its products $RS$ with and fixed element $R\in G$. The other partners have the same $j$ as above, but different $k$.

Now, let's do this for the $s$-orbital of the $4$ hydrogen atoms, $\psi_1$, $\psi_2$, $\psi_3$, and $\psi_4$ located in the $\langle1,1,1\rangle$, $\langle1,-1,-1\rangle$, $\langle-1,1,-1\rangle$, and $\langle-1,-1,1\rangle$ directions respectively. Let's start with a table of the operators of $T_d$, the representation matrices of $D^{(T_2)}$ engendered by $\{x,y,z\}$, and the transformed $s$-orbitals: $$\begin{array}{cccccc}E&C_{3(111)}&C_{3(111)}^2&C_{3(1\bar1\bar1)}&C_{3(1\bar1\bar1)}^2&C_{3(\bar11\bar1)}\\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}& \begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}& \begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}& \begin{bmatrix}0&0&-1\\-1&0&0\\0&1&0\end{bmatrix}& \begin{bmatrix}0&-1&0\\0&0&1\\-1&0&0\end{bmatrix}& \begin{bmatrix}0&0&1\\-1&0&0\\0&-1&0\end{bmatrix}\\ \psi_1&\psi_1&\psi_1&\psi_4&\psi_3&\psi_2\\ \psi_2&\psi_3&\psi_4&\psi_2&\psi_2&\psi_4\\ \psi_3&\psi_4&\psi_2&\psi_1&\psi_4&\psi_3\\ \psi_4&\psi_2&\psi_3&\psi_3&\psi_1&\psi_1\\ \hline \end{array}$$ $$\begin{array}{cccccc}C_{3(\bar11\bar1)}^2&C_{3(\bar1\bar11)}&C_{3(\bar1\bar11)}^2&C_{2(100)}&C_{2(010)}&C_{2(001)}\\ \begin{bmatrix}0&-1&0\\0&0&-1\\1&0&0\end{bmatrix}& \begin{bmatrix}0&0&-1\\1&0&0\\0&-1&0\end{bmatrix}& \begin{bmatrix}0&1&0\\0&0&-1\\-1&0&0\end{bmatrix}& \begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}& \begin{bmatrix}-1&0&0\\0&1&0\\0&0&-1\end{bmatrix}& \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}\\ \psi_4&\psi_3&\psi_2&\psi_2&\psi_3&\psi_4\\ \psi_1&\psi_1&\psi_3&\psi_1&\psi_4&\psi_3\\ \psi_3&\psi_2&\psi_1&\psi_4&\psi_1&\psi_2\\ \psi_2&\psi_4&\psi_4&\psi_3&\psi_2&\psi_1\\ \hline \end{array}$$ $$\begin{array}{cccccc}S_{4(100)}&S_{4(100)}^3&S_{4(010)}&S_{4(010)}^3&S_{4(001)}&S_{4(001)}^3\\ \begin{bmatrix}-1&0&0\\0&0&-1\\0&1&0\end{bmatrix}& \begin{bmatrix}-1&0&0\\0&0&1\\0&-1&0\end{bmatrix}& \begin{bmatrix}0&0&1\\0&-1&0\\-1&0&0\end{bmatrix}& \begin{bmatrix}0&0&-1\\0&-1&0\\1&0&0\end{bmatrix}& \begin{bmatrix}0&-1&0\\1&0&0\\0&0&-1\end{bmatrix}& \begin{bmatrix}0&1&0\\-1&0&0\\0&0&-1\end{bmatrix}\\ \psi_4&\psi_3&\psi_2&\psi_4&\psi_3&\psi_2\\ \psi_3&\psi_4&\psi_3&\psi_1&\psi_1&\psi_4\\ \psi_1&\psi_2&\psi_4&\psi_2&\psi_4&\psi_1\\ \psi_2&\psi_1&\psi_1&\psi_3&\psi_2&\psi_3\\ \hline \end{array}$$ $$\begin{array}{cccccc}\sigma_{110}&\sigma_{101}&\sigma_{1\bar10}&\sigma_{10\bar1}&\sigma_{011}&\sigma_{01\bar1}\\ \begin{bmatrix}0&-1&0\\-1&0&0\\0&0&1\end{bmatrix}& \begin{bmatrix}0&0&-1\\0&1&0\\-1&0&0\end{bmatrix}& \begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}& \begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}& \begin{bmatrix}1&0&0\\0&0&-1\\0&-1&0\end{bmatrix}& \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}\\ \psi_4&\psi_3&\psi_1&\psi_1&\psi_2&\psi_1\\ \psi_2&\psi_2&\psi_3&\psi_4&\psi_1&\psi_2\\ \psi_3&\psi_1&\psi_2&\psi_3&\psi_3&\psi_4\\ \psi_1&\psi_4&\psi_4&\psi_2&\psi_4&\psi_3\\ \end{array}$$ Starting with any of the $4$ $s$-orbitals and any of the $9$ matrix elements we could do $36$ projections if we were so disposed, but here $1$ will suffice. I will choose the row $j=1$ and the column $k=2$ and orbital $\psi_3$. The first $2$ columns of the table have $0$ in the $1,2$ position, but the third has a $1$ there. We have to be careful, though, because the formula requires the inverse operation to the matrix, so we select from the $C_{3(111)}$ column of the $\psi_3$ row because $\left(C_{3(111)}\right)^{-1}=C_{3(111)}^2$. Going through the table in this fashion we get $$\psi_4-\psi_1-\psi_3+\psi_2-\psi_1+\psi_4-\psi_3+\psi_2$$ Normalize to $\phi_2^{(T_2)}=\frac12\left(-\psi_1+\psi_2-\psi_3+\psi_4\right)$ transforms as $y$.

So you can see that you can directly project to partners using the matrices of the irreducible representation rather than projecting through traces and subsequently resolving the partners. Just use the matrix elements rather than the traces in the similar formula.

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  • $\begingroup$ Thanks for the corrections above. A question on your method, though: by choosing j=1, k=2 and phi3, aren't you choosing the projections that give the desired result? If we know beforehand what we want, we can also manipulate the projections resulting from the traditional projection method (which gives the +3, -1,-1,-1 combinations) to show that the canonical orbitals are linear combinations thereof, but if we imagine not knowing what the answer is ahead of time, how would you know in your method which projection would give the orbital that transforms as $y$? $\endgroup$ – Andrew Feb 8 at 21:33
  • $\begingroup$ @Andrew, I chose $j=1$, $k=2$, and $\phi_3$ more or less at random. Choosing $k=1$ would produce the partner that transforms as $x$ or $k=3$ as $z$. Choosing a different $j\ne1$ might produce a result with a different phase or normalization or even a partner from a different set or zero (although not in this case). For a more interesting example of what this method can achieve, try $$\phi(\vec r)=x+xz+2z^3-3x^2z-3y^2z$$ with all $9$ possible projections. You should find that it picks of any $j^{th}$ partner and multiplies by $|G|/n_{\mu}$ and transfers to the $k^{th}$ partner. $\endgroup$ – user5713492 Feb 8 at 23:36
  • $\begingroup$ So after reading your answer it seems you are just cutting to the chase and using what Cotton would call the "full" projection operator, instead of what I did (using the incomplete projection operator, as in summing over the trace).Thank you for calling attention to the distinction. $\endgroup$ – Yajibromine Feb 10 at 18:09
  • $\begingroup$ I was thinking more in terms of Melvin Lax, Symmetry Principles in Solid State and Molecular Physics, John Wiley and Sons, New York 1974, pp. 99-101. This seems to go into more detail than F. A. Cotton, Chemical Applications of Group Theory, John Wiley and Sons, New York 1963. When you want a partner it seems to make sense to project directly to a partner. Of course given one partner here it's easy to get to the others because $C_{3(111)}y=z$ and $C_{3(111)}z=x$. $\endgroup$ – user5713492 Feb 10 at 19:57
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This is an addendum to @Andrew 's answer which I will still keep as the correct one. I went messing about in Mathematica, and because these SALCs are generated from 1s orbitals with no imaginary component, we can actually check the projection method for the correct production of these SALCs above.

And in fact, if you utilize the Schmidt orthogonalization process as I allude to above on the 3 projections $3\chi_i-\chi_j−\chi_k−\chi_l$, you do get the qualitative form of the expected SALCs (the ones with form $\chi_1−\chi_2+\chi_3−\chi_4$ instead)

The actual form of the Schmidt orthogonalized SALCs is incredibly ugly (especially since I used cartesian coordinates), but they reproduce the following density slices:

enter image description here

This is exactly what Andrew's answer from David Bishop's book leads to as well! The two methods appear to be one and the same!

I short, this is done by inputting $$ R_{1s} = 2Z^{3/2}*e^{\frac{-2Z \frac{\sqrt{(x+c_x)^2+(y+c_y)^2+(z+c_z)^2})}{n}}{2}} $$ and $$ Y_{1s} =\frac{1}{4\pi}^{1/2} $$

Such that

$$ \chi_1 = R_{1s}*Y_{1s} $$

Four of these AO's are then placed at the corners of an tetrahedron (using the constant offsets for x, y, and z in the expansion of r), and the combinations thereof which the projection operator indicates are tested.

The first combination is $3\chi_1-\chi_2−\chi_3−\chi_4$ with normalization (divide by $\sqrt{12}$), but the next two require Schmidt orthogonalization w.r.t the first function, ie:

$$ |\chi_2> = |\chi_2> - <\chi_2||\chi_1>|\chi_1> $$

Do this for $\chi_2$ and $\chi_3$ and then plot the density surfaces. They will be the ones shown here, and completely in keeping with the expected $t_2$ SALCs.

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