Question #706c0

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Question #706c0

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Answer 1 · 2017-12-04T06:18:31

I'll help with problems 1 and 2, but not 3, because it would make this too long.

Here are the main points:

The MO diagram can be found here.
Purely from the angular overlap method perspective, square planar is favored because there is $1.33e_(sigma)$ less $sigma$ destabilization.

DISCLAIMER: LONG ANSWER!

$1)$

POINT GROUP AND SYMMETRY ELEMENTS

For an $"AB"_4$ square planar compound, take a right-handed coordinate system where the ligands $B$ lie on the $x$ and $y$ axes.

https://upload.wikimedia.org/

You should work through this and find the symmetry elements belonging to the $D_(4h)$ point group:

(NOTE: You only need to identify $hatC_4(z)$ , $hatC_2'$ , and $hatsigma_h$ to confirm that you have a $D_(4h)$ point group, and then whip out a character table to get the rest of the elements.)

$hatE$ , the identity, because everything has it.
$hatC_4^n(z)$ , the principal axis of 4-fold rotation symmetry. You can use this up to $3$ times before getting the identity back.
$hatC_2'$ , an axis of 2-fold rotation symmetry on the $xy$ plane, along a trans $"B"-"A"-"B"$ bond.
$hatC_2''$ , a rotation axis of 2-fold symmetry on the $xy$ plane, bisecting a cis $"B"-"A"-"B"$ bond.
$hatsigma_v$ , a vertical mirror plane colinear with the $hatC_2'$ axis, along a trans $"B"-"A"-"B"$ bond.
$hatsigma_d$ , a dihedral mirror plane colinear with the $hatC_2''$ axis, bisecting a cis $"B"-"A"-"B"$ bond.
$hatsigma_h$ , a horizontal mirror plane on the $xy$ plane.
$hatS_4$ , an improper rotation axis of 4-fold symmetry, because $hatS_4 = hatC_4hatsigma_h$ , which must be in the point group from the properties of a group (any element in a point group can be generated by multiplication of two other elements in the group).
$hati$ , an inversion center, because all the $"B"$ ligands are identical and there are an even number of them. Thus, $(x,y,z) = (-x,-y,-z)$ for each of them.

CHARACTER TABLE

Its character table, which you should have in front of you, is:

https://www.webqc.org/

I assume you know a few features of the character table, such as:

The sum of the coefficients of the rotation operators $hatR$ gives the order $h$ of the group.
The $A//B$ and $E$ irreducible representations (IRREPs) are one-fold and two-fold degenerate, respectively. This is why the character of $E_g$ under $hatE$ is $2$ and not $1$ .
The "linear" column gives you the orbitals that transform under particular symmetries ( $p_x,p_y,p_z$ ).
The "quadratic" column gives you the orbitals that transform under those particular symmetries ( $overbrace(s)^(x^2 + y^2), d_(z^2), d_(x^2-y^2), d_(xy), [d_(xz),d_(yz)]$ ).

The next thing to do is to generate the reducible representation for the ligand orbitals. Without doing that, we won't know which metal orbitals match.

Since we want only $sigma$ bonding, we assume the ligands $B$ use an $s$ orbital basis and a $p_y$ orbital basis (where $p_y$ points inwards from $B$ towards $A$ ).

However, when doing this for $sigma$ bonding, they both give the same result so we will only show the work once.

GENERATING THE REDUCIBLE REPRESENTATION: $bbs$ OR $bb(p_y)$ ORBITAL BASIS

The reducible representation $Gamma_s$ (as well as $Gamma_(p_y)$ ) is generated by taking every operator in the group and applying it to the four $B$ atoms exactly as they are arranged in the molecule, using spherical orbitals (or dumbbell orbitals pointed inwards, for $p_y$ orbitals).

If the operation returns the orbital unmoved, put $bb1$ in the reducible representation.
If the operation returns the orbital with the opposite phase, put $bb(-1)$ in the reducible representation.
If the operation returns the orbital moved from where it was before, put $bb0$ in the reducible representation.

The results are:

$" "" "hatE" "hatC_4" "hatC_2" "hatC_2'" "hatC_2''" "hati" "hatS_4" "hatsigma_h" "hatsigma_v" "hatsigma_d$
$Gamma_s = 4" "0" "" "0" "color(white)(.)2" "" "0" "" "color(white)(.)0" "0" "color(white)(.)4" "color(white)(.)2" "color(white)(.)0$

$" "color(white)(.,.)hatE" "hatC_4" "hatC_2" "hatC_2'" "hatC_2''" "hati" "hatS_4" "hatsigma_h" "hatsigma_v" "hatsigma_d$
$Gamma_(p_y) = 4" "0" "" "0" "color(white)(.)2" "" "0" "" "color(white)(.)0" "0" "color(white)(.)4" "color(white)(.)2" "color(white)(.)0$

REDUCING TO A SET OF IRREPS: $bbs$ ORBITAL BASIS

Here we seek two or more IRREPs whose line of characters adds up to this. Among them must be the totally symmetric one, $A_(1g)$ , so by subtraction:

$Gamma_s - Gamma_(A_(1g))$

$= 3" "-1" "-1" "1" "-1" "-1" "-1" "3" "1" "-1$

With an even number of orbitals, you can choose their phase so that trans ligands have the opposite phase and cis ligands have same phase. This is antisymmetric with respect to inversion, so $E_u$ (ungerade) is contained in $Gamma_s$ .

$Gamma_s - Gamma_(A_(1g)) - Gamma_(E_u)$

$= 1" "-1" "1" "1" "-1" "1" "-1" "1" "1" "-1$

And by inspection of the character table, this row of characters matches with $B_(1g)$ . So, the IRREPs are:

$color(blue)(Gamma_s = A_(1g) + B_(1g) + E_u)$

$color(blue)(Gamma_(p_y) = A_(1g) + B_(1g) + E_u)$

METAL ORBITAL SYMMETRIES

This isn't as complicated. You can look at the character table and directly read these off to be:

$" "d_(z^2) " "harr A_(1g)$
$" "d_(x^2-y^2) harr B_(1g)$
$" "color(red)(d_(xy)) " "harr color(red)(B_(2g))$ (nonbonding)
$[color(red)(d_(xz), d_(yz))] harr color(red)(E_u)$ (nonbonding)

The orbitals with different symmetries don't interact. So, we get the following interactions:

$"Metal"$ $s$ with $A_(1g)$ , making an $a_(1g)$ bonding and $a_(1g)^"*"$ antibonding MO.

Although $d_(z^2)$ is $A_(1g)$ , it is relatively nonbonding because there are no ligands on the $z$ axis. However, due to the metal $s$ orbital and the ligand $A_(1g)$ orbitals, there is some stabilization even without direct interaction.

$"Metal"$ $d_(x^2-y^2)$ with ligand $B_(1g)$ , making a $b_(1g)$ bonding and $b_(1g)^"*"$ antibonding MO.

$"Metal"$ $color(red)(d_(xy))$ ( $color(red)(B_(2g))$ ) orbital becomes EXACTLY nonbonding due to no matching orbital symmetries.

$"Metal"$ $color(red)(d_(xz), d_(yz))$ ( $color(red)(E_u)$ ) orbitals become EXACTLY nonbonding (ignoring metal $p$ orbitals).

This results in the following orbital diagram without the MOs so far (ignoring metal $p$ and $f$ orbitals for simplicity).

[

When you make the MOs, use relative energy orderings and you should get something like this:

Note that this won't exactly match the full square planar d orbital splitting diagram because we neglected the $pi$ interactions and the metal $p$ orbitals. Those would stabilize the $d_(z^2)$ , destabilize the $d_(xy)$ , and stabilize the $(d_(xz), d_(yz))$ .

$2)$

ANGULAR OVERLAP METHOD

For $sigma$ interactions (Inorganic Chemistry, Miessler et al., pg. 384):

Inorganic Chemistry, Miessler et al., pg. 384

For square planar, ignore positions $1$ and $6$ in the octahedral diagram.
For tetrahedral, use the central diagram.

Since we only consider $sigma$ interactions, and the $sigma$ MOs of the ligands are LOWER in energy than the metal orbitals, they can only destabilize them in energy.

Inorganic Chemistry, Miessler et al., pg. 383

From the table for square planar,

$d_(z^2)$ is destabilized by $1/4e_sigma$ due to ligands $2,3,4,5$ (rows 2 - 5, column 2). This adds up to $color(blue)(e_sigma)$ .
$d_(x^2-y^2)$ is destabilized by $3/4e_sigma$ due to ligands $2,3,4,5$ (rows 2 - 5, column 3). This adds up to $color(blue)(3e_sigma)$ .
The $xy$ , $xz$ , and $yz$ are nonbonding because they have no destabilizing or stabilizing contribution (rows 3 - 5, columns 4 - 6).

From the table for tetrahedral,

$d_(xy)$ , $d_(xz)$ , and $d_(yz)$ are all destabilized by $1/3e_sigma$ due to ligands $7,8,9,10$ (rows 7 - 10, columns 4 - 6). This adds up to $color(blue)(4/3e_sigma)$ for each orbital.
The $z^2$ and $x^2-y^2$ are nonbonding because they have no destabilizing or stabilizing contribution (rows 7 - 10, columns 2 - 3).

Based purely on the angular overlap method, since the ligands are destabilizing the metal $d$ orbitals by

$e_sigma + 3e_sigma = color(blue)(4e_(sigma))$ in a square planar regime

and

$4 xx 4/3e_sigma = color(blue)(5.33e_(sigma))$ in a tetrahedral regime,

the square planar shape is energetically favored. This is an OK approximation because the $"Cl"^(-)$ are weak-field $sigma$ donors with a bit of $pi$ donor behavior.

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