What is the hybridization of copper in tetraamminecopper(II) in water? Why is it NOT sp^3, and how do you prove this using Group Theory?

1 Answer
Aug 26, 2017

It would be sp^2d (not sp^3, and not p^2d^2). The complex is square planar with four electron groups (rather than six).

By the way, a great deal of this answer uses the concept of symmetry elements in chemical group theory. You should read that first if you don't want to be too confused.


GEOMETRY CONSIDERATIONS

Well, tetramminecopper(II) is square planar (and not tetrahedral). How do I know? Well, in water, as it must be (why?), it is more accurately represented as:

["Cu"("NH"_3)_4("H"_2"O")_2]^(2+)

https://chemistry.stackexchange.com/https://chemistry.stackexchange.com/

which is an octahedral complex. Without the waters in consideration, we thus have a square planar complex. As such, since this is not tetrahedral, this is not sp^3 hybridization.

SYMMETRY CONSIDERATIONS

This complex belongs to the D_(4h) point group. You may want to review the symmetry elements here. Some of the elements this point group has are:

  • It has an identity element E, which is just there for completeness.
  • It has a C_4(z) principal rotation axis perpendicular to its plane (i.e. rotate 90^@ and an indistinguishable configuration results).
  • It has (at least) one C_2' rotation axis coplanar with the compound's plane, but perpendicular to the C_4 axis. That is, it is on the xy plane, along the axes.
  • It has (at least) one C_2'' rotation axis on the xy plane, just as the C_2' axes are, but bisecting them instead.
  • It has a sigma_h horizontal mirror plane, coplanar with the compound's plane.
  • It also has vertical mirror planes sigma_v that line up with the coordinate axes (perpendicular to sigma_h), and dihedral mirror planes sigma_d that bisect them.
  • It has the inversion element, i.

http://symmetry.otterbein.edu/http://symmetry.otterbein.edu/

This means it follows this character table:

http://www.webqc.org/http://www.webqc.org/

OBTAINING THE REQUIRED ORBITAL SYMMETRIES

To determine what the hybridization COULD be systematically, we generate a so-called reducible representation Gamma_(NH_3) using the ammine ligands as a spherical basis (since they are all facing inwards and sigma bonding, we can ignore phase).

To do this, operate on the complex using each operation, and examine how the "NH"_3 ligands move due to each operation.

  • Unmoved outer atoms contribute 1.
  • Outer atoms that moved from their original position contribute 0.

" "" "" "" "E" "C_4(z)" "C_2(z)" "C_2'" "C_2''" "i" "S_4" "sigma_h" "sigma_v" "sigma_d

Gamma_(NH_3) = " "4" "0" "" "" "0" "" "2" "" "0" "" "0" "" "0" "4" "" "2" "0

Skipping some steps, we reduce this using a reduction calculator for Gamma_("general") to get this set of irreducible representations:

ul(Gamma_(NH_3)^("red") = A_(1g) + B_(1g) + E_u)

The point of this was to see which symmetries the "NH"_3 use as a group, so that the central atom can match those symmetries for its hybridization.

EXTRACTING COMBINATIONS OF HYBRID ORBITALS

Examine the last two columns of the character table, match up A_(1g), B_(1g), and E_u with the coordinates:

  • A_(1g): x^2+y^2, z^2 includes the s and d_(z^2) orbitals, respectively.
  • B_(1g): x^2 - y^2 includes the d_(x^2-y^2) orbital.
  • E_u: (x,y) includes the p_x and p_y orbitals at the same time.

These are the irreducible representations we got earlier. It turns out that this set of irreducible representations allows the following possibilities for orbital hybridizations:

psi_"hybrid" = c_1overbrace(d_(z^2))^(A_(1g)) + c_2overbrace(d_(x^2 - y^2))^(B_(1g)) + c_3overbrace((p_x + p_y))^(E_u) " "bb((i))

psi_"hybrid" = c_1overbrace(s)^(A_(1g)) + c_2overbrace(d_(x^2 - y^2))^(B_(1g)) + c_3overbrace((p_x + p_y))^(E_u) " "bb((ii))

where the c_i are arbitrary constants such that int_"allspace" psi^"*"psi d tau = 1.

This means we could have had either (i) p^2d^2 or (ii) sp^2d hybridization.

RATIONALIZING WHICH HYBRIDIZATION IS MORE REASONABLE

Since the d_(z^2) orbital is aligned along the z axis (whereas the complex is on the xy plane), the second set of hybrid orbitals is more reasonable.

Thus, we expect ulcolor(blue)(sp^2d) hybridization here.