# 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

It would be

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+)#

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***SYMMETRY CONSIDERATIONS**

This complex belongs to the

- 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# .

This means it follows this **character table**:

**OBTAINING THE REQUIRED ORBITAL SYMMETRIES**

To determine what the hybridization COULD be systematically, we generate a so-called **reducible representation** *spherical* basis (since they are all facing inwards and

To do this, operate on the complex using each operation, and examine how the

- 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 **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

**EXTRACTING COMBINATIONS OF HYBRID ORBITALS**

Examine the last two columns of the character table, match up

#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

**RATIONALIZING WHICH HYBRIDIZATION IS MORE REASONABLE**

Since the

Thus, we expect