# 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 $s {p}^{2} d$ (not $s {p}^{3}$, and not ${p}^{2} {d}^{2}$). 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+)

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 $s {p}^{3}$ hybridization.

SYMMETRY CONSIDERATIONS

This complex belongs to the ${D}_{4 h}$ 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 ${C}_{4} \left(z\right)$ principal rotation axis perpendicular to its plane (i.e. rotate ${90}^{\circ}$ 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 $x y$ plane, along the axes.
• It has (at least) one ${C}_{2} ' '$ rotation axis on the $x y$ 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 ${\Gamma}_{N {H}_{3}}$ 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 ${\text{NH}}_{3}$ ligands move due to each operation.

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

$\text{ "" "" "" "E" "C_4(z)" "C_2(z)" "C_2'" "C_2''" "i" "S_4" "sigma_h" "sigma_v" } {\sigma}_{d}$

${\Gamma}_{N {H}_{3}} = \text{ "4" "0" "" "" "0" "" "2" "" "0" "" "0" "" "0" "4" "" "2" } 0$

Skipping some steps, we reduce this using a reduction calculator for ${\Gamma}_{\text{general}}$ to get this set of irreducible representations:

$\underline{{\Gamma}_{N {H}_{3}}^{\text{red}} = {A}_{1 g} + {B}_{1 g} + {E}_{u}}$

The point of this was to see which symmetries the ${\text{NH}}_{3}$ 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 ${A}_{1 g}$, ${B}_{1 g}$, and ${E}_{u}$ with the coordinates:

• ${A}_{1 g} : {x}^{2} + {y}^{2} , {z}^{2}$ includes the $s$ and ${d}_{{z}^{2}}$ orbitals, respectively.
• ${B}_{1 g} : {x}^{2} - {y}^{2}$ includes the ${d}_{{x}^{2} - {y}^{2}}$ orbital.
• ${E}_{u} : \left(x , y\right)$ 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:

${\psi}_{\text{hybrid}} = {c}_{1} {\overbrace{{d}_{{z}^{2}}}}^{{A}_{1 g}} + {c}_{2} {\overbrace{{d}_{{x}^{2} - {y}^{2}}}}^{{B}_{1 g}} + {c}_{3} {\overbrace{\left({p}_{x} + {p}_{y}\right)}}^{{E}_{u}}$ $\text{ } \boldsymbol{\left(i\right)}$

${\psi}_{\text{hybrid}} = {c}_{1} {\overbrace{s}}^{{A}_{1 g}} + {c}_{2} {\overbrace{{d}_{{x}^{2} - {y}^{2}}}}^{{B}_{1 g}} + {c}_{3} {\overbrace{\left({p}_{x} + {p}_{y}\right)}}^{{E}_{u}}$ $\text{ } \boldsymbol{\left(i i\right)}$

where the ${c}_{i}$ are arbitrary constants such that ${\int}_{\text{allspace" psi^"*}} \psi d \tau = 1$.

This means we could have had either $\left(i\right)$ ${p}^{2} {d}^{2}$ or $\left(i i\right)$ $s {p}^{2} d$ 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 $x y$ plane), the second set of hybrid orbitals is more reasonable.

Thus, we expect $\underline{\textcolor{b l u e}{s {p}^{2} d}}$ hybridization here.