I am totally confused about the bonding in $\ce{[Cu(NH3)4]^2+}$. Here are the thoughts that I had when I first attempted the problem: $\ce{Cu^2+ = [Ar] 3d^9}$ Due to crystal field splitting, $\ce{Cu^2+} = t_{2g}^6e_g^{3}$ Also, we can expect $d_{z^2}$ to contain two paired electrons and $d_...