The deep blue complex [Cu(NH₃)₄]²⁺ is square planar and paramagnetic…

Step 1. The 3d count. $\text{Cu}^{2+}$ has $27 - 18 = 9$ electrons in 3d. Four orbitals hold a pair and one holds a single electron.

Step 2. Freeing an orbital. The theory moves the single 3d electron up into a 4p orbital. Now four 3d orbitals hold pairs and one is empty. That empty 3d orbital, the 4s orbital and two of the 4p orbitals make four $dsp^2$ hybrids. They point to the corners of a square.

Step 3. The unpaired electron. The electron that was moved sits alone in the third 4p orbital. So one electron is unpaired, and $\mu = \sqrt{1 \times 3} = 1.73\ \text{BM}$.

What it shows. An electron in a 4p orbital is far from the nucleus and would be easy to remove. Yet copper(II) complexes are not easily oxidised in this way. The theory reaches the right answer only by a step added by hand.