= Solution
Six neutral <ammonia> donors leave Cu(II), $d^9$, so
$$
\boxed{N_e=9+6(2)=21.}
$$
The donor skeleton has \b[<octahedral molecular geometry> with a pronounced tetragonal Jahn–Teller distortion], normally four shorter equatorial and two longer axial Cu–N distances. The $t_{2g}^6e_g^3$ occupation is orbitally degenerate in the ideal octahedral limit. An <electron> count above 18 is not a contradiction: <ammonia> is primarily a sigma donor and the occupied metal-derived antibonding levels make this a weak-field coordination complex rather than a low-valent carbonyl closed shell.
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