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$\ce{Cu^2+} $ has 9 electrons and a d-orbital and is almost completely filled (except 1 electron vacant) and $\ce{NH3}$ donates an electron eventually forming $\mathrm{sp^3}$ hybrid orbitals.

But it has a square planar geometry hence completely contradicting the above reasoning. Please explain what the hybridization is?

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    $\begingroup$ Hybridisation is determined by geometry. It is purely a mathematical trick to create nicer orbital representations. You cannot assign it without knowing the geometry in advance. $\endgroup$ – bon May 20 '16 at 15:29
  • $\begingroup$ @bon Given that it is sqaure planar,whats the hybridization $\endgroup$ – Gopalkrishna Nayak May 20 '16 at 15:59
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    $\begingroup$ en.wikipedia.org/wiki/Tetraamminecopper%28II%29_sulfate - it isn't square planar $\endgroup$ – Mithoron May 20 '16 at 16:27
  • $\begingroup$ @Mithoron then what is it? $\endgroup$ – Gopalkrishna Nayak May 21 '16 at 7:45
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    $\begingroup$ Do not use hybridisation to describe coordination compounds; it is bad. $\endgroup$ – Jan May 26 '16 at 10:05
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I feel a bit like a broken record. Do not use hybridisation to describe coordination compounds; it is not helpful.

The cation you called tetraamminecopper(II) is more accurately described as tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$, because two aqua ligands are there, just further away. The four ammine ligands donate two electrons each, as do both the aqua ligands, totalling 21 electrons overall. Due to the strongly Jahn-Teller distorted geometry and other reasons, the bonding interactions are overall stronger than antibonding ones making the complex favourable.

The interaction between the metal and ligand involves $\mathrm{e_g, a_{1g}}$ and $\mathrm{t_{2u}}$ orbitals (when viewed as an undistorted octahedral complex). That would include two d-orbitals, three p and one s. If your life depended on it and you are the last person on Earth, you may attempt to extract a hybridisation from that. But don’t do it.

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