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Why does zinc lose electrons while its outermost configuration is 4$\rm{s}^2$ 3$\rm{d}^{10}$ both outermost suborbital are full.

Why does it react or lose electrons?

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You could ask the same question about the alkali earth metals with their $\rm{s}^2$ valence shell. While there is an energy cost to removing electrons, in a spontaneous reaction (e.g. $\ce{Cu^2+_{(aq)} + Zn_{(s)} -> Cu_{(s)} + Zn^2+_{(aq)}}$) there will be a compensatory transfer of energy elsewhere (like the reduction and deposition of $\ce{Cu^2+}$) which leads to a more stable overall system.

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The further out the electron orbital the less energy is associated with a move. The issues is not so much about the stability of the zinc, but the total system. If a lower energy state can be reached by the electron changing orbitals then that is what will happen.

ie it's energetically preferable for 2Zn & O2 to exist as 2ZnO than as the original materials.

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Zinc has quite a stable configuration, but that does not mean it does not want to attain noble gas configuration. You should also observe that $Zn^{++}$ is more stable that ground state of Zn. This is due to the fact that both $e^-$ from $4s^2$ orbital are removed and the Zn ion has a fully-filled d-orbital, which is much more stable than the ground state configuration. Thus Zn reacts to get more stability provided by the fully-filled d-orbital.

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  • $\begingroup$ FYI: Zn has positive ionization energy, therefore $\ce{Zn^{2+}}$ is less stable than $\ce{Zn^0}$ $\endgroup$ – permeakra Aug 8 '16 at 16:10
  • $\begingroup$ You got it the other way round. If the ionization energy is higher (it's always positive, FYI), it indicates that the configuration is more stable. $\endgroup$ – Akshar Gandhi Aug 8 '16 at 16:21
  • $\begingroup$ @downvoter, want to disclose the mistake? $\endgroup$ – Akshar Gandhi Aug 8 '16 at 17:03
  • $\begingroup$ Ionization is a reversible process. If it takes X kJ/mol to go from M</sup>+ to M<sup>2+</sup> then it releases that energy going from M<sup>2+</sup> to M<sup>+</sup>. A large portion of this answer is just balatently wrong. $\endgroup$ – Tristan Maxson Aug 5 '20 at 22:53

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