Can a transition metal ion with a charge of 2+ form without valence shell s and p electrons?

  • $\begingroup$ Do you mean the ion has 0 s and 0 p electrons? Or do you mean the initial atom has no s or p electrons? $\endgroup$ – wes3449 Jan 30 '15 at 14:03
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    $\begingroup$ Your question is unclear. Please be more specific. $\endgroup$ – Yomen Atassi Jan 30 '15 at 19:29

A transition element in its ground state has a configuration of $ (n-1)d^{1-10} ns^{1-2} $
For the atom to have noble gas configuration (which is what I assume you mean) all electrons of the outer orbitals have to be stripped. An oxidation state of +2 implies that only two electrons have been stripped. The lowest number of valence electrons are possessed by the third group elements as far as transition metals are concerned. Their ground state configuration is $ (n-1)d^{1} ns^{2} $ ; removing two electrons gives $ (n-1)d^{1} ns^{0} $ .

Therefore the answer to your question is no, as there has to be at least one electron in the d orbital of a transition metal in +2 state.

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