To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your
answer with examples.
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The elements in the first-half of the transition series exhibit many oxidation states with Mn
exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d5.
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The electronic configurations play a very crucial role in deciding the stability of oxidation states. The presence or absence of complete or half filled d orbitals in that oxidation state decides its stability. Also the ability to give away e- decide the oxidation state which depends on the energy possessed by it and hence forth its electronic configuration,
For example Sc shows +3 oxidation state only [M] 3d1 4s2+1, +2 very unstable due to presence of 1d e- +3 oxidation state is stable with noble gas configuration.
Mn : [Ar] 3d5 4s2 has +2, +3, +4, +5, +6, +7 states but only +2 and +7 states are stable due to the half filled d orbital or completely empty d orbital.
similarly Ti = +2, +3, +4 with +4 most stable due to the same reason.
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