The electronic configuration of manganese is:
1s22s22p63d54s2.
a)What is the maximum oxidation state of manganese?
b)Explain how the electronic configuration shows that Mn is a transition element
Removing 4s2 gives Mn^2+. Removing 4s2 and 3d5 gives Mn^7+.
3d electrons make the transition in row 3 from Ca to Ga.
a) To determine the maximum oxidation state of manganese, we need to look at the valence electrons. In this case, the valence electrons are in the 4s and 3d orbitals.
The electronic configuration of manganese shows that it has 5 electrons in the 3d orbital (3d5) and 2 electrons in the 4s orbital (4s2). The highest possible oxidation state is achieved when all the valence electrons are lost.
Since there are a total of 7 valence electrons (5 from 3d and 2 from 4s), the maximum oxidation state of manganese is +7. This is because if all 7 electrons are lost, manganese would have a completely empty 3d and 4s orbitals.
b) The electronic configuration of manganese, 1s22s22p63d54s2, demonstrates that manganese is a transition element.
Transition elements are characterized by the presence of partially filled d orbitals. In the case of manganese, it has 5 electrons in the 3d orbital, which means that there are unpaired electrons present. This property is commonly observed in transition elements.
Furthermore, transition elements are known for their variable oxidation states, which is also evident in manganese's electronic configuration. With both the 3d and 4s orbitals partially filled, manganese can readily lose or gain electrons to achieve different oxidation states, making it a transition element.