Video Transcript
Given that the electronic
configuration of manganese is brackets Ar 4s2 3d5, what is the maximum oxidation
state for this transition metal?
The oxidation state indicates the
number of valence electrons that have been lost or gained by an atom. Most of the period four transition
metals, which are scandium through copper, have multiple oxidation states. This chart shows some of the more
common oxidation states. However, we notice that the
oxidation states are positive because metals lose electrons to form cations.
In this question, we are given the
electronic configuration of a manganese atom. The valence electrons of the
manganese atom are found in the 4s and 3d subshells. By adding together the
superscripts, we find that the manganese atom has seven valence electrons. But, when looking at the oxidation
states listed in the chart, we can see that it’s not easy to predict the oxidation
states that a transition metal will have based only on the number of valence
electrons.
One thing we do know is that when
writing electronic configurations for period four transition metals, valence
electrons are removed from the 4s subshell first before the 3d subshell. So, for example, the Mn2+ cation
forms when two valence electrons are lost from the 4s subshell. And the electronic configuration of
the ion is therefore brackets Ar 3d5. If a manganese atom lost all seven
of its valence electrons, then both the 4s and 3d subshells would be empty.
We notice in the table that
positive seven is not only the highest oxidation state that manganese can have, but
it’s also the highest oxidation state commonly found in all of the period four
transition metals. And even though iron, cobalt,
nickel, and copper have a greater amount of valence electrons than manganese, their
most common oxidation states are not larger than positive seven.
In conclusion, when considering its
electronic configuration, the maximum oxidation state for manganese is positive
seven.
