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Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.
[tex]Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}[/tex]
Explanation:
In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.
The electronic configuration in terms of noble gases for the given elements are as follows.
- Atomic number of Fe is 26.
[tex]Fe^{3+} - [Ar] 3d^{5}[/tex]
So, there is only 1 unpaired electron present in [tex]Fe^{3+}[/tex].
- Atomic number of Mn is 25.
[tex]Mn^{4+} - [Ar]3d^{3}[/tex]
So, there are only 3 unpaired electrons present in [tex]Mn^{4+}[/tex].
- Atomic number of V is 23.
[tex]V^{3+} - [Ar] 3d^{2}[/tex]
So, there are only 2 unpaired electrons present in [tex]V^{3+}[/tex].
- Atomic number of Ni is 28.
[tex]Ni^{2+} - [Ar] 3d^{8}[/tex]
So, there will be 2 unpaired electrons present in [tex]Ni^{2+}[/tex].
- Atomic number of Cu is 29.
[tex]Cu^{+} - [Ar] 3d^{10}[/tex]
So, there is no unpaired electron present in [tex]Cu^{+}[/tex].
Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.
[tex]Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}[/tex]
Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.
[tex]Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}[/tex]
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