The spin multiplicity of `Fe^(3+)`(`Ec=[Ar]3d^5)` in its ground state

Explain the following observations : (i) Copper atom has completely field d orbitals (3d^(10)) in its ground state, it is regarded as a transition element. (ii) Cr^(2+) is a stronger reducing agent than Fe^(2+) in aqueous solutions.

Copper atom has completely filled d orbitals (3d^(10)) in its ground state. How can we say that it is a transition element

The one which has incompletely filled d-orbitals in its ground state or in any one of its oxidation state is known as

which os the electrons confiuration for an Fe (III) ion in its ground state?

The ground state configuration of Fe^(3+) ion in gaseous state is: (At. No. of Fe = 26)

An element has an electronic configuration [Ar]^(3d) in its + 3 oxidation state. What will be the electronic configuration of its atom ?