Question

Rank these transition metal ions in order of decreasing number of unpaired electrons.

a. Fe^3
b. Mn^4+
c. V3+
d. Ni^2+
e. Cu^+

Answer 1

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

`Mn^(4+) > V^(3+) = Ni^(2+) > Fe^(3+) > Cu^(+)`

Step-by-step 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.

`Fe^(3+) - [Ar] 3d^(5)`

So, there is only 1 unpaired electron present in
`Fe^(3+)`.

• Atomic number of Mn is 25.

`Mn^(4+) - [Ar]3d^(3)`

So, there are only 3 unpaired electrons present in
`Mn^(4+)`.

• Atomic number of V is 23.

`V^(3+) - [Ar] 3d^(2)`

So, there are only 2 unpaired electrons present in
`V^(3+)`.

• Atomic number of Ni is 28.

`Ni^(2+) - [Ar] 3d^(8)`

So, there will be 2 unpaired electrons present in
`Ni^(2+)`.

• Atomic number of Cu is 29.

`Cu^(+) - [Ar] 3d^(10)`

So, there is no unpaired electron present in
`Cu^(+)`.

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

`Mn^(4+) > V^(3+) = Ni^(2+) > Fe^(3+) > Cu^(+)`

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

`Mn^(4+) > V^(3+) = Ni^(2+) > Fe^(3+) > Cu^(+)`