Question

Rank these transition metal ions in order of decreasing number of unpaired electrons. If two ions have the same number of unpaired electrons

Fe^3 , Mn^4+ , V3+ , Ni^2+ , Cu^+

Answer 1

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

Step-by-step explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

Fe: 26:
`1s^2s^22p^63s^23p^64s^23d^(6)`

`Fe^(3+):23:1s^2s^22p^63s^23p^63d^(5)` : 5 unpaired electrons

Mn: 25:
`1s^2s^22p^63s^23p^64s^23d^(5)`

`Mn^(4+):23:1s^2s^22p^63s^23p^63d^(3)` : 3 unpaired electrons

V: 23:
`1s^2s^22p^63s^23p^64s^23d^(3)`

`V^(3+):23:1s^2s^22p^63s^23p^63d^(2)` : 2 unpaired electrons

Ni : 28 :
`1s^2s^22p^63s^23p^64s^23d^(8)`

`Ni^(2+):23:1s^2s^22p^63s^23p^63d^(8)`: 2 unpiared electrons

Cu : 29 :
`1s^2s^22p^63s^23p^64s^13d^(10)`

`Cu^(+):23:1s^2s^22p^63s^23p^63d^(10)`: 0 unpaired electrons

Thus the order of decreasing number of unpaired electrons:

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