Question

Two elements in the second transition series (y through cd ) have three unpaired electrons in their 3+ ions. what elements fit this description?

Answer 1

Answer: The elements that have 3 unpaired electrons in their +3 oxidation states are Molybdenum and Palladium.

Explanation: The elements in the second transition series are:

Yttrium (Y), Zirconium (Zr), Niobium (Nb), Molybdenum (Mo), Technetium (Tc), Ruthenium (Ru), Rhodium (Rh), Palladium (Pd), Silver (Ag) and Cadmium (Cd).

The electronic configuration of these elements are:

`Y=[Kr]5s^24d^1` ;
`Y^(3+)=[Kr]`

`Zr=[Kr]5s^24d^2` ;
`Zr^(3+)=[Kr]4d^1`

`Nb=[Kr]5s^14d^4` ;
`Nb^(3+)=[Kr]4d^2`

`Mo=[Kr]5s^14d^5` ;
`Mo^(3+)=[Kr]4d^3`

`Tc=[Kr]5s^14d^6` ;
`Tc^(3+)=[Kr]4ds^4`

`Ru=[Kr]5s^14d^7` ;
`Ru^(3+)=[Kr]4d^5`

`Rh=[Kr]5s^14d^8` ;
`Rh^(3+)=[Kr]4d^6`

`Pd=[Kr]5s^04d^(10)` ;
`Pd^(3+)=[Kr]4d^7`

`Ag=[Kr]5s^14d^(10)` ;
`Ag^(3+)=[Kr]4d^8`

`Cd=[Kr]5s^24d^(10)` ;
`Cd^(3+)=[Kr]4d^9`

From the above configurations of the elements in their +3 ionic state, the elements having 3 unpaired electrons are Molybdenum and Palladium.