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:
[tex]Y=[Kr]5s^24d^1[/tex] ; [tex]Y^{3+}=[Kr][/tex]
[tex]Zr=[Kr]5s^24d^2[/tex] ; [tex]Zr^{3+}=[Kr]4d^1[/tex]
[tex]Nb=[Kr]5s^14d^4[/tex] ; [tex]Nb^{3+}=[Kr]4d^2[/tex]
[tex]Mo=[Kr]5s^14d^5[/tex] ; [tex]Mo^{3+}=[Kr]4d^3[/tex]
[tex]Tc=[Kr]5s^14d^6[/tex] ; [tex]Tc^{3+}=[Kr]4ds^4[/tex]
[tex]Ru=[Kr]5s^14d^7[/tex] ; [tex]Ru^{3+}=[Kr]4d^5[/tex]
[tex]Rh=[Kr]5s^14d^8[/tex] ; [tex]Rh^{3+}=[Kr]4d^6[/tex]
[tex]Pd=[Kr]5s^04d^{10}[/tex] ; [tex]Pd^{3+}=[Kr]4d^7[/tex]
[tex]Ag=[Kr]5s^14d^{10}[/tex] ; [tex]Ag^{3+}=[Kr]4d^8[/tex]
[tex]Cd=[Kr]5s^24d^{10}[/tex] ; [tex]Cd^{3+}=[Kr]4d^9[/tex]
From the above configurations of the elements in their +3 ionic state, the elements having 3 unpaired electrons are Molybdenum and Palladium.
