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Verdich [7]
3 years ago
5

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

his description?

Chemistry
2 answers:
Tanzania [10]3 years ago
7 0
Molybdenum and Rhodium, I believe. All of the elements in these 2 groups follow unusual orbital filling rules.

Hope this helped!
kotykmax [81]3 years ago
4 0

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.

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41. A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
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