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gavmur [86]
1 year ago
7

Use electron configurations to account for the stability of the lanthanide ions Ce4⁺ and Eu2⁺.

Chemistry
1 answer:
Inga [223]1 year ago
4 0

Eu2+ is stable because of the two electrons in its outermost orbit and after losing 4 electrons. Ce +4 is stable

What are lanthanide ions?

All lanthanide elements form trivalent cations, Ln3+, whose chemistry is largely determined by the ionic radius, which decreases steadily from lanthanum to lutetium. These elements are called lanthanides because the elements in the series are chemically similar to lanthanum.

The Ce metal has the following electronic configuration: [Xe] 4f¹5d¹6s²

It obtains noble gas configuration after losing 4 electrons. Ce +4 is stable, so.

Eu's electronic setup is as follows.

[Xe] 4 f ⁷6s²

[Xe] 4 f ⁷

Eu2+ is stable because of the two electrons in its outermost orbit. Additionally stable is its +3 oxidation state. Ce⁺²

To learn more about lanthanides click on the link below:

brainly.com/question/12756990

#SPJ4

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1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm
sashaice [31]
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm


=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


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2 X(2+) + O2(g) ----> X2O2 = 2XO     or


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In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X


Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol


That does not correspond to any of the metal with valence 1+


So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.
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