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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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Why doesn’t neon readily form an ionic bond?
Goryan [66]

By definition of noble gases, neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

<h3>Noble gases</h3>

Noble gases are not very reactive, that is, they practically do not form chemical compounds. This means that they do not react with other substances, nor do they even react between atoms of the same gas, as is the case with diatomic gases such as oxygen (O₂).

The chemical stability of the noble gases and therefore the absence of spontaneous evolution towards any other chemical form, implies that they are already in a state of maximum stability.

All chemical transformations involve valence electrons, they are involved in the process of covalent bond formation and the formation of ions. Therefore, the practically null reactivity of the noble gases is due to the fact that they have a complete valence shell, which gives them a low tendency to capture or release electrons.

Since the noble gases do not react with the other elements, they are also called inert gases.

<h3>Neon</h3>

Neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

Learn more about noble gases:

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6 0
3 years ago
Read 2 more answers
The electrolysis of molten NaCl is an industrial process. What does the electrolysis produce?
saveliy_v [14]

Answer:

it is D

Explanation:

NaOH and chlorine gas

Electrolysis of Aqueous NaCl Since water can be both oxidized and reduced, it competes with the dissolved Na+ and Cl– ions. Rather than producing sodium, hydrogen is produced. ... The net process is the electrolysis of an aqueous solution of NaCl into industrially useful products sodium hydroxide (NaOH) and chlorine gas.

3 0
3 years ago
Enter the net ionic equation for this reaction. Express your answer as a net ionic equation. Identify all of the phases in your
Lisa [10]

Answer:

2H+(aq) + 2OH-(aq) → 2H2O(l)

Explanation:

Step 1: The balanced equation

2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)

This equation is balanced, we do not have the change any coefficients.

Step 2: The netionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)

After canceling those spectator ions in both side, look like this:

2H+(aq) + 2OH-(aq) → 2H2O(l)

6 0
3 years ago
Why is the r.a.m. Value used?
Andrew [12]

Answer:

The r.a.m. value is used to determine how many isotopes an element has.

Explanation:

5 0
3 years ago
WILL MARK BRAINLIEST!!!!
Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0
3 years ago
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