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Zanzabum
3 years ago
8

Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital

s. How many electrons would such an atom have? Write its electron configuration using noble -gas notation for the previous noble gas, radon .
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
7 0

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a <u>completely filled 5f, 6d, 7s and 7p orbitals</u>.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the <u>number of electrons</u> = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = <u>118</u>.

<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

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Use the following equation to answer question 1-4. Make sure you balance first.
worty [1.4K]
<h3>Answer:</h3>

5.2 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

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

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

  1. Set up:                             \displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})
  2. Multiply/Divide:               \displaystyle 5.2 \ mol \ H_2O
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