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alexandr1967 [171]
3 years ago
5

Suppose you take a trip to a distant universe and find that the periodic table there is derived from an arrangement of quantum n

umbers different from the one on Earth. The rules in that universe are: principal quantum number ????=1,2,... (as on Earth); angular momentum quantum number ℓ=0,1,2,...,????−1 (as on Earth); magnetic quantum number ????ℓ=0,1,2,...,ℓ (only positive integers up to and including ℓ are allowed); spin quantum number ????s=−1,0,+1 (that is, three allowed values of spin). Assuming that the Pauli exclusion principle remains valid in the distant universe, what is the maximum number of electrons that can populate a given orbital there? maximum number electrons per orbital: Write the electronic configuration of the element with atomic number 8 in the periodic table. Superscript numbers where appropriate but omit parentheses. electronic configuration:
Mathematics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

  • <em>Maximun number of electrons per orbital:</em> <u>3</u>
  • <em>Electron configuration of the element with atomic number 8</em>:

           <u>1s³ 2s³ 2p²</u>

Explanation:

<u>1) Pauli's exclusion principle.</u>

Pauli's exclusion principle states that none two electrons of an atom may have the same set of quantum numbers.

Since the real rule (in our normal universe) is that the spin quantum number can only have two values (s = +1/2 or -1/2), that implies that only two electrons can populate a given orbital here.

<u>2) Rules in the distant universe.</u>

  • The rules for the principal quantum number (n), and the angular momentum quantum number (ℓ), are the same of the Earth.

  • The rule for the magnetic quantum number (mℓ) is different than in the Earth:

        - In the Earth: mℓ = from - ℓ to + ℓ

        - In the distant universe: mℓ = from 0 to ℓ

The implication of this is that there will be only two p orbitals in the distant universe, correponding to ℓ = 0 and ℓ = 1,  instead of three p orbitals as in the Earth.

  • The rule for the spin (ms) number is different than in Earth:

        - In Earth: s = +1/2 or -1/2 (two possibilities)

        - In the distant universe: ms = -1, 0, - 1

Then in each s or p orbital there will be 3 electrons.

<u>3) Electron configuration in the distant universe</u>

Hence, for the element with atomic number 8, which means that the number of electrons is 8, the configuration is:

  • 1s³ (because 3 electrons can populate this orbital)
  • 2s³ (because 3 electrons can populate the second s orbital
  • 2p² (because the 2 remaining electrons can be placed in the orbitals p: remember that in this distant universe there are two p orbitals, so you can accomodate until 6 electrons in them, 2 × 3 = 6.

<u>Conclusion:</u>

  • Maximun number of electrons per orbital: 3
  • Electron configuration of the element with atomic number 8:

             1s³ 2s³ 2p²

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Answer:

Smallest surface area is of Cuboid B i.e 440 cm²

So, The company will choose cuboid B

Step-by-step explanation:

We need to find the surface area of all cuboids.

Surface Area of Cuboid A:

Length = 6

Breadth = 25

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The formula used is: Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)

Putting values and finding surface area:

Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)\\Surface \ Area \ of \ Cuboid=2((6 \times 25)+(25 \times 4)+(6 \times 4))\\Surface \ Area \ of \ Cuboid=2(150+100+24)\\Surface \ Area \ of \ Cuboid=2(274)\\Surface \ Area \ of \ Cuboid=548\: cm^2

So, Surface Area of Cuboid A = 548 cm²

Surface Area of Cuboid B:

Length = 10

Breadth = 6

Height = 10

The formula used is: Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)

Putting values and finding surface area:

Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)\\Surface \ Area \ of \ Cuboid=2(10 \times 6)+(6 \times 10)+(10 \times 10))\\Surface \ Area \ of \ Cuboid=2(60+60+100)\\Surface \ Area \ of \ Cuboid=2(220)\\Surface \ Area \ of \ Cuboid=440\: cm^2

So, Surface Area of Cuboid B = 440 cm²

Surface Area of Cuboid C:

Length = 2

Breadth = 20

Height = 15

The formula used is: Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)

Putting values and finding surface area:

Surface \ Area \ of \ Cuboid=2((Length\times Breadth)(Breadth \times Height)+(Length \times Height)\\Surface \ Area \ of \ Cuboid=2((2 \times 20)+(20 \times 15)+(2 \times 15))\\Surface \ Area \ of \ Cuboid=2(40+300+30)\\Surface \ Area \ of \ Cuboid=2(370)\\Surface \ Area \ of \ Cuboid=740\: cm^2

So, Surface Area of Cuboid C = 740 cm²

So, We get:

Surface Area of Cuboid A = 548 cm²

Surface Area of Cuboid B = 440 cm²

Surface Area of Cuboid C = 740 cm²

The company wants to choose the design having smallest surface area.

So, smallest surface area is of Cuboid B i.e 440 cm²

So, The company will choose cuboid B

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