Question

please answer
Identify the element whose last electron would have the following four quantum numbers:
8. 3, 1,-1, +1/2
9. 4,2, +1, +1/2
10. 6, 1, 0, -1/2
11. 4,3, +3, -1/2
12. 2, 1, +1, -1/2
I

Answer 1

Answer:

See Explanation

Explanation:

8.  3, 1, -1, +1/2 => 3pₓ¹ => Aluminum

9.  4, 2, +1, +1/2 => 4d₁¹ => Chromium

10. 6, 1, 0, -1/2 => 6p₀² => Argon

11.  4, 3, +3, -1/2 => 4f₊₃² => Lutetium

12. 2, 1, +1, -1/2  => 2p₊₁² => Neon

Answer 2

8. Aluminum- Atomic number-13

9. Yttrium-Atomic number-39

10. Thalium-Atomic number-81

11. Cerium-Atomic number-58

12. Boron-Atomic number-5

Firstly, lets see four quantum numbers:

The set of four quantum numbers are n, l, ml and ms which is principal, azimuthal, magnetic and spin quantum numbers respectively.

l= 0= s orbital

l= 1= p orbital

l= 2=d orbital

l= 3= f orbital

8. 3,1,-1,+1/2;

n=3, l=1, ml=-1 and ms=+1/2= $3p^1$ orbital

so on writing electronic configuration:

The element is- Al- 1s² 2s² 2p⁶ 3s² 3p¹

9. 4,2,+1,+1/2;

n=4,l=2,ml=+1,ms=+1/2=$4d^1$ orbital

so on writing electronic configuration:

The element is-Y -[Kr]4d¹5s²

10.6,1,0,-1/2;

n=6,l=1,ml=0,ms=-1/2= $6p^1$ orbital

so on writing electronic configuration:

The element is- Tl- [Xe]4f¹⁴5d¹⁰6s²6p¹

11. 4,3, +3, -1/2;

n=4,l=3,ml=+3,ms=-1/2=$4f^1$ orbital

so on writing electronic configuration:

The element is-Ce- [Xe]4f¹5d¹

12. 2, 1, +1, -1/2;

n=2,l=1,ml=+1,ms=-1/2= $2p^1$ orbital

so on writing electronic configuration:

The element is-B-1s² 2s², 2p¹

So the given elements are.

8. Aluminum- Atomic number-13

9. Yttrium-Atomic number-39

10. Thalium-Atomic number-81

11. Cerium-Atomic number-58

12. Boron-Atomic number-5

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