Respuesta :
The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.
The element are : K, Ne, Xe, Bh and Ce
Further explanation
In an atom, there are energy levels in the shell and sub shell.
This energy level is expressed in terms of electron configurations.
Writing the electron configuration starts from the lowest to the highest subshell's energy level. There are 4 sub-shells in an atom's shell, namely s, p, d, and f.
Each sub-shell also has orbitals drawn in the form of a square box in which there are electrons symbolized by half arrows.
Each orbital in an atom consists of 4 quantum numbers
- n is the principal quantum number ---> describes the size of the orbitals
- l is the angular momentum / azimuthal quantum number -> describes the shape of the orbitals.
- ml, the magnetic quantum number -> describes the orientation of orbitals in space
- ms, the electron-spin quantum number ---> describes the spin of electrons
Value of n: positive integer
value of l: s = 0, p = 1, d = 2, f = 3, ... (n-1)
ml value: between -l to + l
ms value: +1/2 or -1/2
Determining the values of n and l can be obtained from the valence electrons of each element
From attached picture :
1. electron configurations of element ¹⁹K
[Ar] 4s¹ ⇒ n = 4, l = 0
2. electron configurations of element ¹⁰Ne
[He] 2s²2p⁶ ⇒ n = 2, l = 1
3. electron configurations of element ⁵⁴Xe
[Kr]4d¹⁰5s²5p⁶ ⇒ n = 5, l = 1
4. electron configurations of element ¹⁰⁷Bh
[Rn]5f¹⁴6d⁵7s²⇒ n = 7, l = 0
5. electron configurations of element ⁵⁸Ce
[Xe]4f¹5d¹6s² ⇒ n = 6, l = 0
Learn more
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