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What is the maximum number of electrons in an atom that can have the following quantum numbers?A) n= 4, Ms = -1/2B)n=5, I=3C) n = 2, / = 1, m = -1D) n=6, I=3, mI=+3, ms=-1/2

User Izabel
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Answer: considering only n = 4 and Ms = -1/2, the atom could present a maximum of 36 electrons.

Step-by-step explanation:

The question requires us to determine the maximum number of electrons for an atom with the following quatum numbers: n = 4, Ms = -1/2

The principal quantum number (n) refers to the size of an orbital. In other words, it describes how many "shells" the atom can present. If n = 4, it means that the atom presents 4 shells.

The spin quantum number (Ms) describes the angular momentum of an electron. In a simple representation, Ms describes if the arrow representing the electron points up or down. The spin quantum number can present magnitude 1/2 and direction + or -. Let's assume that +1/2 refers to "spin-up" and -1/2 refers to "spin-down".

Considering the electronic distribution given by the Linus Pauling diagram, an atom with n = 4 (and no angular quantum number, l, specified) could present the following configuration:


1s^2\text{ 2s}^2\text{ 2p}^6\text{ 3s}^2\text{ 3p}^6\text{ 4s}^2\text{ 3d}^(10)\text{ 4p}^6

The total amount of electrons, considering only n = 4, is 36 electrons.

If we take the spin quantum number into consideration, the total amount of electrons the atom could present would not change - Ms = -1/2 means that the spin of the electron is pointing down and this electron is in an orbital which is already filled with one electron.

Therefore, considering only n = 4 and Ms = -1/2, the atom could present a maximum of 36 electrons.

User Salomon Zhang
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