In the n = 5 principal shell, what is the maximum number of electrons that can have m1= -2
Hint:
For n=5, l can be 0, 1, 2,3, and 4.
For 1=1 you know ml can be -1,0,+1 so for
2, 3, and 4, there must be how many -2s for mL. And if you can put two electrons into each of those orbitals, how many total is that?
To determine the maximum number of electrons that can have m1= -2 in the n = 5 principal shell, we need to first understand the quantum numbers that describe the position and orientation of electrons in an atom.
In this case, m1 represents the magnetic quantum number, which specifies the orientation of the electron's orbital in three-dimensional space. The magnetic quantum number can range from -l to +l, where l is the azimuthal quantum number.
The azimuthal quantum number (also known as the orbital quantum number) determines the shape of the electron's orbital. It is given by the formula l = n - 1, where n is the principal quantum number.
For n = 5, the azimuthal quantum number would be l = 5 - 1 = 4. Therefore, the magnetic quantum number (m1) can have values from -4 to +4.
To find the maximum number of electrons that can have m1 = -2, we need to consider the Pauli exclusion principle, which states that no two electrons in an atom can have the same set of quantum numbers.
In the n = 5 principal shell, there are multiple orbitals with different values of m1. The number of orbitals in a shell is given by 2l + 1.
For l = 4, the number of orbitals is 2(4) + 1 = 9. This means that there are 9 orbitals available in the n = 5 shell.
Since each orbital can hold a maximum of 2 electrons (one with spin-up and one with spin-down), the maximum number of electrons with m1 = -2 would be 2 electrons per orbital x 9 orbitals = 18 electrons.
Therefore, the maximum number of electrons that can have m1 = -2 in the n = 5 principal shell is 18 electrons.