The quantum numbers for the last valence electron in nickel are (3, 2, 0, -½).
The electron configuration of nickel is [Ar]4s²3d⁸
The last electron added is a 3d electron, so n = 3 and l = 2.
We construct a table of quantum numbers.
Element n l mₗ mₛ
Sc 3 2 2 +½
Ti 3 2 1 +½
V 3 2 0 +½
Cr 3 2 -1 +½
Mn 3 2 -2 +½
Fe 3 2 2 -½
Co 3 2 1 -½
Ni 3 2 0 -½
The quantum numbers for the last electron in nickel are (3, 2, 0, -½).
The four sets of quantum numbers for nickel (n, l, [tex]m_{l}[/tex], [tex]m_{s}[/tex]) are (3, 2, 0, -1/2).
To find the quantum numbers of Ni, we need to look at its electron configuration. The atomic number of Ni is 28, so its electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² → [Ar] 3d⁸ 4s²
The last electron is in the d-shell, so let's find the quantum numbers.
Since the shell of the last electron is 3 (3d⁸), this number 3 corresponds to the quantum number n (n=3).
This quantum number is related to the shape of the orbital and is given by n-1. The types of shapes for the values of l are:
l = 0 corresponds to the orbital's shape "s"
l = 1 corresponds to the orbital's shape "p"
l = 2 corresponds to the orbital's shape "d"
l = 3 corresponds to the orbital's shape "f "
The shell of the last electron of Ni is 3 (3d⁸), so l is:
[tex] l = n - 1 = 3 - 1 = 2 [/tex]
This quantum number varies from -l to +l. For Ni, l = 2, so the possible values of [tex]m_{l}[/tex] are:
[tex] m_{l} = - 2, -1, 0, +1, +2 [/tex]
This means that we have 5 orbitals, which can have a maximum of 2 electrons (Pauli exclusion principle). Electron filling in those orbitals goes in order from -2 to +2, so for Ni, the last electron is in the 0 orbital.
Hence, the value of [tex]m_{l}[/tex] is 0.
This number can have two possible values: +1/2 (spin-up) or -1/2 (spin-down). For the filling of the electrons in the orbitals, we start for the positive values (+1/2, spin-up) and then we culminate with the negative ones (-1/2, spin-down), to complete the filling of two electrons per orbital. Hence, for Ni, the value of [tex]m_{s}[/tex] is -1/2 (spin-down).
Therefore, the four sets of quantum numbers (n, l, [tex]m_{l}[/tex], [tex]m_{s}[/tex]) for nickel are (3, 2, 0, -1/2).
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