Question

Please follow the Aufbau principle, Pauli exclusion principle and Hund’s rule to determine the four quantum numbers (principle, angular, magnetic, spin) for the final electron of Scandium.
A. $3,2, - 2, + \dfrac{1}{2}$
B. $5,1,0, - \dfrac{1}{2}$
C. $4,0,0, - \dfrac{1}{2}$
D. $4,1, - 1, - \dfrac{1}{2}$

Answer 1

Hint: The Aufbau principle, also known as the Aufbau rule, states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. Pauli's Exclusion Principle states that no two electrons in the same atom can have identical values for all four of their quantum numbers. According to Hund’s rule every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Complete answer:
Scandium is a chemical element with the symbol \[Sc\] and atomic number \[21\] . A silvery-white metallic d-block element, it has historically been classified as a rare-earth element, together with yttrium and the lanthanides. The electronic configuration of Scandium is written as:
$Sc = [Ar]4{s^2}3{d^1}$
As we can observe from the above electronic configuration, the outermost electron of Scandium lies in the \[3d\] orbital and not in the \[4s\] orbital. The quantum numbers are used to describe the address of an electron inside an atom and is used to mark the path or trajectory of the electron inside the atom. There are a total of four different quantum numbers that help to describe the exact location of an electron inside the atom. They are principal quantum number ($n$ ), angular or azimuthal quantum number ($l$ ), magnetic quantum number ($m$ ) and spin quantum number ($s$ ). For scandium, the quantum numbers for the outermost electron, which lies in the 3d orbital, are:
$n = 3$
$l = 2$
$m = - 2, - 1,0, + 1, + 2$ (any one of them)
$s = + \dfrac{1}{2}or - \dfrac{1}{2}$ (any one)

Thus, the correct option is A.

Note:
In cases where (\[n\; + \;l\] ) is the same for two orbitals (e.g., $3d$ and $4s$ ), the ($n + l$ ) rule says that the orbital with lower n has lower energy. In other words, the size of the orbital has a larger effect on orbital energy than the number of planar nodes. This is the reason why we consider the $3d$ orbital in place of the $4s$ orbital while considering the outermost electron for Scandium.