Question

Quantum numbers \[n=3,l=1,{{m}_{l}}=0,{{m}_{s}}=\pm \dfrac{1}{2}\] are given. It represents the last electron to be added to complete the ground state electron configuration of an element. Which of the following could be the symbol of this element?
A) Na
B) Si
C) Th
D) V

Answer 1

Hint: We need to understand the relation between the quantum numbers and the electronic configuration to solve this problem. The number which suits the given quantum state will be the valence electron of the element. So, we can know the element easily.

Complete Solution :
We are the quantum state of a particular electron in an element given by \[n=3,l=1,{{m}_{l}}=0,{{m}_{s}}=\pm \dfrac{1}{2}\]. It is given that this electron constitutes the valence electron in the system. So, the total number which we get from the quantum numbers will be the atomic number of the element.
We need to understand the different quantum numbers to solve this problem. There are mainly four kinds of quantum numbers which are described below.
1. Principal quantum number: It is the quantum number denoted by ‘n’ which gives the shell number of the electron. For the ‘n’ being 1, 2, 3 the respective shells will be 1, 2, 3 itself.

2. Azimuthal quantum number: The quantum number which describes the subshell number of the electron is called so. It is represented by ‘l’. For the ‘l’ being 0, 1, 2, 3 the respective sub-shells are s, p, d, f.

3. Orbital quantum number: It is the quantum number which denotes the angular momentum possessed by the electron. It is represented by \[{{m}_{l}}\]. The \[{{m}_{l}}\] is related to the azimuthal quantum as –
\[-l<{{m}_{l}}<+l\]
i.e., for ‘l’ is 2, \[{{m}_{l}}\] can be -1, 0 or 1.

4. Spin quantum number: This is the fourth quantum number which makes each electron in the element unique. It gives the direction of the spinning of the electron as \[{{m}_{s}}=\pm \dfrac{1}{2}\]

Now, we can study the quantum number given to us –
\[n=3,l=1,{{m}_{l}}=0,{{m}_{s}}=\pm \dfrac{1}{2}\]
From this we can write the valence shell configuration of the element as –
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{2}}\]
The total number of electrons is 14. The element with atomic number 14 is Silicon. The required answer is Si.
So, the correct answer is “Option B”.

Note: The electronic configuration can be easily made up from the quantum states of the electron under consideration. The principal quantum gives the least possible idea about the location of the electron, but without which we cannot find the exact position.