Question

Atomic number of sodium is 11. Write down the four quantum numbers of the electron having the highest energy.

Answer 1

Hint: Quantum numbers are the set of numbers that decides the position and energy of the electron in an atom. There are four quantum numbers: Principal quantum number (n), Azimuthal quantum number (l), spin quantum number ($\text{ }{{\text{m}}_{l}}$), and spin quantum number $\text{ }{{\text{m}}_{s}}$. In sodium, the last shell electron enters in the $\text{3s}$ orbital. This is the highest energy electron of a sodium atom.

Complete step by step solution:
An atom is completely described by the four quantum numbers: principal quantum number (n), azimuthal quantum number (l), magnetic moment ($\text{ }{{\text{m}}_{l}}$), and spin quantum number$\text{ }{{\text{m}}_{s}}$.
Let’s first write down the subshell electronic configuration of the sodium atom. We know that the atomic number of sodium is 11. So, the subshell electronic configuration is as follows:
$\text{Na = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{1}}}$
According to the Aufbau principle, we know that electrons are filled in orbitals of an atom according to their energy. The electron having higher energy will be filled last. So, we know that the electron in the 3s orbital has the highest energy.
 Now we have to write the quantum numbers for this electron.
(1) Principal quantum number (n):
The principal quantum number describes the number of shells. Here, we are interested in the highest energy level electron. In sodium, the highest electron enters in the $\text{3s}$ orbital. Therefore the principal quantum number is in the third shell,$\text{(n = 3)}$.
(2) Azimuthal quantum number (l):
The orbital momentum quantum number (l) determines the shape of the subshell. The ‘l’ value is less than the principal quantum number$\text{(n}-\text{1)}$. For sodium, $\text{(n = 3)}$therefore the ‘l’ value is calculated as:
$\begin{align}
& l=\text{ (n}-1)=3-1=2 \\
& l=\text{ (n}-2)=3-2=1 \\
& l=\text{ (n}-1)=3-3=0 \\
\end{align}$
Therefore, the azimuthal quantum number (l) for the third shell of sodium atom is 0, 1, 2.
Here,
$\begin{align}
& l=0\text{ }\Rightarrow \text{ s} \\
& l=1\text{ }\Rightarrow \text{ p} \\
& l=2\text{ }\Rightarrow \text{ d} \\
\end{align}$
Since the highest shell electron of sodium is in the $\text{3s}$shell, therefore the ‘l’ value for $\text{3s}$ an electron is $l=0$
(3) Magnetic quantum number$\text{ }({{\text{m}}_{l}})\text{ }$:
The quantum number describes the orbitals and their orientation in the subshell. These values depend on the azimuthal quantum number. For a certain ‘l’ value the $\text{ }({{\text{m}}_{l}})\text{ }$values range from$-l\text{ to }+l$. Therefore, the above ‘l’ values, the $\text{ }({{\text{m}}_{l}})\text{ }$values for the third shell of sodium is:
$\begin{align}
& {{m}_{l}}=\text{ }\pm \text{ }l \\
& \text{For }l=2\text{ , }{{m}_{l}}\text{ = }\pm \text{2 or }-2\text{ , +2} \\
& \text{For }l=1\text{ , }{{m}_{l}}\text{ = }\pm 1\text{ or }-1\text{ , +1} \\
& \text{For }l=0\text{ , }{{m}_{l}}\text{ = 0 } \\
& \therefore \text{ }{{m}_{l}}\text{ = }-2\text{ , }-1\text{ , 0 , +1 , +2} \\
\end{align}$
Therefore, for $\text{(n = 3)}$ the magnetic quantum numbers are: $-2\text{ , }-1\text{ , 0 , +1 , +2}$
Since the sodium has the $l=0$ , thus the magnetic quantum number for $\text{3s}$ an electron is $\text{ }{{({{\text{m}}_{l}})}_{\text{Na}}}=\text{ 0}$
Since the $\text{ 3s }$electron of the sodium
(4) Electron spin quantum number ($\text{ }{{\text{m}}_{s}}$):
The electron spin quantum number $\text{ }{{\text{m}}_{s}}$ does not depend on another quantum number. It exhibits the direction of the electron spin. The spin has a value of $+\frac{1}{2}$ or $-\frac{1}{2}$. That means, when the $\text{ }{{\text{m}}_{s}}$ value is positive, the electron has an upper spin, and when a negative electron has down spin.
Therefore, an electron in the third shell of an electron may be ‘spin up’ or ‘spin down’.
$\text{ (}{{\text{m}}_{s}}{{)}_{\text{Na}}}\text{ = }\pm \text{ }\frac{1}{2}$
So, the quantum numbers for the electron of sodium atom having the highest energy are 3, 0, 0, and $+\frac{1}{2}$or$-\frac{1}{2}$.

Note: The electrons are filled on the basics of the Pauli Exclusion Principle, Hund's rule. The s orbitals hold the 2 electrons, p orbitals hold the 6 electrons, d electron holds the 10 electrons and f electron holds the 14 electrons. Since the magnetic quantum number decides the total number of subshells present in the shell. Remember that the quantum number is interdependent.