Chapter 44, Problem 18SP

To determine

The first four univalent atoms, if there were no $m_l$ quantum numbers.

Answer to Problem 18SP

Solution:

$\text{H}$, $\text{Li}$, $\text{N}$, and $\text{Al}$.

Explanation of Solution

Introduction:

The value of the principle quantum number $\left(n\right)$ tells about the shell of the electron. For example, $K,L,M,N\mathrm{....}$

The value of the orbital quantum number shows the subshells for the electron. The range of the orbital quantum number $\left(l\right)$ goes from $0$ to $\left(n-1\right)$.

The range of the magnetic quantum number $\left(m_l\right)$ goes from $-l$ to $0$ to $+l$.

The range of the spin quantum number $m_s$ is $\pm \frac{1}{2}$.

Pauli exclusion principle says that there could be $2$ electrons in $n=1$ shell, $8$ electrons in $n=2$, and $18$ electrons in $n=3$ shell according to the expression $2n^2$.

When a single electron is present in the outer Bohr’s orbit, it is easier to remove it from the atom, and that atom is known as univalent atom.

Explanation:

Find the quantum numbers for one electron, two electrons, and so on, to find the first four univalent elements, if $m_l$ quantum numbers are not present for electrons.

Find the quantum numbers for electrons, if there were no $m_l$ quantum numbers.

$\begin{array}{lllll}\text{Electron 1}:\hfill & n=1,\hfill & l=0,\hfill & m_s=+\frac{1}{2},\hfill & \left(\text{Univalent}\right)\hfill \\ \text{Electron 2}:\hfill & n=1,\hfill & l=0,\hfill & m_s=-\frac{1}{2},\hfill & \hfill \\ \text{Electron 3}:\hfill & n=2,\hfill & l=0,\hfill & m_s=+\frac{1}{2},\hfill & \left(\text{Univalent}\right)\hfill \\ \text{Electron 4}:\hfill & n=2,\hfill & l=0,\hfill & m_s=-\frac{1}{2},\hfill & \hfill \\ \text{Electron 5}:\hfill & n=2,\hfill & l=1,\hfill & m_s=+\frac{1}{2},\hfill & \hfill \\ \text{Electron 6}:\hfill & n=2,\hfill & l=1,\hfill & m_s=-\frac{1}{2},\hfill & \hfill \\ \text{Electron 7}:\hfill & n=3,\hfill & l=0,\hfill & m_s=+\frac{1}{2},\hfill & \left(\text{Univalent}\right)\hfill \\ \text{Electron 8}:\hfill & n=3,\hfill & l=0,\hfill & m_s=-\frac{1}{2},\hfill & \hfill \\ \text{Electron 9}:\hfill & n=3,\hfill & l=1,\hfill & m_s=+\frac{1}{2},\hfill & \hfill \\ \text{Electron 10}:\hfill & n=3,\hfill & l=1,\hfill & m_s=-\frac{1}{2},\hfill & \hfill \\ \text{Electron 11}:\hfill & n=4,\hfill & l=0,\hfill & m_s=+\frac{1}{2},\hfill & \left(\text{Univalent}\right)\hfill \end{array}$

Observe from the table that for all electrons, $m_l$ is absent. So, the maximum number of electrons filled in the subshell will be $2$. For $\text{Electron 1}$, there is only one electron in the $n=1$ shell. As, one electron exists in the hydrogen atom. Thus, hydrogen $\left(\text{H}\right)$ will be the first univalent atom.

For $\text{Electron 3}$, two electrons exist in the $n=1$ shell, but the third electron will be in the $n=2$ subshell. Thus, one electron will exist in the new shell corresponding to $n=2$ shell. As, three electrons exist in lithium atom. So, the second univalent atom will be lithium $\left(\text{Li}\right)$.

For $\text{Electron 7}$, two electrons will be in the $n=1$ shell, four electrons will be in the $n=2$ shell, and one electron will be in the new shell corresponding to $n=3$, as there were no $m_l$ quantum numbers. As, seven electrons are present in nitrogen. So, the third univalent atom will be nitrogen $\left(\text{N}\right)$.

Understand that, if there were no $m_l$ quantum numbers, then the maximum number of electrons in $p$ subshell will be $2$ including the spin of electrons.

For $\text{Electron 11}$, two electrons will be in the $n=1$ shell, four electrons will be in the $n=2$ shell, four electrons will be in the $n=3$ shell, and one last electron will be in the $n=4$ shell, as there were no $m_l$ quantum numbers. As, eleven electrons are present in aluminum. So, the fourth univalent atom will be $\text{N}$.

Conclusion:

Hence, the first four univalent atoms, if there were no $m_l$ quantum numbers are $\text{H}$, $\text{Li}$, $\text{N}$, and $\text{Al}$.