Animated Solution for Chemistry - Atomic Structure: In an atom, the total number of electrons having quantum numbers $n=4$, $|m_l|=1$ and $m_s=-\frac{1}{2}$ is

Imagine you are a cosmic detective tasked with finding a very specific group of electrons within an atom. To find them, you are given a precise "quantum address" consisting of three clues: the principal quantum number $n=4$, a strict condition on the magnetic quantum number $|m_l|=1$, and a specific spin quantum number $m_s=-\frac{1}{2}$.

Our mission is to systematically decode this address and count exactly how many electrons live there. Let's break it down step by step.

The first clue is $n=4$. This tells us we are looking exclusively within the fourth principal shell of the atom. But a shell is like a large apartment building; we need to know which specific floors (subshells) exist inside it.

The rules of quantum mechanics dictate that the azimuthal quantum number, $l$, can take any integer value from $0$ up to $n-1$.

For $n=4$, the possible values for $l$ are $0,1,2,$ and $3$. These correspond to the 4s, 4p, 4d, and 4f subshells, respectively. Our target electrons must be hiding somewhere within these subshells.

Now, let's look at the second, trickier clue: $|m_l|=1$. The absolute value bars are a classic trap! Mathematically, if the magnitude of $m_l$ is $1$, it means the actual value of $m_l$ can be either $+1$ or $-1$.

We need to scan through all our available subshells and find every single orbital that has an $m_l$ value of exactly $+1$ or $-1$.

Let's conduct our search subshell by subshell:

1. The 4s Subshell ($l=0$): For an s-subshell, the only possible value for $m_l$ is $0$. Since $0$ is neither $+1$ nor $-1$, there are no matching orbitals here.

2. The 4p Subshell ($l=1$): For a p-subshell, $m_l$ can be $-1,0,$ or $+1$. Look at that! We have exactly two orbitals that match our criteria: the one with $m_l=-1$ and the one with $m_l=+1$.

3. The 4d Subshell ($l=2$): For a d-subshell, $m_l$ ranges from $-2$ to $+2$ (specifically $-2,-1,0,+1,+2$). Scanning through, we again find exactly two orbitals that fit the bill: $m_l=-1$ and $m_l=+1$.

4. The 4f Subshell ($l=3$): For the massive f-subshell, $m_l$ ranges all the way from $-3$ to $+3$. Even among these seven orbitals, there are still only two orbitals that have an $m_l$ of $-1$ or $+1$.

Tallying them all up, we found $2+2+2=6$ orbitals in total across the entire fourth shell that satisfy the condition $|m_l|=1$.

We have found the 6 rooms (orbitals), but we need to count the specific residents (electrons). Our final clue is the spin quantum number: $m_s=-\frac{1}{2}$.

According to Pauli's Exclusion Principle, a single orbital can hold a maximum of two electrons, and they must have opposite spins (one with $m_s=+\frac{1}{2}$ and one with $m_s=-\frac{1}{2}$).

This means that inside each of our 6 selected orbitals, there is exactly one electron spinning with $m_s=-\frac{1}{2}$.

To find the total number of electrons, we simply multiply the number of valid orbitals by the number of valid electrons per orbital:

The final answer is 6.