We need to count electrons in an atom that satisfy all of the following:

• principal quantum number: $n=4$
• magnetic quantum number magnitude: $|m_l|=1$
• spin quantum number: $m_s=-\tfrac{1}{2}$

For $n=4$, the allowed subshells are:

• $l=0$ (4s)
• $l=1$ (4p)
• $l=2$ (4d)
• $l=3$ (4f)

Now apply the condition $|m_l|=1$.

• For $l=0$: possible $m_l=0$ only, so not allowed.
• For $l=1$: possible $m_l=-1,0,+1$ so allowed values with $|m_l|=1$ are $m_l=\pm 1$.
• For $l=2$: possible $m_l=-2,-1,0,+1,+2$ so allowed values are $m_l=\pm 1$.
• For $l=3$: possible $m_l=-3,-2,-1,0,+1,+2,+3$ so allowed values are $m_l=\pm 1$.

Thus, for each of the subshells $l=1,2,3$, there are exactly $2$ orbitals satisfying $|m_l|=1$.

Each orbital can contain two electrons with spins: $m_s=+\tfrac{1}{2},\quad m_s=-\tfrac{1}{2}$ But we only want electrons with: $m_s=-\tfrac{1}{2}$ So each such orbital contributes exactly 1 electron.

Total count:

• from 4p: $2$ electrons
• from 4d: $2$ electrons
• from 4f: $2$ electrons

Therefore, $2+2+2=6$

Final answer: $\boxed{6}$

Comparison with stored correct answer:

• Stored correct answer = $6$
• Derived answer = $6$
• They match.