1. Write the ground state electronic configuration of Ge ($Z=32$):

2. Find orbitals with $m_l=0$ in each subshell and count only those that are completely filled.

For a subshell with azimuthal quantum number $l$, the possible values of $m_l$ are:

So each subshell has exactly one orbital with $m_l=0$.

Now check each occupied subshell:

• $1s^2$: for $s$, $l=0$, so only $m_l=0$ exists. This orbital is completely filled. Count = 1
• $2s^2$: completely filled $m_l=0$ orbital. Count = 1
• $2p^6$: $p$ has $m_l=-1,0,+1$; the $m_l=0$ orbital is filled in $p^6$. Count = 1
• $3s^2$: completely filled $m_l=0$ orbital. Count = 1
• $3p^6$: $m_l=0$ orbital completely filled. Count = 1
• $3d^{10}$: $d$ has $m_l=-2,-1,0,+1,+2$; in $d^{10}$ all are completely filled, so the $m_l=0$ orbital is completely filled. Count = 1
• $4s^2$: completely filled $m_l=0$ orbital. Count = 1
• $4p^2$: here $p$ subshell is not completely filled. By Hund’s rule, the two electrons occupy two different $p$ orbitals singly, so the $m_l=0$ orbital is not certainly completely filled. Hence do not count it.

3. Total count:

4. Final answer:

5. Comparison with stored correct answer:

Stored correct answer = 7, which matches the derived answer.