Ge (Z = 32) in its ground state electronic configuration has x completely filled orbitals with m$$_l$$ = 0. The value of x is _________.

The magnetic quantum number $$m_l$$ tells us the orientation of an orbital.
For any subshell with azimuthal quantum number $$l$$, the possible $$m_l$$ values are $$-l,\;-(l-1),\dots ,0,\dots ,+(l-1),\; +l$$.
Hence every subshell that has $$l \ge 0$$ always contains exactly one orbital with $$m_l = 0$$.

The question asks for the number of such $$m_l = 0$$ orbitals that are completely filled (contain two electrons) in the ground-state configuration of germanium $$(Z = 32)$$.

Step 1: Write the ground-state electronic configuration of Ge (using the Aufbau, Hund and Pauli rules).
$$\text{Ge}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^2$$

Step 2: Examine each occupied subshell.
For every subshell, decide whether its $$m_l = 0$$ orbital is fully occupied (2 electrons) or not.

s-subshell ⇒ $$l = 0$$ ⇒ only one orbital ($$m_l = 0$$). It has 2 e⁻ ⇒ count 1.

Same reasoning ⇒ count 1 more (total 2).

p-subshell ⇒ three orbitals ($$m_l = -1,0,+1$$). Six electrons fill all three, so the $$m_l = 0$$ orbital is filled ⇒ count 1 (total 3).

s-subshell ⇒ count 1 (total 4).

p-subshell completely filled ⇒ count 1 (total 5).

s-subshell ⇒ count 1 (total 6).

d-subshell ⇒ five orbitals ($$m_l = -2,-1,0,+1,+2$$). Ten electrons fill all five, so the $$m_l = 0$$ orbital is filled ⇒ count 1 (total 7).

This subshell contains only two electrons. According to Hund’s rule they occupy two different orbitals with parallel spins, leaving the third orbital half-filled. Thus the $$m_l = 0$$ orbital holds at most one electron and is not completely filled ⇒ do not count.

Step 3: Add all contributions.
Total completely filled $$m_l = 0$$ orbitals $$= 7$$.

Hence, $$x = 7$$.