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

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Step 1: Understanding the Concept: Each subshell $l$ has $2l+1$ orbitals with magnetic quantum numbers $m_l$ ranging from $-l$ to $+l$. For every subshell ($s, p, d, f$), exactly one orbital has $m_l=0$. Step 2: Detailed Explanation: The ground state electronic configuration of Ge ($Z=32$) is: $$ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^2 $$ Let's identify the completely filled orbitals with $m_l=0$: - $1s^2$: One $s$ orbital ($m_l=0$). Completely filled. (1) - $2s^2$: One $s$ orbital ($m_l=0$). Completely filled. (1) - $2p^6$: Three $p$ orbitals ($m_l = -1, 0, +1$). The $m_l=0$ orbital is completely filled. (1) - $3s^2$: One $s$ orbital ($m_l=0$). Completely filled. (1) - $3p^6$: Three $p$ orbitals ($m_l = -1, 0, +1$). The $m_l=0$ orbital is completely filled. (1) - $4s^2$: One $s$ orbital ($m_l=0$). Completely filled. (1) - $3d^{10}$: Five $d$ orbitals ($m_l = -2, -1, 0, 1, 2$). The $m_l=0$ orbital is completely filled. (1) - $4p^2$: According to Hund's rule, the 2 electrons occupy separate orbitals (e.g., $m_l = -1$ and $m_l = 0$). The orbital with $m_l=0$ contains only 1 electron and is thus not completely filled. Sum of completely filled orbitals with $m_l=0$: $ 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7 $. Step 3: Final Answer: The value of $x$ is 7.

Ge (\(Z=32\)) in its ground state electronic configuration has \(x\) completely filled orbitals with \(m_l=0\). The value of \(x\) is \(\dots\dots\dots\).

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Correct Answer: 7

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