question 23 (1 point)
which of the following is the electron configuration for the valence shell of oxygen?
$\bigcirc$ a) $\frac{\uparrow\downarrow}{2s}$ $\frac{\uparrow\downarrow}{2p}$ $\downarrow$ $\uparrow$
$\bigcirc$ b) $\frac{\uparrow}{2s}$ $\frac{\uparrow\downarrow}{2p}$ $\frac{\uparrow\downarrow}{}$ $\uparrow$
$\bigcirc$ c) $\frac{\uparrow}{2s}$ $\frac{\uparrow}{2p}$ $\frac{\uparrow\downarrow}{}$ $\frac{\uparrow\downarrow}{}$
$\bigcirc$ d) $\frac{\uparrow\downarrow}{2s}$ $\frac{\uparrow\downarrow}{2p}$ $\uparrow$ $\uparrow$
$\bigcirc$ e) $\frac{\uparrow\uparrow}{2s}$ $\frac{\uparrow\downarrow}{2p}$ $\uparrow$ $\uparrow$

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# Brief Explanations:
1. First, recall the electron configuration of oxygen. The atomic number of oxygen is 8. The electron configuration is $1s^2 2s^2 2p^4$. So the valence shell is the $n = 2$ shell, with $2s^2 2p^4$ electrons.
2. For the $2s$ orbital: it can hold 2 electrons, so it should be filled with a pair (↑↓).
3. For the $2p$ orbitals: there are three $2p$ orbitals. According to Hund's rule, electrons fill orbitals singly first with parallel spins before pairing. The $2p$ subshell has 4 electrons. So first, each of the three $2p$ orbitals gets one electron (parallel spins), and then one orbital gets a second electron (paired). So the $2p$ electrons will have two orbitals with one unpaired electron and one orbital with a pair? Wait, no: 4 electrons in $2p$: first, three orbitals get one electron each (total 3), then the fourth electron pairs with one of them. So the $2p$ configuration is two electrons paired in one orbital and one unpaired in each of the other two? Wait, no, Hund's rule: electrons fill degenerate orbitals (same energy) singly with parallel spins before pairing. So for $2p^4$:
   - The three $2p$ orbitals: let's denote them as $2p_x$, $2p_y$, $2p_z$.
   - First, each gets one electron: $2p_x^1$, $2p_y^1$, $2p_z^1$ (parallel spins, say all ↑).
   - Then the fourth electron pairs with one of them, say $2p_x^2$ (now ↑↓), leaving $2p_y^1$ and $2p_z^1$ (still ↑).
   - So the $2s$ is ↑↓, and the $2p$ has one orbital with ↑↓ and two orbitals with ↑. Wait, but looking at the options:
   - Option d: $2s$ is ↑↓, $2p$ has two orbitals with ↑ (unpaired) and one with ↑↓? Wait, no, the diagram for option d: $2s$ is ↑↓, $2p$ has two ↑ (unpaired) and one ↑↓? Wait, the options are represented with the orbitals. Let's check the number of electrons:
   - $2s$ has 2 electrons (↑↓), $2p$ has 4 electrons. Let's count the electrons in each option:
     - Option a: $2s$ (2) + $2p$ (3? Wait, the diagram: $2s$ is ↑↓ (2), $2p$ has three electrons? No, the $2p$ part: the first $2p$ orbital is ↑↓, then two single? Wait, no, the way the options are drawn: each "box" is an orbital. For $2p$, there are three orbitals. Let's count the electrons:
       - Option d: $2s$ (↑↓: 2) + $2p$: first $2p$ orbital ↑↓ (2), then two orbitals with ↑ (1 each). So total $2 + 2 + 1 + 1 = 6$? No, wait, no: the $2p$ subshell has three orbitals. So in option d: $2s$ is one orbital (↑↓: 2), $2p$ has three orbitals: first $2p$ orbital ↑↓ (2), second $2p$ orbital ↑ (1), third $2p$ orbital ↑ (1). So total electrons: $2 + 2 + 1 + 1 = 6$? No, that can't be. Wait, no, the valence shell for oxygen is $2s^2 2p^4$, so total 6 electrons? Wait, no: $2s^2$ is 2, $2p^4$ is 4, total 6. Oh right! The valence shell is $n = 2$, so $2s^2 2p^4$, total 6 electrons.
       - So let's count the electrons in each option:
         - Option a: $2s$ (↑↓: 2) + $2p$: first $2p$ orbital ↑↓ (2), then two orbitals with ↓ and ↑ (1 each). Total: $2 + 2 + 1 + 1 = 6$? Wait, no, the $2p$ orbitals: how many? Three. So option a: $2s$ (1 orbital, 2 electrons), $2p$ (three orbitals: first ↑↓, second ↓, third ↑). So electrons: $2 + 2 + 1 + 1 = 6$? But the spins: in $2p$, the second orbital has ↓, third has ↑. Not parallel.
         - Option d: $2s$ (↑↓: 2), $2p$ (three orbitals: first ↑↓ (2), second ↑ (1), third ↑ (1)). So total electrons: $2 + 2 + 1 + 1 = 6$. And the $2p$ electrons: two orbitals with ↑ (parallel spins) and one with ↑↓. This follows Hund's rule: the two unpaired electrons have parallel spins, and one orbital has a pair.
         - Let's check the electron count: $2s^2$ (2) + $2p^4$ (4) = 6. Option d: $2s$ (2) + $2p$ (4: 2 + 1 + 1) = 6. And the $2p$ electrons: two unpaired (↑) and one paired (↑↓), which is correct for $2p^4$ (since 4 electrons: three orbitals, so one orbital with two, two with one each, and the unpaired ones have parallel spins).
# Answer:
d) $\frac{\uparrow\downarrow}{2s}\ \frac{\uparrow\downarrow}{2p}\ \frac{\uparrow}{}\ \frac{\uparrow}{}$

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