9. which of the following is the ground - state electron configuration for the na^+ ion?
(a) 1s^22s^22p^5
(b) 1s^22s^22p^6
(c) 1s^22s^22p^63s^1
(d) 1s^22s^22p^63s^2
10. which of the following ground - state electron configurations represents the atom that has the lowest first - ionization energy?
(a) 1s^22s^1
(b) 1s^22s^22p^2
(c) 1s^22s^22p^6
(d) 1s^22s^22p^63s^1
11. 1s^22s^22p^63s^23p^6
how many unpaired electrons are in the atom represented by the electron configuration above?
(a) 0
(b) 1
(c) 2
(d) 3
12. which of the following is the electron configuration of an excited atom that is likely to emit a quantum of energy?
(a) 1s^22s^22p^63s^23p^1
(b) 1s^22s^22p^63s^23p^5
(c) 1s^22s^22p^63s^1
(d) 1s^22s^22p^63s^13p^1
13. which of the following represents the electron configuration of an oxygen atom in the ground state?
(a)
(b)
(c)
(d)

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Accepted Answer

### Question 9
# Brief Explanations:
Sodium (Na) has an atomic number of 11, so its ground - state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$. When Na forms the $Na^{+}$ ion, it loses one electron from the 3s orbital. So the electron configuration of $Na^{+}$ should be $1s^{2}2s^{2}2p^{6}$. Option A has an incorrect number of electrons in the 2p orbital. Option C is the electron configuration of neutral Na. Option D has an incorrect number of electrons in the 3s orbital.
# Answer:
B. $1s^{2}2s^{2}2p^{6}$

### Question 10
# Brief Explanations:
First - ionization energy is the energy required to remove the outermost electron from an atom. Atoms with larger atomic radii and more loosely held outermost electrons have lower first - ionization energy. The electron configuration $1s^{2}2s^{2}2p^{6}3s^{1}$ (option D) corresponds to a sodium atom. Sodium has a relatively large atomic radius and its outermost electron is in the 3s orbital, which is more loosely held compared to the electrons in the atoms with the other configurations. Option A is He (noble gas, high ionization energy), option B is C (smaller radius, higher ionization energy than Na), option C is Ne (noble gas, very high ionization energy).
# Answer:
D. $1s^{2}2s^{2}2p^{6}3s^{1}$

### Question 11
# Explanation:
## Step 1: Analyze the electron configuration
The electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. We need to look at the last sub - shell that is not completely filled (or check all sub - shells for unpaired electrons). The $1s$, $2s$, $2p$, $3s$, and $3p$ sub - shells:
 - $1s^{2}$: The two electrons are paired (spin - up and spin - down).
 - $2s^{2}$: The two electrons are paired.
 - $2p^{6}$: All six electrons are paired (three orbitals, each with two paired electrons).
 - $3s^{2}$: The two electrons are paired.
 - $3p^{6}$: All six electrons are paired (three orbitals, each with two paired electrons).
## Step 2: Determine the number of unpaired electrons
Since all electrons in each sub - shell are paired, the number of unpaired electrons is 0.
# Answer:
A. 0

### Question 12
# Brief Explanations:
An excited atom that is likely to emit a quantum of energy will have an electron in a higher - energy orbital that can fall back to a lower - energy orbital. In a ground - state electron configuration, electrons fill the lowest - energy orbitals first. For an atom with atomic number (let's assume it's a chlorine - like atom, but we can also use the Aufbau principle), the ground - state configuration of a chlorine - like atom would have electrons filling lower - energy orbitals. Option A: $1s^{2}2s^{2}2p^{6}3s^{2}3p_{x}^{1}$ (assuming the last electron is in a 3p orbital) is a ground - state - like configuration. Option B: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$ is a ground - state configuration (for Cl). Option C: $1s^{2}2s^{2}2p^{6}3s^{1}$ is a ground - state configuration (for Na). Option D: $1s^{2}2s^{2}2p^{6}3s^{1}3p^{1}$ has an electron in the 3p orbital while the 3s orbital is not completely filled (it has only 1 electron instead of 2). This means the electron in the 3p orbital is in an excited state and can fall back to the 3s orbital, emitting a quantum of energy.
# Answer:
D. $1s^{2}2s^{2}2p^{6}3s^{1}3p^{1}$

### Question 13
# Brief Explanations:
Oxygen has an atomic number of 8, so its electron configuration is $1s^{2}2s^{2}2p^{4}$. According to Hund's rule, electrons in the same sub - shell (2p in this case) will occupy separate orbitals with parallel spins before pairing up. The 2p sub - shell has 3 orbitals. For 4 electrons in the 2p sub - shell, two electrons will be in separate orbitals with parallel spins, and one orbital will have a pair of electrons.
 - Option A: The 2p sub - shell has 2 electrons, which is incorrect (O has 4 electrons in 2p).
 - Option B: The 1s, 2s sub - shells are filled correctly, and the 2p sub - shell has 3 unpaired electrons (total 4 electrons in 2p: 3 unpaired and 1 paired? No, Hund's rule: first, electrons fill orbitals singly with parallel spins. For 4 electrons in 3 orbitals: 3 orbitals, 3 electrons with parallel spins (unpaired) and 1 orbital with a pair. Wait, no: 4 electrons in 3 orbitals: 2 orbitals with 1 electron (unpaired) and 1 orbital with 2 electrons (paired)? Wait, no, the correct filling for $1s^{2}2s^{2}2p^{4}$ is: 1s (paired), 2s (paired), 2p: two orbitals with 1 electron (unpaired) and one orbital with 2 electrons (paired). Looking at the options:
Option B shows 1s (paired), 2s (paired), 2p with 3 unpaired electrons (total 5 electrons in 2p? No, O has 4 in 2p). Wait, no, let's count the electrons:
 - Option A: 1s (2) + 2s (2)+2p (2) = 6 electrons (wrong, O has 8).
 - Option B: 1s (2)+2s (2)+2p (3) = 7 electrons (wrong).
 - Option C: 1s (2)+2s (2)+2p (3) = 7 electrons (wrong). Wait, no, maybe I made a mistake. Wait, O's electron configuration: $1s^{2}2s^{2}2p^{4}$. So 1s: 2, 2s: 2, 2p: 4. The 2p sub - shell has 3 orbitals. So, according to Hund's rule, we first put one electron in each of the three 2p orbitals (parallel spins), then we pair one of them. So we have two unpaired electrons in the 2p sub - shell.
Looking at the options:
 - Option B: The 1s and 2s are filled (2 electrons each), and the 2p has 3 unpaired electrons? No, wait, the diagram for option B: 1s (paired), 2s (paired), 2p has 3 arrows (unpaired) and one empty? No, the number of electrons: 1s (2) + 2s (2)+2p (3) = 7? No, O has 8. Wait, maybe the diagrams are representing the electron - spin and orbital filling. Let's re - evaluate:
The correct electron filling for O: $1s^{2}2s^{2}2p^{4}$. So 1s: paired, 2s: paired, 2p: two electrons in separate orbitals (unpaired) and one orbital with a pair. So the 2p sub - shell has 2 unpaired electrons.
Option B: The 1s and 2s are filled, and the 2p has 3 unpaired electrons? No, wait, the diagram for option B: 1s (two arrows, paired), 2s (two arrows, paired), 2p (three arrows, unpaired) and one empty orbital? No, the total number of electrons: 2 (1s)+2 (2s)+3 (2p) = 7, which is wrong. Wait, maybe I messed up. Let's count the number of electrons in each option:
 - Option A: 1s (2) + 2s (2)+2p (2) = 6 (wrong, O has 8).
 - Option B: 1s (2)+2s (2)+2p (3) = 7 (wrong).
 - Option C: 1s (2)+2s (2)+2p (3) = 7 (wrong). Wait, no, maybe the 2s sub - shell in option C is wrong. Wait, the correct option is B? Wait, no, let's check the electron configuration again. O has 8 electrons: $1s^{2}2s^{2}2p^{4}$. The 2p sub - shell has 3 orbitals. So, electrons fill the 2p orbitals as follows: first, each of the 3 orbitals gets one electron (parallel spins), that's 3 electrons. Then the fourth electron pairs with one of the electrons in a 2p orbital. So in the 2p sub - shell, we have 2 unpaired electrons (in two separate orbitals) and one orbital with a pair of electrons.
Looking at the options:
 - Option B: The 1s and 2s are filled (2 electrons each), and the 2p sub - shell has 3 unpaired electrons? No, that would be 5 electrons in 2p. Wait, I think I made a mistake. Let's count the number of electrons in each option:
 - Option A: 1s (2) + 2s (2)+2p (2) = 6 (incorrect, O has 8).
 - Option B: 1s (2)+2s (2)+2p (3) = 7 (incorrect). Wait, no, the 2s sub - shell in option C has 3 electrons? No, 2s can hold a maximum of 2 electrons. Oh! Option C has 3 electrons in 2s, which is incorrect. Option D has paired electrons in all 2p orbitals, which violates Hund's rule. Option B: 1s (2), 2s (2), 2p (3) – no, 2s can only have 2 electrons. Wait, no, the diagram for option B: 1s has two arrows (paired), 2s has two arrows (paired), 2p has three arrows (unpaired) – but O has 4 electrons in 2p. Wait, maybe the question is about the orbital - filling diagram. The correct orbital - filling diagram for O ($1s^{2}2s^{2}2p^{4}$) is: 1s (paired), 2s (paired), 2p: two orbitals with 1 electron (unpaired) and one orbital with 2 electrons (paired). So among the options, option B has 1s (paired), 2s (paired), and 2p with 3 unpaired electrons? No, that's not right. Wait, maybe I misread the options. Let's re - look:
Option B: 1s (two arrows), 2s (two arrows), 2p (three arrows) – total electrons: 2 + 2+3 = 7? No, O has 8. Wait, the 2p sub - shell in O has 4 electrons. So 1s (2), 2s (2), 2p (4). So in the 2p sub - shell, with 3 orbitals, we have 4 electrons: two orbitals with 1 electron (unpaired) and one orbital with 2 electrons (paired). So the correct diagram should have 1s (paired), 2s (paired), and 2p with two unpaired electrons (in two separate orbitals) and one orbital with a pair. Option B: the 2p sub - shell has 3 unpaired electrons (so 3 electrons) and one orbital empty? No, that's 3 electrons in 2p. Wait, I think the correct answer is B. Because:
 - Option A: 2p has 2 electrons (wrong, O has 4 in 2p).
 - Option B: 1s (2), 2s (2), 2p (3) – no, 2p should have 4. Wait, no, maybe the diagram is showing the spin - up and spin - down. Let's count the number of electrons:
 - Option A: 1s (2) + 2s (2)+2p (2) = 6 (wrong).
 - Option B: 1s (2)+2s (2)+2p (3) = 7 (wrong). Wait, this is confusing. Wait, oxygen's electron configuration is $1s^{2}2s^{2}2p^{4}$. The 2p sub - shell has 3 orbitals. So, according to Hund's rule, electrons fill the orbitals singly first. So for 4 electrons in 3 orbitals: 3 orbitals with 1 electron (parallel spins) and then one orbital with a second electron (paired). So the number of unpaired electrons is 2. Looking at the options:
 - Option B: The 2p sub - shell has 3 unpaired electrons? No, that would be 3 electrons. Wait, maybe the option B has 4 electrons in 2p: 3 unpaired and 1 paired? No, 3 unpaired electrons would mean 3 electrons, and 1 paired would be 4. Oh! Yes, 3 unpaired electrons (3 orbitals, each with 1 electron) and 1 paired electron (in one of the orbitals) – total 4 electrons. So option B: 1s (2), 2s (2), 2p (3 unpaired + 1 paired) – that's 4 electrons in 2p. So option B is correct. Option A has 2 electrons in 2p, option C has 3 electrons in 2s (which is wrong, 2s can hold 2), option D has paired electrons in all 2p orbitals (violates Hund's rule).
# Answer:
B (the orbital - filling diagram with 1s paired, 2s paired, and 2p with 3 unpaired electrons and 1 paired electron (total 4 electrons in 2p))

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Explanation:

Question 9

Step 1: Analyze the electron configuration

Step 2: Determine the number of unpaired electrons

Answer:

Question 10