7. which element has its last electron in the $4d^7$ sublevel? explain how you determined it.
8. write the electron configuration of the ion $fe^{3+}$ and explain how it differs from fe.
9. which atom is more stable: n, $o^{2-}$, or ne? defend your reasoning using orbital theory.
10. how many orbitals exist in the n = 4 shell? how many total electrons can it hold?
11. predict the most likely ion formed by arsenic and justify using periodic trends.

Question

Accepted Answer

### Question 7
# Explanation:
## Step1: Find noble gas core
The noble gas before the 4d block is Kr (atomic number 36, electron configuration: $[Kr]$).
## Step2: Add 4d⁷ electrons
Add the 7 electrons in the 4d sublevel: $[Kr]4d^7$.
## Step3: Count total electrons
Total electrons = $36 + 7 = 43$. The element with atomic number 43 is Technetium (Tc).
# Answer:
Technetium (Tc). The noble gas core before the 4d sublevel is Kr (36 electrons). Adding 7 electrons in the 4d sublevel gives a total of 43 electrons, which matches the atomic number of technetium.

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### Question 8
# Explanation:
## Step1: Find neutral Fe configuration
Iron (Fe, atomic number 26) has an electron configuration of $[Ar]3d^64s^2$.
## Step2: Remove 3 electrons for Fe³⁺
When forming $Fe^{3+}$, electrons are lost first from the 4s sublevel, then the 3d sublevel: remove 2 from 4s, 1 from 3d.
# Answer:
Electron configuration of $Fe^{3+}$: $[Ar]3d^5$.
Neutral Fe has the configuration $[Ar]3d^64s^2$; $Fe^{3+}$ has lost 3 electrons (2 from the 4s sublevel and 1 from the 3d sublevel) to form a positively charged ion, resulting in a half-filled, more stable 3d⁵ sublevel.

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### Question 9
# Brief Explanations:
Stability is determined by having a full valence electron shell (octet rule) and minimized electron-electron repulsion. Neon (Ne) has a full valence shell ($2s^22p^6$) with no unpaired electrons and no excess charge. Nitrogen (N) has a $2s^22p^3$ configuration (half-filled p sublevel, but not a full octet). $O^{2-}$ has the same electron configuration as Ne, but carries a -2 charge, which introduces electrostatic repulsion from excess electrons, making it less stable than neutral Ne.
# Answer:
Neon (Ne) is the most stable. Ne has a full, neutral valence electron shell ($2s^22p^6$) with no charge-related repulsion. N lacks a full octet, and $O^{2-}$ has the same electron configuration as Ne but has a negative charge that creates destabilizing electron-electron repulsion.

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### Question 10
# Explanation:
## Step1: Calculate total orbitals for n=4
For a shell with principal quantum number $n$, the number of orbitals is $n^2$.
Expression: $4^2 = 16$
## Step2: Calculate total electrons for n=4
Each orbital holds 2 electrons, so total electrons = $2n^2$.
Expression: $2	imes4^2 = 32$
# Answer:
There are 16 orbitals in the n=4 shell, and it can hold a total of 32 electrons.

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### Question 11
# Brief Explanations:
Arsenic (As) is in group 15 of the periodic table. Group 15 elements have 5 valence electrons. To achieve a stable noble gas electron configuration (octet), they tend to gain 3 electrons to fill their valence shell, forming a -3 charged ion.
# Answer:
The most likely ion formed by arsenic is $As^{3-}$. Arsenic is in group 15 of the periodic table, with 5 valence electrons. Following periodic trends, group 15 elements gain 3 electrons to reach a full octet (matching the electron configuration of the noble gas krypton), forming a negatively charged ion with a -3 charge.

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QUESTION IMAGE

Question

Explanation:

Question 7

Step1: Find noble gas core

Step2: Add 4d⁷ electrons

Step3: Count total electrons

Step1: Find neutral Fe configuration

Step2: Remove 3 electrons for Fe³⁺

Answer:

Question 8