Electron Configuration Practice worksheet for learning atomic structure and electron arrangements.

Here are the step-by-step solutions to the electron configuration practice problems.

Part A: For each element

1. Nitrogen
* Number of electrons: Nitrogen is element #7, so it has 7 electrons.
* Configuration: We fill orbitals starting from the bottom up (Aufbau principle).
* 1s gets 2 ($1s^2$)
* 2s gets 2 ($2s^2$)
* That leaves 3 electrons for the 2p orbital. According to Hund's rule, we put one in each box before pairing. So, three up-spins in the 2p section.
* Full config: $1s^2\ 2s^2\ 2p^3$

2. Silicon
* Number of electrons: Silicon is element #14.
* Configuration:
* 1s ($2$), 2s ($2$), 2p ($6$) = 10 electrons used (Neon core).
* Next is 3s ($2$) -> Total 12.
* Remaining 2 go into 3p.
* Full config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^2$

3. Neon
* Number of electrons: Neon is element #10.
* Configuration:
* 1s ($2$), 2s ($2$), 2p ($6$). Total is 10.
* Full config: $1s^2\ 2s^2\ 2p^6$

4. Copper
* Number of electrons: Copper is element #29.
* Configuration: This is a famous exception to the standard rules!
* Standard filling would be: $...4s^2\ 3d^9$.
* However, a full d-subshell is more stable. One electron moves from 4s to 3d.
* Correct config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^1\ 3d^{10}$

5. Zirconium
* Number of electrons: Zirconium is element #40.
* Configuration:
* Noble gas core [Kr] accounts for 36 electrons ($1s^2...4p^6$).
* Next is 5s ($2$) -> Total 38.
* Remaining 2 go into 4d.
* Full config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^2$

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Part B: Identify the element

6. $1s^2 2s^2$
* Total electrons: $2 + 2 = 4$. Element #4 is Beryllium (Be).

7. $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$
* Count the electrons: $2+2+6+2+6+2+2 = 22$. Element #22 is Titanium (Ti).

8. $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^4$
* Count the electrons: $2+2+6+2+6+2+10+4 = 34$. Element #34 is Selenium (Se).

9. $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$
* Count the electrons: $2+2+6+2+6+2+10+6+1 = 37$. Element #37 is Rubidium (Rb).

10. $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^9$
* Let's sum these up carefully. The noble gas Xenon (Xe) ends at $5p^6$ and has 54 electrons.
* Add the rest: $2 (6s) + 14 (4f) + 9 (5d) = 25$.
* Total: $54 + 25 = 79$. Element #79 is Gold (Au).

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Part C: Write the ground state electron configuration

11. Iron (Fe)
* Atomic number: 26.
* Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d.
* Config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^6$

12. Sulfur (S)
* Atomic number: 16.
* Filling order: 1s, 2s, 2p, 3s, 3p.
* Config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^4$

13. Krypton (Kr)
* Atomic number: 36.
* Filling order: 1s through 4p.
* Config: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6$

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Part D: Shorthand (noble gas) notation

14. Chlorine (Cl)
* Atomic number: 17.
* Previous noble gas: Neon (Ne, atomic number 10).
* Electrons remaining: 7. They go into 3s and 3p.
* Notation: [Ne] 3s^2 3p^5

15. Antimony (Sb)
* Atomic number: 51.
* Previous noble gas: Krypton (Kr, atomic number 36).
* Electrons remaining: 15. Fill 5s, then 4d, then 5p.
* Notation: [Kr] 5s^2 4d^{10} 5p^3

16. Lanthanum (La)
* Atomic number: 57.
* Previous noble gas: Xenon (Xe, atomic number 54).
* Electrons remaining: 3. It goes into 6s and 5d (it skips 4f here as it starts the transition series).
* Notation: [Xe] 6s^2 5d^1

Final Answer:
A.
1. 7; $1s^2\ 2s^2\ 2p^3$
2. 14; $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^2$
3. 10; $1s^2\ 2s^2\ 2p^6$
4. 29; $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^1\ 3d^{10}$
5. 40; $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^2$

B.
6. Be
7. Ti
8. Se
9. Rb
10. Au

C.
11. $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^6$
12. $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^4$
13. $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6$

D.
14. [Ne] $3s^2\ 3p^5$
15. [Kr] $5s^2\ 4d^{10}\ 5p^3$
16. [Xe] $6s^2\ 5d^1$