Question

directions: in a group of 3 - 4 people, complete one section of the practice below at a time. when your group is finished, bring it to the teacher to check. if you are right, your group will get to put a sticky note on any spot on the board. first team with 4 in a row on the board wins!
i. configuration writing practice
a. write the required electron configuration for each neutral atom.

1. longhand - na:
2. shorthand - pb:
3. longhand - sr:
4. shorthand - u:
5. longhand - n:

stop and bring your work to mrs. muller to check for accuracy and earn your first spot pick!
b. for the following electron configurations determine the possible elements they may represent.

1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴:
2. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶:
3. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁰:

9.1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵:

1. kr 5s² 4d¹⁰ 5p³:
2. ar 4s¹:

12.xe 6s² 4f¹⁰:

1. ne 3s² 3p¹:

stop and bring your work to mrs. muller to check for accuracy and earn your next spot pick!
c. in the space below, write the full (unabbreviated) electron configurations of the following elements:

1. sodium
2. iron
3. bromine
4. barium
5. palladium

stop and bring your work to mrs. muller to check for accuracy and earn your next spot pick!

Explanation:

Step1: Recall electron - configuration rules

Electrons fill orbitals in order of increasing energy (1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p). The s - orbital can hold 2 electrons, p - orbital can hold 6, d - orbital can hold 10 and f - orbital can hold 14.

Step2: Find atomic numbers

Use the periodic table to find the atomic number of each element. The atomic number is equal to the number of electrons in a neutral atom.

Step3: Write long - hand electron configurations

1. Sodium (Na), atomic number = 11. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{1}$
2. Strontium (Sr), atomic number = 38. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}$
3. Nitrogen (N), atomic number = 7. Long - hand: $1s^{2}2s^{2}2p^{3}$
4. Sodium (Na), atomic number = 11. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{1}$
5. Iron (Fe), atomic number = 26. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
6. Bromine (Br), atomic number = 35. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$
7. Barium (Ba), atomic number = 56. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}$
8. Palladium (Pd), atomic number = 46. Long - hand: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{0}4d^{10}$

Step4: Write short - hand electron configurations

1. Lead (Pb), atomic number = 82. Short - hand: $[Xe]4f^{14}5d^{10}6s^{2}6p^{2}$
2. Uranium (U), atomic number = 92. Short - hand: $[Rn]5f^{3}6d^{1}7s^{2}$

Step5: Determine elements from electron configurations

1. Count the electrons: $2 + 2+6 + 2+6 + 2+10 + 4=34$. Element is Selenium (Se).
2. Count the electrons: $2 + 2+6 + 2+6 + 2+10 + 6=36$. Element is Krypton (Kr).
3. Count the electrons: $2 + 2+6 + 2+6 + 2+10 + 6+2 + 10+6 + 2+10=70$. Element is Ytterbium (Yb).
4. Count the electrons: $2 + 2+6 + 2+6 + 2+5=23$. Element is Vanadium (V).
5. Krypton (Kr) core + $5s^{2}4d^{10}5p^{3}$. Counting the extra electrons: $2 + 10+3 = 15$ more than Kr. Element is Antimony (Sb).
6. Argon (Ar) core + $4s^{1}$. Element is Potassium (K).
7. Xenon (Xe) core + $6s^{2}4f^{10}$. Element is Erbium (Er).
8. Neon (Ne) core + $3s^{2}3p^{1}$. Element is Aluminum (Al).

Answer:

1. $1s^{2}2s^{2}2p^{6}3s^{1}$
2. $[Xe]4f^{14}5d^{10}6s^{2}6p^{2}$
3. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}$
4. $[Rn]5f^{3}6d^{1}7s^{2}$
5. $1s^{2}2s^{2}2p^{3}$
6. Selenium (Se)
7. Krypton (Kr)
8. Ytterbium (Yb)
9. Vanadium (V)
10. Antimony (Sb)
11. Potassium (K)
12. Erbium (Er)
13. Aluminum (Al)
14. $1s^{2}2s^{2}2p^{6}3s^{1}$
15. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
16. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$
17. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}$
18. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{0}4d^{10}$