Question

b. for the following electron configurations determine the possib

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 :

Explanation:

Step1: Recall electron - configuration rules

The electron - configuration notation follows the order of filling of atomic orbitals (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, etc.). The number of electrons in each orbital is given as a superscript. The total number of electrons in an atom is equal to its atomic number.

Step2: Count electrons for 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴

$2 + 2+6 + 2+6 + 2+10 + 4=34$. The element with atomic number 34 is Selenium.

Step3: Count electrons for 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

$2 + 2+6 + 2+6 + 2+10 + 6 = 36$. The element with atomic number 36 is Krypton.

Step4: Count electrons for 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁰

$2+2 + 6+2+6+2+10+6+2+10+6+2 + 10=70$. The element with atomic number 70 is Ytterbium.

Step5: Count electrons for 1s²2s²2p⁶3s²3p⁶4s²3d⁵

$2+2 + 6+2+6+2+5=25$. The element with atomic number 25 is Vanadium.

Step6: Expand [Kr]5s²4d¹⁰5p³

Krypton has 36 electrons. Adding $2 + 10+3$ gives $36+15 = 51$. The element with atomic number 51 is Antimony.

Step7: Expand [Ar]4s¹

Argon has 18 electrons. Adding 1 gives 19. The element with atomic number 19 is Potassium.

Step8: Expand [Xe]6s²4f¹⁰

Xenon has 54 electrons. Adding $2+10$ gives $54 + 12=66$. The element with atomic number 66 is Erbium.

Step9: Expand [Ne]3s²3p¹

Neon has 10 electrons. Adding $2 + 1$ gives 13. The element with atomic number 13 is Aluminum.

Answer:

1. Selenium
2. Krypton
3. Ytterbium
4. Vanadium
5. Antimony
6. Potassium
7. Erbium
8. Aluminum