Question

in the space below, write the longhand electron configurations of the following elements:

1. oxygen
2. iodine
3. argon

Explanation:

1) Oxygen

Step1: Determine atomic number

Oxygen has atomic number 8, so 8 electrons.

Step2: Fill orbitals (1s, 2s, 2p)

1s holds 2, 2s holds 2, remaining 4 in 2p.
Configuration: $1s^2 2s^2 2p^4$

Step1: Atomic number of Iodine

Iodine (I) has atomic number 53.

Step2: Use noble gas (Kr) for shorthand base, then fill remaining

Kr is $[Kr] = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$. Remaining electrons: 53 - 36 = 17. Fill 5s, 4d, 5p: 5s², 4d¹⁰, 5p⁵.
Longhand: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^5$

Step1: Atomic number of Argon

Argon (Ar) has atomic number 18.

Step2: Fill orbitals (1s, 2s, 2p, 3s, 3p)

1s², 2s², 2p⁶, 3s², 3p⁶ (sum: 2+2+6+2+6=18).
Configuration: $1s^2 2s^2 2p^6 3s^2 3p^6$

Answer:

$1s^2 2s^2 2p^4$

2) Iodine