Question

question 6. how many electrons must be gained by each of the following atoms to achieve a stable electron configuration?
a. chlorine (cl).
b. nitrogen (n).
c. oxygen (o).
d. iodine (i).
e. sulfur (s).
f. phosphorus (p).

Explanation:

Step1: Recall stable electron configuration (noble gas, octet rule: 8 valence electrons, except H/He).

For Chlorine (Cl):

• Atomic number 17. Electron config: \(1s^2 2s^2 2p^6 3s^2 3p^5\). Valence electrons: 7 (3s²3p⁵). To reach 8 (octet), needs \(8 - 7 = 1\) electron.

Step2: For Nitrogen (N):

• Atomic number 7. Electron config: \(1s^2 2s^2 2p^3\). Valence electrons: 5 (2s²2p³). Needs \(8 - 5 = 3\) electrons.

Step3: For Oxygen (O):

• Atomic number 8. Electron config: \(1s^2 2s^2 2p^4\). Valence electrons: 6 (2s²2p⁴). Needs \(8 - 6 = 2\) electrons.

Step4: For Iodine (I):

• Iodine is in Group 17 (halogens), same as Cl. Electron config ends with \(ns^2 np^5\). Valence electrons: 7. Needs \(8 - 7 = 1\) electron.

Step5: For Sulfur (S):

• Atomic number 16. Electron config: \(1s^2 2s^2 2p^6 3s^2 3p^4\). Valence electrons: 6 (3s²3p⁴). Needs \(8 - 6 = 2\) electrons.

Step6: For Phosphorus (P):

• Atomic number 15. Electron config: \(1s^2 2s^2 2p^6 3s^2 3p^3\). Valence electrons: 5 (3s²3p³). Needs \(8 - 5 = 3\) electrons.

Answer:

a. Chlorine (Cl): 1
b. Nitrogen (N): 3
c. Oxygen (O): 2
d. Iodine (I): 1
e. Sulfur (S): 2
f. Phosphorus (P): 3