Question

1. mg

e⁻ = __ p⁺ = n⁰ = ____
electron configuration:
no. of shells = ____ valence e⁻ = ____

1. mg²⁺

e⁻ = __ p⁺ = n⁰ = ____
electron configuration
no. of shells = ____ valence e⁻ = ____

Explanation:

For 3. Mg:

Step1: Determine protons, electrons, neutrons

Magnesium (Mg) has atomic number 12, so \( p^+ = 12 \). Neutral atom, so \( e^- = 12 \). Atomic mass ~24, so \( n^0 = 24 - 12 = 12 \).

Step2: Electron configuration

Electron configuration: \( 1s^2 2s^2 2p^6 3s^2 \).

Step3: Shells and valence electrons

Number of shells: 3 (n=1,2,3). Valence electrons: 2 (from \( 3s^2 \)).

Step1: Determine protons, electrons, neutrons

Protons (\( p^+ \)) same as Mg: 12. \( \text{Mg}^{2+} \) loses 2 electrons, so \( e^- = 12 - 2 = 10 \). Neutrons (\( n^0 \)): \( 24 - 12 = 12 \) (same as Mg).

Step2: Electron configuration

Electron configuration: \( 1s^2 2s^2 2p^6 \).

Step3: Shells and valence electrons

Number of shells: 2 (n=1,2). Valence electrons: 8 (stable octet in \( 2p^6 \)).

Answer:

\( e^- = 12 \), \( p^+ = 12 \), \( n^0 = 12 \)
Electron Configuration: \( 1s^2 2s^2 2p^6 3s^2 \)
No. of Shells = 3, Valence \( e^- = 2 \)

For 4. \( \text{Mg}^{2+} \):