Question

section 3: ions and electron configurations

1. write the electron configuration (shorthand) for:

• mg²+:
• o²-:

Explanation:

Step1: Determine the electron configuration of Mg

Magnesium (Mg) has an atomic number of 12. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}\). When it forms \(Mg^{2 +}\), it loses 2 electrons from the 3s orbital.

Step2: Write the shorthand electron configuration of \(Mg^{2+}\)

The shorthand electron configuration of \(Mg^{2+}\) is \([Ne]\) (since neon (\(Ne\)) has an electron configuration of \(1s^{2}2s^{2}2p^{6}\) and \(Mg^{2+}\) has the same electron configuration as \(Ne\))

Step3: Determine the electron configuration of O

Oxygen (O) has an atomic number of 8. Its electron configuration is \(1s^{2}2s^{2}2p^{4}\). When it forms \(O^{2-}\), it gains 2 electrons. The electron configuration becomes \(1s^{2}2s^{2}2p^{6}\)

Step4: Write the shorthand electron configuration of \(O^{2-}\)

The shorthand electron configuration of \(O^{2-}\) is \([Ne]\) (because it has the same electron configuration as \(Ne\))

Answer:

• \(Mg^{2+}\): \([Ne]\)
• \(O^{2-}\): \([Ne]\)