Question

part d
al³+
express your answer in complete form, in order of increasing orbital. for example, 1s²2s² would be entered as 1s^22s^2.
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part e
what do all the electron configurations have in common?
all of the obtained electron configurations are isoelectronic with group 1a elements. noble gases. group 6a elements. group 7a elements. group 2a elements.

Explanation:

Step 1: Determine the number of electrons in \(Al^{3+}\)

Aluminum (\(Al\)) has an atomic number of \(13\), meaning it has \(13\) electrons in its neutral state. When it forms \(Al^{3+}\), it loses \(3\) electrons. So, \(13 - 3=10\) electrons.

Step 2: Write the electron configuration for \(10\) electrons

The electron configuration for \(10\) electrons follows the Aufbau principle. The order of filling orbitals is \(1s\), then \(2s\), then \(2p\).

• The \(1s\) orbital can hold \(2\) electrons: \(1s^{2}\)
• The \(2s\) orbital can hold \(2\) electrons: \(2s^{2}\)
• The \(2p\) orbital can hold \(6\) electrons: \(2p^{6}\)

Combining these, the electron configuration of \(Al^{3+}\) is \(1s^{2}2s^{2}2p^{6}\)

For part E:
Noble gases have full - filled valence shells. \(Al^{3+}\) (with electron configuration \(1s^{2}2s^{2}2p^{6}\)) is isoelectronic (has the same number of electrons) with neon (\(Ne\), which is a noble gas with electron configuration \(1s^{2}2s^{2}2p^{6}\))

Answer:

Part D: \(1s^{\wedge}22s^{\wedge}22p^{\wedge}6\)
Part E: noble gases.