A neutral atom has two electrons in its first energy shell, eight electrons in its second energy shell, and three electrons in its third energy shell. Which electrons will be lost when the atom reacts to become stable?
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A
Electrons in the first shell
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B
Electrons in the second shell
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C
Electrons in the third shell
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D
Electrons in the second or third shell
The electrons that will be lost are those in the third shell.
This electron configuration (2, 8, 3) is that of aluminum (Al). Chemical stability for main-group elements is often achieved by attaining a full valence shell, typically an octet, which resembles the electron configuration of a noble gas. For metals, this is most easily done by losing valence electrons.
A) Electrons in the first shell
The first shell (n=1) is full with its maximum of 2 electrons, achieving a stable, low-energy configuration identical to helium. These are core electrons, held very tightly by the nucleus due to their proximity and lack of shielding. Removing core electrons requires an enormous amount of energy and would destabilize the atom, not promote stability.
B) Electrons in the second shell
The second shell (n=2) in this atom has 8 electrons, which is a full octet (like neon). This is already a stable, low-energy configuration for that shell. Losing electrons from this complete shell would disrupt its stability, require significant energy input, and would not lead to a more favorable electron configuration.
C) Electrons in the third shell
The third shell is the valence shell, containing only 3 electrons. For a metal like aluminum, the most energetically favorable path to stability is to lose these three valence electrons. By losing them, the atom sheds its incomplete outer shell. What remains is the stable, full electron configuration of the second shell (2,8), which now becomes the outermost shell. This results in the formation of a cation with a +3 charge (Al³⁺), which is isoelectronic with the noble gas neon.
D) Electrons in the second or third shell
While it is theoretically possible to remove electrons from any shell, the principle of chemical reactivity is based on achieving stability with the least energy expenditure. Removing electrons from the already-stable, tightly bound second shell is highly unfavorable compared to removing the more loosely held valence electrons from the third shell. Atoms react via their valence electrons because those are the ones involved in bonding and energy changes that lead to stable arrangements.
Conclusion:
Chemical reactivity is driven by valence electrons—those in the outermost shell. The atom described achieves a stable noble gas electron configuration most efficiently by losing the three electrons in its incomplete third (valence) shell, thereby exposing the full, stable octet of the second shell.

Topic Flashcards
Click to FlipWhat is the total number of electrons in a neutral atom with the configuration 2, 8, 3, and what element is it?
13 electrons. The element is Aluminum (Al).
Which shell contains the valence electrons for the atom with configuration 2, 8, 3?
The third shell. Valence electrons are those in the outermost energy level.
When aluminum (2, 8, 3) reacts to become a stable ion, what noble gas electron configuration does it achieve?
The neon configuration (2, 8). It loses 3 electrons to become Al³⁺, which is isoelectronic with neon.
Why are electrons from the first or second shell NOT typically lost during chemical reactions for stability?
They are core electrons. The first and second shells are already full and stable. Removing these tightly held electrons requires immense energy and destabilizes the atom.
What is the charge on the ion formed when the atom with configuration 2, 8, 3 loses electrons to become stable?
+3. It loses its three valence electrons from the third shell.