[allforza2]

I had a homework assignment to find the most probable way that certain elements would get 8 valence electrons through bonding. My teacher said that gallium could lose its outer 3 electrons to become stable. But isn't 8 electrons the most stable configuration? The next lower shell has 18. Wouldn't gaining 5 or Covalent Bonding be more likely?

[Dan]

But isn't 8 electrons the most stable configuration? The next lower shell has 18. Wouldn't gaining 5 or Covalent Bonding be more likely?

Though still a simplification, it is better to say that the ion will be most stable with a noble gas configuration - i.e. full outer shell. What constitutes a full outer ahell? Do you think it is always 8 electrons - what's better, Li⁺ (2 electrons) or Li^7- (8 electrons)?

[allforza2]

I understand that 2 electrons in the first shell make the ion stable. It takes less energy to take away one atom compared to adding 7. Like how Hydrogen is found usually in a diatomic form. so basically a full shell is stable. The electron count doesn't matter just that it's a full shell?

[Dan]

The electron count doesn't matter just that it's a full shell?

Generally, the key is a full shell. It's not that the electron count is not important, it's that the electron count will correspond to a full shell, and the electron count of a full shell is not always 8 - it could be 2, 8, 18 or 32.

Be aware, this simplified picture does not hold for the transition metals and the heavier elements.