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Author Topic: Ionic Bonding Question  (Read 1034 times)

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allforza2

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Ionic Bonding Question
« on: December 09, 2012, 04:59:20 PM »

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?
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Dan

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Re: Ionic Bonding Question
« Reply #1 on: December 09, 2012, 09:18:06 PM »

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 Li7- (8 electrons)?
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allforza2

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Re: Ionic Bonding Question
« Reply #2 on: December 10, 2012, 01:01:18 PM »

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?
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Dan

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Re: Ionic Bonding Question
« Reply #3 on: December 10, 2012, 09:24:05 PM »

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.
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