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Home > Science > Chemistry   »   Ionizatiopn energy.

 
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Old Mar 15, 2007, 12:42 PM
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Ionizatiopn energy.

i know all about ionization energy so no need to explain to me.

The first six ionization energies of another element M are:

'a line graph showing low ionzation of the 2 outermost electrons. After that line appears to be going into a new subshell as the the next four electrons are considerably higher above in ionzation energy.'

Explain why M cannot have an atomic number less than 12.

-----considering the graph shows only the last 6 electrons energies.

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Old Mar 15, 2007, 01:05 PM   #2  
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I have no idea what you're going on about mate. Is there meant to be a picture?
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Old Mar 15, 2007, 02:13 PM   #3  
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there is a graph
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Old Mar 15, 2007, 02:25 PM   #4  
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M must have an atomic number greater than 12 for the following reason:

M must have two electrons in its outermost shell. This is indicated by the two outermost electrons having a smaller ionization energy.

For there to be two electrons there, the previous shells must all be full. The first two shells have two and eight electrons when full, respectively. Therefore, in order to have two electrons in the outermost shell, and at least six overall, there must be at least two full shells (10 electrons) plus the two electrons in the outermost, making the minimum total 12.
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Old Mar 15, 2007, 10:26 PM   #5  
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I am never able to see the graph when people attatch it in a certain way and i dont know why
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Old Mar 16, 2007, 04:07 AM   #6  
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I couldn't see the graph either, but I could visualize it from the description.
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Old Mar 17, 2007, 02:51 AM   #7  
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Nonsansa? That is actually what I wrote in the equation, but it's still an assumption. We are assuming that there are only 3 total shells, with 2 electrons in the first and last, and 8 electrons in the middle shell.

Where is the evidence to back up the answer?
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Old Mar 17, 2007, 10:22 AM   #8  
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No assumptions involved.

The difference in the energies can only be explained by having the two outermost electrons in a different shell than the four previous ones. Thus, the outermost shell must have exactly two.

For there to be any electrons in a shell, all previous shells must be full.

The first valence shell can two electrons. For there to be electrons in any other shell, it must have these two in the first shell.

The second shell can have up to eight electrons. For the third shell to have any, these eight must be present.

Therefore, in order to have two on the outermost shell, and greater than six total, you must have at least 12: 2 in the first shell, 8 in the second shell, and 2 in the third.

You can have more than 12, by filling up the outermost shell and adding two electrons to the next shell.
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Old Mar 17, 2007, 12:13 PM   #9  
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Quote:
Originally Posted by nosnosna
For there to be any electrons in a shell, all previous shells must be full.
That is not true.

It works out that way for the first three (or two, depending on how you look at it) shells but none after that.

For example...

Calcium has 2 valence electrons which are located in the 4th shell. The 3rd shell has 8 electrons in it which is what we consider a "full" outer shell but the 3rd level can actually hold 18 electrons and once you hit the D block starting with Scandium the 3rd level continues filling.

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Nosnosna agrees: I knew I'd forget something important :)
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Old Mar 17, 2007, 12:27 PM   #10  
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I stand corrected. I had forgotten that the d and f sub-levels behave differently.

My explanation does stand for why the atom in question must have an atomic number of 12 or greater.
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