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March 4, 2015

March 4, 2015

Posted by **Nicole** on Wednesday, November 30, 2011 at 4:23pm.

I'm really having trouble figuring this out!

- Chemistry -
**DrBob222**, Wednesday, November 30, 2011 at 6:52pmRemember these rules.

6p means n = 6 and p means l = 1.

m_{l}= -l to +l in increments of 1 and that includes zero. In practice all that means is that m_{l}can be -1, or 0, or +1.

Then remember that m_{s}can be +/- 1/2 for each m_{l}value.

- Chemistry -
**Nicole**, Wednesday, November 30, 2011 at 11:05pmI'm really bad at this, I still dont understand.

- Chemistry -
**DrBob222**, Wednesday, November 30, 2011 at 11:25pmThere isn't anything to understand. There are four (4) quantum numbers and they have rules to follow:

1. The value of n may be 1,2,3,4,5 or any whole number larger than 0 but may not be zero.

2. The value of l may be any whole number but may not be larger than n-1. Therefore, for n = 1, l may be zero. For n = 2, l may be 0 or 1. For n = 3, l may be 0, 1, or 2, etc. If l = 0 we call it an s electron. If l = 1 we call it a p electron. If l = 2 we call it a d electron and if l = 3 we call it an f electron.

3.The value of m_{l}may have values from -l to +l, all in whole numbers, including zero.

4. The value of m_{s}may be +/- 1/2.

Your question doesn't need all of this to answer it but I thought it might be useful to write ALL of the rules, then you apply these rules to the question.

6 is the n value.

p means l = 1

So we can have

n = 6

l = 1

m_{l}= -1

M_{s}+1/2 and -1/2 (Two electrons here with values of n, l, m the same and the only difference is m_{s}is +1/2 for 1 electron and -1/2 for the other.)

n = 6

l = 1

m_{l}= 0

m_{s}= +1/2 and -1/2 (Two more electrons here.)

n = 6

l = 1

m_{l}= +1

m_{s}= +1/2 and -1/2 (Two more electrons)

There are six electrons possible. That's all of the p electrons an atom can have for any given value of n

- Chemistry -
**Nicole**, Wednesday, November 30, 2011 at 11:32pmYeah, I figured it out after I said I didn't understand. It's actually really easy. Thanks for all your help though!

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