anonymous
  • anonymous
Hi, Can someone help me with the definitions of nuclear charge, orbital charge and overall charge, in relation to Beryllium?
Biology
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I am thinking that the orbital charge has to do with the number of electrons. Does the overall charge deal with the protons and neutrons?
aaronq
  • aaronq
Nuclear charge =Z= number of protons Overall charge is dependent on how many electrons it has lost or gained. (which would be called an ion, not an atom). I've never heard the term orbital charge, do you know what thats referring to?
anonymous
  • anonymous
Thanks!

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anonymous
  • anonymous
It is in my study guide put together by my teacher.
anonymous
  • anonymous
I will ask him about it.
anonymous
  • anonymous
The answer that is in the answer key shows it to be zero.
anonymous
  • anonymous
Never mind, I meant -4.
anonymous
  • anonymous
I figured it meant the electrons added up. -2+-2 = -4 ?
anonymous
  • anonymous
Since the electrons orbit, idk.
aaronq
  • aaronq
oh, i get what he means. Still never heard of it, but i guess it's an exercise for you to get acquainted with the number of electrons present in different species. (ps. electrons don't really "orbit" nuclei, analogously to planets orbiting the sun)
anonymous
  • anonymous
right... Can you set up a few equations to show me examples of how we are getting these numbers?
aaronq
  • aaronq
For nuclear charge, there is no equation, it's essentially counting how many protons an atom has (which it's given by it's identity, the atomic number). Unless you're dealing with effective nuclear charge \(Z_{eff}\), which is the actual charge experienced by a valence electron due to it's repulsion from core electrons. Overall charge just deals with electrons, if it gained it becomes negatively charged and viceversa, it becomes positive if it lost.
anonymous
  • anonymous
The nuclear charge of beryllium is 4 because it has 4 protons in the nucleus. The overall charge is 4 because there are 4 total electrons. The orbital charge is -4 because of the two electrons. I think I just confused myself. The answer that the book shows for beryllium is 0 and I am not sure why that is.
anonymous
  • anonymous
I can understand that beryllium will lose the two electrons is the second shell.
anonymous
  • anonymous
I mean 0 for the overall charge.
aaronq
  • aaronq
if beryllium is in it's elemental state (meaning equal number of protons to electrons) then it will be neutral (an overall charge of 0).
anonymous
  • anonymous
OK I see.
anonymous
  • anonymous
I want to ask you something in general. If there are not an equal number of protons and electrons then the overall charge wouldn't be 0? Give me some examples of this situation please.
anonymous
  • anonymous
Thanks for your patience.
aaronq
  • aaronq
no problem, really. okay, you have to know that atoms NEVER lose protons in chemical reactions (only in nuclear reactions). So, the only variable is the number of electrons. so if chlorine, (Z=17),(group 7) gains 1 electron (to achieve an octet) it becomes -1 anion, \(Cl^-\), the overall charge (called "formal charge") is written as an exponent. Potassium, K, (Z=19), (group 1) loses a proton (again, to achieve an octet) it becomes a 1+ cation, \(K^+\). What will be the charge on Mg, (group 2) if becomes an ion?
anonymous
  • anonymous
Gaining would mean taking a negative charge and losing would mean a positive charge. It depends on how many electrons as to the number of plus or minus. I would say +2 for Mg.
anonymous
  • anonymous
There are 8 electons in the second shell so it is stable.
anonymous
  • anonymous
As long as it gives up the two electrons.
aaronq
  • aaronq
Great. thats correct. You can generalize: group 1: always +1 group 2: always +2 group 6: always -2 group 7: always -1
anonymous
  • anonymous
It makes sense.
aaronq
  • aaronq
Good stuff, you should get a periodic table and study where the elements are, because they are positioned in such a way, that you can estimate what kind of ions atoms will be.
anonymous
  • anonymous
Good point. Come to think of it, I haven't seen one in all of the 5 books I am using. I do remember always seeing one in a chemistry or biology book, in the past classes I took. I also need to know the roman numerals.
aaronq
  • aaronq
yeah, roman numerals are important, at least up to 10 (X). you should google one and print it out. It comes in handy
anonymous
  • anonymous
I will, thanks! Take care.
aaronq
  • aaronq
no problem. thanks, you too.

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