-2
$\begingroup$

New to chemistry so forgive the fundamental question. What exactly determines the element's charge? Oxygen has 2- and hydrogen has 1+. Oxygen wants to receive 2 electrons to complete a shell when hydrogen wants to donate 1 electron. For example why wouldn't hydrogen want to gain 1 to complete its shell? What exactly is the methodology to this of any given element (non transitional obviously). I understand that the groups of the periodic can be used and memorized but whats the actual scientific theory as to why a given element donates rather than accepts and vise versa.

$\endgroup$

closed as too broad by Jan, Jon Custer, Klaus-Dieter Warzecha, Todd Minehardt, jerepierre Jun 6 '16 at 17:06

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ There's no such thing as "element's charge" and saying that hydrogen wants to get rid of electrons is nonsense. $\endgroup$ – Mithoron Jun 6 '16 at 12:41
  • $\begingroup$ @Mithoron then when we say oxygen is 2- what is that referring to? Its ionic form? Can u better explain? $\endgroup$ – Atticus283blink Jun 6 '16 at 18:18
  • $\begingroup$ You can say for example that Na2O consists of sodium cations and oxygen anions, but it's simplification. $\endgroup$ – Mithoron Jun 6 '16 at 21:55
1
$\begingroup$

Since you are new to chemistry, I assume the concept you are searching for is electronegativity.

Each element has an electronegativity, which in short is its tendency to attract electrons. Hence, since oxygen has electronegativity 3.4 and carbon has electronegativity 2.6, the $\ce{CO2}$ molecule can be seen as one $\ce{C^{4+}}$ and two $\ce{O^{2-}}$.

Of course this is a gross oversimplification and not actually how the $\ce{CO2}$ molecule works, so don't get attached to this view as you learn more chemistry.

To go forward, I suggest you check the different kinds of chemical bonds.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.