Okay, so I'm new to chemistry and absolutely confused. Can anyone please explain to me how to solve the following?? Mg - ____ + 2e- ....Help would be very much appreciated!!

Question

Okay, so I'm new to chemistry and absolutely confused. Can anyone please explain to me how to solve the following?? Mg -> ____ + 2e- ....Help would be very much appreciated!!

anonymous · Answer

$$mg \rightarrow mg ^{2+} + 2 e$$

anonymous · Answer

thats awesome thank you sooooo much. I still have three more to figure out, could you explain how to do it?

anonymous · Answer

i'm sowry.. i don't know how to explain it...

anonymous · Answer

thats okay, thanks for the help you did give me though. :)

anonymous · Answer

we must make it neutral :) do u understand?

anonymous · Answer

well, that i get, im just not sure how to do so. For example: Ba -> Ba^(2+) + _______ ??? How would i get rid of the two? is that what yhou mean im supposed to do? sorry, chemistry is very confusing to me...

anonymous · Answer

ah.. how to explain it. if 2e = debt (-2) so, we must pay it by +2. just like that. but, we must write it on the right-top the element (Mg)

unklerhaukus · Answer

well, magnesium has 12 protons - by definition,
to be a neutral atom the positive charges of the protons is canceled out by negatively charged electrons , 
(-)elections and (+)protons have a charge that is equal in magnitude and opposite in sign,

$$ _{12}\text{Mg}^0 \longrightarrow \_\_ + 2e^- $$

$$ _{12}\text{Mg}^0 -2e^- = \_\_ $$

taking away electrons from magnesium will effect the overall charge 

                       $0-(-2)=2$

the nucleus of the atom will remain the same

anonymous · Answer

yup, u must pay it by ...... electron(s). to make it neutral :)

anonymous · Answer

So, would it be Ba^(2-) ?? Sorry for asking so many questions. I just want to be sure.

unklerhaukus · Answer

$$_{12}\text{Mg}^0 -2\times_0e^{-1} =~ _{12}\text{Mg}^{2+}$$

anonymous · Answer

Thanks everybody =D

anonymous · Answer

uwc :)