Jump to content


  • Content Count

  • Joined

  • Last visited

Community Reputation

-9 Poor

About XVV

  • Rank

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

  1. I feel no joy by coming here. My question is not ill posted. Your mind is c*m filled. It is the right answer In the bracket beside it, I wrote (enters).
  2. This answer is not the best but it is still understandable. It helps. I have got no issues with you. What I meant was ''no person online on this website can give my answer''.
  3. Energy isn't always released when an electron is added to an atom. It depends on the kind of atom you are adding the electron to. Energy is released if the electronegative atom attains a more stable state by accepting the electron (by say attaining a octet configuration in the valence shell). Stable states have less energy when compared to other states and this difference in energy is released when an atom accepts an electron. On the other hand, it actually requires energy to add an electron to an electropositive atom which has an extra shell of electrons which makes it unstable. To add another electron we would have to overcome the repulsions due the already present electrons and hence energy would have to be supplied rather than it being released. If by how an electron entere you mean that how does an atom gain or lose electrons,its to complete its octet configuration.For example when Mg reacts with o2 As O has 2 less electrons to reach its octet whilst Mg has 2 extra electrons.Mg gives its electrons to O for both of them to complete their octet.This is known as an Ionic bond.Hope it helps This answer I got from another website. Which is far better than yours.
  4. First of all, I am not a drama queen. I didn't repost the question on all websites simultaneously. I just posted this question on 3 websites in a span of a week bcz of not getting answers and then I posted on this website. Luckily, I suddenly got understandable answers from all those websites just today. Anyway I just now know that no one can give answers on this website. your reply proves that I was right about saying that there are more stupid insults instead of good answers.
  5. No you are wrong. Only you are not understanding the question. I have asked this question on 3 more websites. They understood the question and nicely explained the answer instead of just finding faults in the questions if they can't figure out its answer.
  6. I know all this. But no one is answering my question accordingly.
  7. I am asking of Electron Affinity, not ionization energy. How does an electron add up(enters) in the valance shell of an atom? Why is energy released when an electron adds up in the valance shell of an isolated atom. This is what I am asking. Can someone please answer accordingly?
  8. XVV

    True vacuum

    Yeah, I am new to this website. This website is administrated by some clowns who think they are 'out of the world' or something by saying some stupid jokes that they think of as insults . I have asked some questions but there were more stupid insults and jokes than answers. I support you on this topic.
  9. How does an electron add up(enters) in the valance shell of an atom? Why is energy released when an electron adds up in the valance shell of an isolated atom.
  10. Can someone give an optimum interpretation of my question?
  11. yes beside that one yes this does help but I don't understand that for example, in s = v.t why does v+v+v+v..........t does not make a sense
  12. Can you give an example with a figure?
  13. What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, s=v.t? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean?
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.