Write an equation for the first electron affinity of helium. Would you predict a positive or a negative energy value for this process? Is it possible to directly measure the first electron affinity of helium?

Video transcript

welcome back everyone in this example. We need to write the equation for the second electron affinity of bromine. And we need to identify whether this process will have a negative or positive energy value. And can the second electron affinity of romaine be directly measured. So what we're going to begin with is writing out our electron affinities. So we should begin with our adam bromine which we should recognize is in group seven A. So it's a halogen recall that that is the halogen group and we should recall that it's definitely going to be a gas since it's a halogen. So recall that for the first electron affinity we're going to be gaining an electron here. Just one electron which will form brahman as an an ion. Because recall that when we have extra electrons than our stable normal atom here, We would have a charge now. So we have a negative charge when we have more electrons than needed. So because we just added one electron, we just have a -1 charge. We still have roaming in a gas form. And this is going to be our first electron affinity. Reform mean as an an eye on So now because we need to give the equation for the second electron affinity. We're going to begin with what we produced from our first electron affinity. So that was B R minus, We would be gaining a second electron here. And this is going to give us a -2 charge now since we now added a second electron. And so this is our second electron affinity equation. So as we highlighted above this is going to be brimming in its most stable form because it's neutral As a halogen here with just 35 electrons. Because we should recall that the atomic number for Romain is 35. However, with the first electron affinity we now have 36 electrons. And with the second electron affinity we now have a total of electrons because we added to our total. And so when we gain one electron we would say that it's relatively stable because we should recognize that because bro mean when it's neutral is in group seven A. When it forms the B. R minus an eye on it moves to group eight A. Which we should recall is our noble gas group. And because it's in the noble gas group with the -1 charge, it's going to be relatively stable. But when we get to our second electron affinity, we would say that it's least stable because we will no longer be in group eight a. So we're not going to be in the noble gas group nor will we be in our halogen group because we're definitely not neutral. We're only a minus two and I in charge here. So we would recognize that with a minus two and I in charge, we would move to group one a. And we would need energy to be raised to force this process Of creating this -2 Roman an ion. And that is why this is the least stable form of bromine to exist in the second electron affinity of romaine. And so we would say that for our final answers we have our equation highlighted for the second electron affinity of bromine. This will require a positive value of energy because this process must be forced because this is the least stable form of bromine. And because this is the least stable and energy is required, we would say that therefore The 2nd electron affinity of brahman cannot be measured or it cannot be directly measured because energy is needed to force this process. So this would be and sorry about that. This would be our third highlighted answer here. So everything highlighted in yellow represents our final answer, which will correspond to choice C as the correct choice in our multiple choice options. So I hope everything. Our view was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.