What is the electron configuration for helium (He)? 1s¹

Question

What is the electron configuration for helium (He)? 1s¹ 1s² 1s²2s¹ 1s²2s²

Accepted Answer

welcome back everyone in this example, we need to identify our electron configuration of the atom zirconium. So we want to recall zirconium position on our periodic table. We see that it corresponds to the atomic number which we recall is represented by the symbol Z equal to 40. And that is also located across period five in Group four B. Which we should recognize as our transition metal D block of our periodic tables. Because we recognize that we have a neutral atom of zirconium given from the prompt. We would say that therefore we have 40 protons and electrons for our atom of zirconium. And we should recall that we're going to be distributing these electrons in our atomic orbital's to make up our configuration of zirconium. But before we write out that configuration, we want to recall that our s orbital which is lowest in energy has a total of just one orbital and can hold a maximum of two electrons. Moving on up in energy. We have our p orbital's which we should recall consists of three orbital's which can hold a maximum of six electrons. Then up in energy we have our d orbital's Which we should recall has a total of five orbital's which can hold a maximum of electrons. And then lastly, that leaves us with our f orbital's Which we should recall consists of a total of seven orbital's which can hold a maximum of 14 electrons. So we're going to start at atomic number one on our periodic table to begin our configuration where hydrogen is located and we would fully fill in our s block, which begins at the first energy level. So we have beginning our configuration one S. And then we recall that the S block holds a maximum of two electrons moving forward. In our configuration, we would move up to the second period of our periodic table where we hit the S. Block, which we would also fill in another two electrons moving further in our configuration. We're going to get to the second period of R. P block, which we recall holds a maximum of six electrons moving forward in our configuration. We would then go down to the third period of our periodic table where we hit our S block which can again hold a maximum of two electrons moving forward in our configuration, we would then hit the third period of R. P block, which we recall again holds a maximum of six electrons moving forward. In our configuration, we go to the fourth period of our periodic table, hitting our S block, which can hold a maximum of those two electrons. Then moving forward. We finally hit our D block, which we recall at the fourth period of our periodic table, begins at the third energy level, meaning we would have three D. We want to fully fill in our D block so that we have it's 10 electrons filled in. Moving on up in energy, we get to the fourth period of our P block which we recall hold a maximum of six electrons moving down or moving up in energy. We then get to the fifth period of R. S. Block which again holds a maximum of two electrons. And as we stated, zirconium is located in our D. Block and it has atomic number 40. So at five S. Two we would realize that we are at our Adam strontium which corresponds to atomic number 38. So at the fifth period of our periodic tables, we hit our D block, which we should recall begins at the fourth energy level. So we would have four D. And once we're in the D block, we recognize that the zirconium is only a count of two units in our D block, meaning that we would only fill in for two electrons to get to our atom zirconium. And this again would give us our configuration for our zirconium 40. And so this entire configuration here would be our final answer to make up our electron configuration of our neutral atom of zirconium. So this completes this example. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

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