How many valence electrons does lithium (II) have available fo...

Question

How many valence electrons does lithium (II) have available for bonding?1234

Accepted Answer

welcome back everyone in this example, we have the element vanadium. We need to identify its electron configuration and its number of valence electrons in a neutral atom of vanadium. So we want to recall that for neutral atoms only our number of protons are always equivalent to our number of electrons. And we can also recognize that our atomic number of our atom tells us our number of protons, so they're equal to one another. So if we find vanadium on our periodic table represented by the symbol V, we see that it's located in our transition metal D block in Group five B Where it corresponds to the Atomic # 23. So recall that atomic number is also represented by the symbol Z. So we would say that Z is equal to 23 and therefore we would have 23 protons for our atom of Vanadium Vanadium. And because this is a neutral atom according to our prompt, we know that We have also an equivalent of 23 electrons. And so this is what we will use to fill in our orbital's for our electron configuration. Before we get into the configuration, we want to remember that we have precisely three orbital's involved where we're going to be filling in these 23 electrons for our electron configuration. And the three orbital's would be our s orbital, which is our smallest orbital and only consist of one orbital. We next have our p orbital to fill in for which we should recall have a total of three orbital's and then we have our D orbital because as we stated, vanadium is in the D block of our periodic table. The block of P. T. And so recall. The D block has a total of five orbital's that you would fill in electrons for now because each orbital can hold a maximum of two electrons per orbital. We would say that the s orbital holds a maximum of two electrons Whereas we would count a total of 246 electrons for the P orbital max And then we can fill in a maximum of 246,810 electrons for our deep orbital's. And this is what these orbital's hold. So these hold a maximum of 26 and 10 electrons that we can fill in for the configuration of vanadium. So we want to refer to our periodic tables. Starting out with hydrogen in group one A or atomic number or at atomic number one. We would pass from hydrogen to helium which is our first period of our periodic table and would cover our first orbital at the first energy level being our one S two orbital. And we're saying to because we want to fully fill in this orbital with two electrons, the maximum that it can hold so that we can continue on in our configuration to vanadium. So moving forward, we would go to the second period or second energy level of our periodic table where we would go to the to s to orbital that we're filling in to pass from lithium to beryllium, moving into our P block at the second period of our periodic table, we would go through the two P. And then in the exponent we should have six because we want to fully fill in from boron to neon to get through our configuration to our adam vanadium where we go into the third period of our periodic table at the third energy level where we fill in for the S orbital, its maximum of two electrons moving through our third period of our periodic table will take us into our P block directly. So we would have three P where we fill in a maximum of its six electrons in each of the three orbital's. They're taking us from aluminum all the way through argon, moving into our fourth period of our periodic table. We're at the fourth energy level. So sorry, we have four s as our next orbital where we fill in its two electrons to take us from potassium to calcium. And then finally we hit our P block, which begins at the third energy level, recall at the fourth period of our periodic table. And so we want to continue in our D block until we hit our adam vanadium, which is a count of three. And so because we know that our D block begins at the third energy level of our fourth period on the periodic table, we would have three D. And we would count for a total of three units to hit our adam vanadium meaning we're only going to fill in three electrons in our D orbital's. And so our D orbital for vanadium specifically is going to be filled in with two electrons of opposite spins in the first orbital. And then we have a single electron in the second orbital of our D block. And it can be in any direction spin looking up for our three electrons that we filled in. So this is going to be our extended configuration of vanadium. But if we want to make a shortcut for ourselves, we're going to recognize that we want to find the noble gas that proceeds vanadium on the periodic table and that would be are gone. And so we can say that all of our electrons covering from one us all the way up to three P six where argon is located at on our periodic tables would cover our configuration corresponding to the noble gas argon. Where we can say we have argon in brackets and then we have for us to three D three. As also another way to express our configuration for neutral vanadium. So now that we have our configuration, we need to determine our number of valence electrons. So we want to recognize that valence electrons means our outer most, sorry outermost electrons. Those are our valence electrons, meaning they should be in the highest energy levels or outermost energy levels. So everything in the lower energy levels would be our core electrons That we filled in our configuration, whereas everything in the higher energy levels being our for us and 3D orbital's are going to be our valence electrons. And so adding these together are exponents here. We would have two plus three, which would give us a total of five valence electrons. And so for our two final answers, we have our configuration for vanadium and we have its five valence electrons that we determined from the configuration. So everything highlighted in yellow are our two final answers. If you have any questions, leave them down below and I'll see everyone in the next practice video.

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How many valence electrons does lithium (II) have available for bonding?1234

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