Consider the following XF 4 ions: PF 4 - , BrF 4 - , ClF 4 + , and AlF 4 - . (a) Which of the ions have more than an octet of electrons around the central atom?

welcome back everyone in this example, we need to identify which of the following X A three minus ions have an expanded octet with more than eight electrons on the central atom. So we have the ions where we begin with nitrate, we have carbonate, we have 48 and then we have chlorate. So we need to begin by drawing out our lewis structures of each of these ions to determine how they bond in structures. So beginning with our nitrate which we have n 03 minus. We want to first calculate total valence electrons before we can draw its lewis structure. And so we're going to look at every atom in this compound. We have nitrogen and oxygen where we have three oxygen atoms. So we'll say times three and then let's go ahead and begin with nitrogen. Where on our product tables we find it in group five A and recall that group number corresponds to valence electrons. So we would say are one nitrogen atom is multiplied by five valence electrons. Again, the three here is our three atoms of oxygen from our formula here. In in the subscript we can see and on our proud of cables when we find oxygen, we see that it's in group six A corresponding to six valence electrons. So we multiply by six valence electrons here. So we want to add these together and then also recognize that we have this minus one and ion charge. And recall that a negative charge means we gain one electron. So we would say plus one electron. So taking the sum here, We have three oxygen atoms times six valence electrons which will give us 18 electrons added to five. That will give us 23 and then adding this one electron here, we will have a total of 24 valence electrons total. So this is what we will use for our structure. And according to the formula we have nitrogen in the center surrounded by three oxygen atoms, so 12 and three. And we're going to begin by connecting all of these atoms with sigma bonds. So we call that sigma bonds are just another word for saying single bonds. And recall that one sigma bond represents two electrons. We have 246 electrons that we've used up so far. So we would subtract six electrons that we've used And that leaves us with 18 valence electrons left for our structure. The next thought we want to have is of any bonding preferences and the octet of our central atom. And so right now nitrogen does not have a complete octet. It has three bonds with a total of six electrons surrounding itself shared in covalin bonds here and we want to make this more stable. So recalling upon oxygen's bonding preference which is to have two bonds and two lone pairs around itself, we're going to add a second bond here between this nitrogen and the top oxygen atom which is going to one fulfill the octet for nitrogen because it now has 2468 electrons shared in co valent bonds. And when we add in our lone pairs on the oxygen, which is two sets of lone pairs, we will make this oxygen stable and happy because it's bonded according to its bonding preference. And so therefore it has a formal charge of zero. And so in adding these details, we've used a total of 24 and then six more electrons. So we're going to subtract that so we can keep track of what we've used so far, Leaving us with 12 electrons to continue to use in our structure. Now, as we stated, nitrogen octet is full. It cannot form anymore. Covalin bonds with these central or with the outer atoms which are our two bottom oxygen atoms here. And so we can't form an expanded octet, particularly because nitrogen is not able to violate the octet rule, nitrogen is only a period two element. Recall that to have an expanded octet, you have to be Between period three and below of our periodic table. So that means that we have no choice but to add in the rest of our electrons as lone pairs on our bottom oxygen atoms. So we would have 2468, 10 and then 12. So adding these lone pairs gives us formal charges of now minus one on both of these oxygen atoms here. And that is because we should recognize that again according to our periodic tables because oxygen is in group six A. It prefers to have six valence electrons. But in this case we have a total of 1234567 valence electrons. Where one is being shared in a co valent bond with the central atom nitrogen which is one more than what oxygen prefers to have, giving us that formal charge of minus one on both of these atoms and using up the last of our electrons to complete our structure. Now this is our complete lowest structure. And as we can see, we don't have an expanded octet on oxygen or sorry, on nitrogen. So we'll say no expanded on because nitrogen cannot violate the octet rule because it's only period too on the periodic table. So let's move on to our next structure which is for the carbonate, an ion. So we have C 032 minus for carbonate. And we want to follow the same process which is calculating total valence electrons first. We have one carbon atom and three oxygen atoms recognize that on our product table, carbon is located in group four A. So we would recall that that corresponds to four valence electrons. And then oxygen again as we stated as a group six A corresponding to six valence electrons. Next we have this minus to an ion charge corresponding to a gain of two electrons. And so taking the sum here between everything, we have three times six. Again giving us 18 plus four giving us 22 then plus the two from the anti on charge gives us a total of 24 valence electrons total for our structure. According to our formula, we have carbon in the center surrounded by three oxygen atoms. We're going to begin by making our base connections so 12 and three. And now we want to and sorry let's make this oxygen leader. Now we want to recall again upon bonding preferences, carbon has a bonding preference to have four bonds to be stable and complete the octet rule where as oxygen has a bonding preference of having two bonds and two lone pairs on itself to be stable because it prefers to have six valence electrons total. So we would like before make a second bond between the top oxygen and the middle carbon here where we would fill in too long pairs on this oxygen giving it a formal charge of zero because it has a total of six valence electrons where two are being shared co violently with carbon and two are on the oxygen as lone pairs. So this also keeps our carbon atom happy because it has a full octet where we have 2468 valence electrons being shared. Or sorry, eight total electrons being shared in covalin bonds with the other atoms in the molecule. Now we just have to recognize that because carbon is happy. And also because carbon is also a period two element. It will not have an expanded octet. which leaves us with the only option of adding in the rest of our electrons as lone pairs on the bottom two oxygen atoms. So just like before we have three sets of lone pairs on these bottom oxygen atoms giving the so we can classify that this carbon does not have an expanded octet, Carbon cannot violate the octet rule. And again, that is due to the fact that we recognize that it's a period two element on the periodic table. So following the same process for our next ion, given we have our berate ion where we have B minus calculating total valence electrons. First, we recognize we have one braun atom and three oxygen atoms On our periodic tables. We would find boron located in group three a on the periodic table corresponding to three valence electrons. Whereas oxygen is again in group six A corresponding to six valence electrons giving us six times three atoms which is 18 electrons total. And then we have a three minus an ion charge which we recall means we gain three electrons. Now taking the sum of everything, we have 18 electrons plus three plus another three gives us a total of 24 valence electrons total for our lower structure. Yet again, boron is in the center surrounded by three oxygen atoms. We make our base connections to these oxygen atoms and the boron using up a total of six of our 24 valence electrons. Now, as we stated in our calculation of total valence electrons boron is in group three A on the periodic table and it has three valence electrons. Right now, boron, our central atom born, our central atom has three valence electrons connecting it to our three oxygen atoms. Where born has 11 valence electron of itself contributed to each of these three sigma bonds here. And so we have a total of 123 of its total three valence electrons being shared in a bond. So this bron is stable and happy. And Also because it's in group two or sorry, Period two on our periodic tables, we cannot use boron as having an expanded octet. It must be period three or below. And so therefore we can't add any other bonds to our central atom between our outer atoms. So we're going to have to use the rest of our electrons as lone pairs on each of the oxygen atoms. So right now with these three bonds, we have a total of six valence electrons used up and that's going to leave us with a total of 18 valence electrons left, which we'll use as lone pairs on our oxygen atoms. So we have a total of three lone pairs on each oxygen atom. So we have 2468, 10, 12, 14, 16 and 18. And because these oxygen atoms now have a total of 1234567 valence electrons around themselves where they would prefer to have a total of only six. These oxygen's are not as happy so they're a little bit unstable and have a net formal charge of -1 each. And this would complete our structure for this board an ion. And we can confirm that because this is not an expanded octet due to the fact that boron is a period to element on the periodic table, so it cannot violate the octet rule. So moving on to our last ion given we have the chlorate an ion, so that would be C. And sorry about that. We have C L 03 minus calculating total valence electrons. First we have one chlorine atom and again three oxygen atoms, recognizing that on our periodic table, chlorine is located in group seven A corresponding to seven valence electrons. And oxygen, as we stated earlier is located in group six A corresponding to six valence electrons. So that gives us six valence electrons times three atoms for a total of 18 valence electrons. And according to our minus one charge, we recall that we would gain one electron based on that minus one and ion charge. So taking the sum of everything, we would have 18 plus seven giving us 25 plus that one extra valence electron giving us a total of 26 valence electrons total for our lewis structure. According to our formula, we have chlorine in the center surrounded by three oxygen atoms Making our base connections. We can make one bond to each of the or from each of the outer atoms to the central atom chlorine using up a total of six of our valence electrons. Leaving us with 20 left Now, we want to focus on bonding preferences and whether we have a full octet on our central atom and right now with only three bonds around itself, chlorine does not have a full octet. So we refer to the periodic table to see if chlorine is in period three or below. And we would see that chlorine Is a period three element. And so therefore it can violate the octet rule. So it can have an expanded octet. And so as we stated, chlorine should have seven valence electrons because it's located in Group seven A of our periodic table. Right now, it only has a total of 123. So we need to create more bonds from the central atom to the outer atoms to give it a total of seven valence electrons so that our chlorine is stable. So we can add a total of two pi bonds or double bonds to two of our oxygen atoms. So we'll do another bond here and then another pi bond here. That's going to give us a total of 12345 directly attached valence electrons to our chlorine atoms. So it needs to more and because adding a bond here would only give it a total of six and we need seven. We're going to choose to add a lone pair on our central atom chlorine, which will give it a total of now. 1234567 directly attached valence electrons in our structure of our chlorate an ion. So as we've added everything, we can count a total of 2468, 10, 12. Total of our valence electrons use. So we have 20 minus 12. And that difference gives us of eight valence electrons left to use in our structure which we can just add as lone pairs on our oxygen atoms. So we would go a ahead and add a total of two lone pairs on this top oxygen atom so that it's stable and bonded according to its bonding preference, which is to have two bonds and two lone pairs. So it has a net formal charge of zero. Moving on to this oxygen atom here, we can also add two more valence electrons here. So we have used a total of um and sorry, this was a mistake in my behalf of my calculations here. So we're going to subtract a total of 12 or sorry, 2468, 10, 12 of our valence electrons used initially Leaving us with 14 valence electrons left where so far we've added a total of 2468 And we're going to keep going as lone pairs on the third oxygen atom where we have 10, 12 and then 14 of our valence electrons used. So we would have all of our valence electrons used, leaving us with none left completing our structure, recognizing that this third oxygen atom over here has a total of 1234567 directly attached valence electrons, which is one more than I would prefer to have because oxygen again has six valence electrons when it's happy and stable. So this has a formal charge of minus one and all of the other atoms in the structure have a formal charge of zero because they're bonded according to their preference or able to have an expanded octet in the case of chlorine and our answer was to choose the atom that can have an expanded octet or sorry, the ion. And so we would confirm that the correct answer choice to complete this example is going to be our chlorate, an ion that is able to have an expanded octet. So chlorate is our final answer to complete this example. If you have any questions, leave them down below and I'll see everyone in the next practice video

Ch.9 - Molecular Geometry and Bonding Theories

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Brown 14th Edition

Ch.9 - Molecular Geometry and Bonding Theories

Problem 87a

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Chapter 9, Problem 87a

Consider the following XF₄ ions: PF₄^-, BrF₄^-, ClF₄⁺, and AlF₄^-. (a) Which of the ions have more than an octet of electrons around the central atom?

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Determine the number of valence electrons for each central atom: P, Br, Cl, and Al.

Add or subtract electrons based on the charge of the ion: add one electron for negative charges and subtract one for positive charges.

Calculate the total number of electrons around the central atom by adding the electrons from the fluorine atoms (each F contributes one electron to the central atom).

Check if the total number of electrons around the central atom exceeds 8, which would indicate more than an octet.

Identify which ions have central atoms with more than 8 electrons, indicating an expanded octet.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Octet Rule

The octet rule is a chemical principle that states atoms tend to bond in such a way that they have eight electrons in their valence shell, achieving a stable electron configuration similar to that of noble gases. This rule is particularly relevant for main group elements, which typically form bonds to fulfill this requirement. However, some elements, especially those in the third period and beyond, can expand their octet due to available d-orbitals.

Expanded Octet

An expanded octet occurs when an atom has more than eight electrons in its valence shell. This phenomenon is common in elements from the third period of the periodic table and beyond, such as phosphorus, sulfur, and chlorine, which can utilize their d-orbitals to accommodate additional electrons. Understanding which elements can expand their octet is crucial for predicting the bonding and structure of certain molecules and ions.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They are essential for visualizing the arrangement of electrons and determining whether an atom has achieved an octet or an expanded octet. By drawing Lewis structures for the given ions, one can easily identify the number of electrons surrounding the central atom and assess compliance with the octet rule.