(b) Which of these compounds or ions is an exception to the octet...

Question

(b) Which of these compounds or ions is an exception to the octet rule: borohydride (BH4-), borazine (B3N3H6, which is analogous
to benzene with alternating B and N in the ring), or boron
trichloride

Accepted Answer

well everyone. So in this video we're gonna take a look at different compounds and we're seeing which one has an exception for the octet rule. So, how we're gonna do this is by drawing out the lower structure and just visually see it. So first step is to calculate the total number advanced electrons. Let's start with this I cl compound. All right. So our iodine is going to give us seven vans electrons because it's a group 78 and our chlorine animal also give us seven vans electrons because it's also in Group 70. So 7-plus 7 is equal to 14 vance electrons. Alright now, drawing the Lewis structure, we just have two atoms. So the the iodine will be connected to the chlorine and fulfilling its octet will give each atom three lone pairs. So 12, there you go. 12 and three. So that's going to be our lowest structure for our first molecule. Then our second one here is going to be our b E C L two. So RB That's going to be two. Vans electrons sentencing her to a then we have our core in being in group 78 and we have two of those. So two times seven is equal to 14, giving us a sum of a total of 16 valence electrons. And during out the lewis structure then our B is going to be our central atom connected to two chlorine atoms. Let's fill fulfill it. The chlorine, Architect. And that's going to be equal to 16 valence electrons that we have used. So we have six electrons here. Six here. That's 12. And then we have two per bond. All right now moving on to the third compound, we have R n C L three so nitrogen is in group 58. That's going to give us five valence electrons. And for chlorine we have three florins each given seven valence electrons. That's three times seven is 21 doing the math here adding five and 21. We get a total of 26 valence electrons. Now drawing up the lowest structure, Our nitrogen will be our central atom connected to three Florin Adams. And of course we have to fulfill its octet here for the chlorine. Alright, so each coin item gets three lone pairs and the final two valence electrons that we have calculated from here will go on to our nitrogen. So now that we have drawn out all of our lower structures, we can see here that everything looks correct. The iodine and the pouring has its octet. So this is good. And moving on to this here, we can see the corn is very happy. It has its octet. However, the central item does not, it only has 1234 electrons. Now this one an exception. So we just put a little X here And last CRNCL three. Our central item is very happy. It has its full octet as well as each of its chlorine atoms. So this one gets a checkmark. So the question is asking us which compound or ion has an exception to the octet rule, and it's going to be our BCL two compound. Let's go ahead and highlight that. And this is going to be our final answer for the problem. Thank you all so much for watching.

(b) Which of these compounds or ions is an exception to the octet rule: borohydride (BH4-), borazine (B3N3H6, which is analogous to benzene with alternating B and N in the ring), or boron trichloride

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