5

6) ?-pealyn ? Ha Ni - 72-penlya< NHa t na (ox_NaNHa)F-buku NBS9) /-melhylcyckhexen? + BHs 5 #un paoX 8nd bajen)aorna tc,...

Question

6) ?-pealyn ? Ha Ni - 72-penlya< NHa t na (ox_NaNHa)F-buku NBS9) /-melhylcyckhexen? + BHs 5 #un paoX 8nd bajen)aorna tc,

6) ?-pealyn ? Ha Ni - 7 2-penlya< NHa t na (ox_NaNHa) F-buku NBS 9) /-melhylcyckhexen? + BHs 5 #un paoX 8nd baje n)aorna tc,



Answers

$$ 6 r^{2} t+7 r t^{2}+t^{3} $$

So in problem with the three of chapter six compounds of P block elements to So in this question uh the oxidation state of xenon atom in the following compounds are respectively. So first is given Xia four. So let us consider the oxidation the oxidation state of xenon in this case is X. So the mm. So the flooring is definitely it is minus one. So uh this non comes out of B plus four by using the oxidation number rule this some of oxidation number of all the elements that is equal to zero. So zero on here is plus four. So now in the next case we have H. X. E. O. For minors. So let us see uh the non observation status X. So for hydrogen, it is one then the non four X for oxygen for atom and that is minus two for the oxidation state of oxide oxygen in this case, so that is a color minus one. So X comes out to be plus six. So here this is known is plus six. And uh finally, in case of any for X +06. Again, we can consider the oxidation state of the union is X. And uh the sodium is plus one here minus six. This is uh basically plus plus eight. Okay, so this is for item of sodium and this is the one is the oxidation state of sodium X 406 atom of oxygen and minus two for the oxidation state of oxygen. So the sum of oxidation number of all the element that is equal to zero. So X comes out to be plus gate, so therefore plus four plus six and plus four plus six plus eight are the oxidation state in the following xenon compounds. So option is the right answer.

This problem is asking you to use sigma tint notation to write the sun. So you have Sigmund imitation noted by your Sigma symbol. And if you look at this, you'll notice that what's changing here is the seven is staying constant even though you have this cap in here. And just so you know, that just means the pattern continues from 1 to 2. Everybody between going to six, going by integers. And so the seven stays the same and the finder staying the same. And so what's changing is this one over 62 over six, all the way up to six over six. So you pick an index and we'll use, um, I have a good one. So you're gonna go from 1 to 6, which represents your starting at one and ending at six. And so you would have seven i over six plus five as your method for writing the some using sigma notation

With potassium Fellows I night Sweet potassium Barbados I night with potassium pharaoh Sign I'd solution. See you two platforms. See You. two platforms. See you too. F E. CN six chocolate colored PPT. Chocolate Good lad PPT. So according to the option in this problem, Option C option C Eat correct. And said for this problem with potassium Ferdowsi night solution, see you two plus form. See you to Ephesians six chocolate colored PPT. So I hope you understand the solution of this problem.

In this problem, the yellow solution that is, I need to see a shadow for in pageants of as tours of four as well as as to will give orange solution we teach. I need to see our 247 and this is orange in color so this is orange solution. And according to the option option C. H, correct answer for this problem. Option C. Each correct and said for this problem, I hope you understand the reaction of this problem.


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