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Question

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Answers

$\mathrm{FcCl}_{\text {j }}$ solurion addcd ro $\mathrm{K}_{i}\left[\mathrm{Fc}(\mathrm{CN})_{6}\right]$ givcs (a) while with KSCN gives (b). (a) and (b) respecrively arc (a) $\mathrm{Fe}_{1}\left[\mathrm{Fe}(\mathrm{CN})_{\mathrm{G}}\right]_{3}, \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{SCN})_{6}\right]$ (b) $\mathrm{\Gammae}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}, \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CNS})_{6}\right]$ (c) $\mathrm{Fc}_{3}\left[\mathrm{Fc}(\mathrm{CN})_{i}\right]_{2}, \mathrm{Fc}(\mathrm{CNS})_{3}$ (d) $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}, \mathrm{KFe}(\mathrm{CNS})_{3}$

So here, the question is which molecule have is have english tendency to form hydrogen born. So at I as at least, and nancy to form hydrogen born, it's born has won. It is because electronic city, because of low electronic nativity is because of low electronic nativity of I only in I and its side is also large, and I eyes off. Oh, option D is correct. And option he is correct, actually has less tendency to form hydrogen more.

In this problem, I'm writing the reaction. Just look at it carefully. Any energy for at be your ford four H two will react to form any and at four at people for plus four as to and and ME energy for add bill for will react to form and the beauty plus an STD plus as to So according to the option. In this problem option. See each correct. Was there not that we have any Beauty which reacts with metallic oxide to give colored or to post fails.

In this problem I can write the reaction and CST CH two managed to in perchance of action or two 0-5° integrate will give this compound CS three CH 20 H. And this component to denso. PBL three will give CST CH two beyond and this compound in pageants of GCN we'll give CST CH two M. C. And this component presence of AL I am at full well finally give CST CH two NHCS three. This is compounded by this is component so according to the option, option C it correct here, option siege, correct answer.

In this problem we have to ride the gondola or two minus. So just look at it carefully. JK sigma s sigma 22 sigma. Do as two Take my Star to add to Sigma two. period too 5 2, PX two By two, P.Y two by star, who P x two by style two py two py one. So bond order Is equal to 8 -5 x two, which is equal to 1.5 according to the option option. Age, correct here.


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