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Gruana Cav Fccrsatg | # zols Fot ca Cb 1.(6 fls) XJn 4 uliLew/s xniculrc For $ cupanJ Ihat hus Ihe formuls €,Il" Anat nuaralle enald neanaC omalthtCe-(9 ...

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Gruana Cav Fccrsatg | # zols Fot ca Cb 1.(6 fls) XJn 4 uliLew/s xniculrc For $ cupanJ Ihat hus Ihe formuls €,Il" Anat nuaralle enald neanaC omalthtCe-(9 Nts) Drju 4 !ulid orbitil piturc o sur mokrule Hrurn Meblen (uneelyAan= turs and |om: puirs 4 Well

Gruana Cav Fccrsatg | # zols Fot ca Cb 1.(6 fls) XJn 4 uliLew/s xniculrc For $ cupanJ Ihat hus Ihe formuls €,Il" Anat nuaralle enald neanaC omalthtCe -(9 Nts) Drju 4 !ulid orbitil piturc o sur mokrule Hrurn Meblen (uneelyAan= turs and |om: puirs 4 Well



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licre the ycllow and orangc precipirares arc, rcspectively (a) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{-}$ (b) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}$ (c) $\mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$ (d) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$

K two Cr awful gives a yellow precipitated it's a yellow policy heated on reaction with on reaction with Be a two plus, as well as B B two plus, therefore, option C and option the uh, correct answer for this problem, Option C and option B. R. Correct answer for this problem.

For 40 minutes are dominated the hybrid of nickel in these molecules. Okay, so I really will be the hospital and the hospital. Yes, because of a carbon and water is strong pairing will be possible Due to bidding. It is an inner order complex, so both have the sp two Hi variation to this, the correct option.

Explanation for the given question. Yeah, we have. I don't see if it can be well, particularly by infiltration with K two CR 207 using K three FCC and all six as an external indicator. How do you detect the end point here? The answer for the question is yeah. Option B secret adoption here we will observe blue blue too colorless columnists problems. This is because island salt iron souls bye by trading with Yeah, K two CR 207 by using F B by using get three F E CN hold six as an external indicator cater. We will observe. Yeah, we were observed blue collar. Mhm empty and point at the end point after the process of mm titillation complete, complete. We will observe no kind of because it has been mhm completely oxidized. Okay.

So any problem 1 75 of chapter six compounds of P block elements to so in this question, majesty following. Uh so uh 40 xenon cetera fluoride, which is xia for So this is the structure of the non tetra fluoride as seen on having the eight valence electron. And uh out of these four are the bond peers and involved and who are the lone lone pairs? So these are the long pair of electrons. So that makes uh it's a scare planner. So it has four bond pair and too long beard. So that makes, its hybridization is basically sp three due to. Okay, so uh shape is this carefully? Okay, so next is the non hexafluoride. X. EF six. So this is the structure in which six bond pair and one lone pair. So this is distorted Octa federal. And makes its hybridization sp three D three. So it's hybridization is sp three D three and its shape is distorted of capital. So the next is ex ceo three. So this is the pyramidal shape and then we have XC offer and this is the battleship. So therefore the correct match becomes for a. That is much with the R. For be that much with the B. Foresee, that is S. And for the that is your


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